Transcript for:
Chemistry Regents Review

hey everyone so this is the chemistry regions full review so this will include everything you need to know for the Regions exam uh so before I start the video I just want to preface a couple things um this video will be divided up into 12 video chapters covering 12 core units um there's going to be around 10ish practice problems for each unit some units have upwards of 15 practice problems others might have like 8ish um it all depends but in total there's over 150 practice problems so definitely a lot of practice to do um I would definitely recommend you take out a reference table because a lot of the questions you do need a reference table in order to solve also try and do the problems on your own before you was the solution that's the only way you're going to improve and actually learn now yeah that's it and good look on your exam and I hope this video helps you guys to the question so base your answers to the following question on the diagram of a molecule of nitrogen shown below so they just gave us this little diagram here two circles connected together uh draw a particle diagram that shows at least six molecules of nitrogen gas um so for that we can just draw these um molecules together and you want them relatively far apart in spread out let's just pretend those two are connected it's kind of hard to draw these molecules with my mouse 1 2 3 4 5 um they won't be in a straight line like this cuz that's actually scientifically wrong um that you want them all spread out and stuff but you can also add arrows just to show that they are you know just random movement very erratic uh B says draw a particle model that shows at least six molecules of liquid nitrogen so liquid nitrogen we're going to want them at least I will draw them in a container and they will be sort of like together right so they're going to be stuck not stuck together entirely but very very close to each other um and this will get a little messy but it's five and six all right those are some horrible drawings but I think you get the point um so let's move on to the next question base your answers to question two through five so these ones right here um on the information below and on your knowledge of chemistry a student prepares two 141 G mixtures A and B each mixture consists of nh4 nh4 Cl sand h2o at 15° C both mixtures are thoroughly stirred and allowed to stand the mass of each component uh used to make the mixtures is listed in the data table below I'm going to start typing my answers cuz I'm sick of drawing with my mouse the first first question says describe one property of sand that would enable the student to separate the sand from the other components in mixture B um so when I hear sand I think filtration and you have your other components with the H2O and nh4cl um and they're allowed to stand so yeah we can definitely uh talk about filtration here and the thing that makes filtration work is their different types of particle size um we can also talk about just solubility um if you were to use distillation but I'll just say particle size theic type particle size of the sand varies number three says determine the temperature at which all the nh4cl in mixture a dissolves to form a saturated solution so for this question you're going to want to have your reference table and let's see if I can find the table so if you go to table G uh find nh4cl and then we want to dissolve what is this for mixture a so mixture a there's 40 G of nh4cl so we just look on the right uh left side at 40 G lining up to the nh4cl we can see it's in between 20 and 30 uh at the temperature so we can just say it's around 25° C which type of mixture is mixture B so mixture B here has 10 G of nh4cl 31 G of sand and 100 G of water um so it's either homogeneus or heterogenous and this one is heterogenous I don't think I spelled that right um but it's heterogenous because it is not the uniform distribution of molecules right because sand has a different density than water uh has a different density than nh4cl it's not going to be an equal distribution versus let's say we had like salt water that salt is going to evenly um dissolve in that water and when you draw the particle diagrams you can see that um number five says State evidence from the table indicating the proportion of the components in a mixture can vary so that was basically my explanation before um but we can say that there are different amounts of um need a mixture can vary okay so there are different amounts of nh4 Cl um and sand in each of the mixtures um so we can see that in mixture a we have 40 G of NH nh4cl versus mixture B we had 10 G um and then since they are the same 141 G in total the sand is going to have to vary based on that let's move on so based your answers um to this paragraph right here a few pieces of dry ice CO2 solid that's kind of important to know it's solid because it's dry ice at -78° C are placed in a flask that contains Air at 21° uh Celsius the flask is sealed by placing an uninflated balloon over the mouth of flask as the balloon inflates the dry ice disappears and there's no liquid observed in the flask all right write the name of the process that occurs as dry ice undergoes a phase change in the flask so we know it starts as a solid I just write this write this to the side so it starts from a solid it says no liquid is observed and the balloon inflates so can you guess it goes from a solid to a gas um because you know the balloon is inflated by gas um so you're just going to have to know this solid to gas directly is called sublimation all right number seven State the direction of heat flow that occurs between dryo uh and the air and the flow so heat flow this is another rule you need to know heat flows from high to low so we just look at which one had the higher heat to start with so the air was at 21° C CO2 dry ice was -78° C so the direction of heat flow is from the air to to the dry ice all right let's move on uh final three questions here so student investigated the heat transfer using a bottle of water the student placed the bottle in a room at 20.5 de C the stud student measured the temperature of the water in the bottle at 7:00 a.m. and again at 3:00 p.m. the data from the investigation uh is shown in the table below so we can see that the mass of water didn't change so nothing really evaporated or anything um but the temperature did change there's a increase of 10° C uh let's see what the question asks here show a numerical setup for calculating the change in thermal energy of the water in the bottle between 7:00 a.m. and 3:00 p.m. all right so we need to know um let's just write our equation it doesn't say to but it's always helpful so the change in energy is equivalent to mass times the specific heat heat capacity of our sub substance uh or in this case water and then we multiply this by the change in temperature uh so we don't need to calculate the actual value just uh the numerical calculation so Mass we know is 800 G because that's just constant in between um and I don't even think you need units here really so we can just write this times the specific heat of water which is represented by C that is 4.18 and then the change in temperature like we said before is 10° C let's see what number N9 is saying so it says State the direction of heat transfer between the surroundings and the water in the bottle between 7 a.m. and 300 p.m. so the water increased uh in temperature and therefore that means the heat must have come from the air so the heat um so heat transfers from air to water all right um oh not from there we can just say the surroundings cuz maybe it was something else uh oops I spell surroundings wrong compare the average kinetic energy of the water molecule in the bottle at 7 7 a.m. to the average kinetic energy of the water molecules in the bottle at 3 p.m. so what is average kinetic energy that is the exact definition of temperature so if you have a increased temperature that means you'll have a increased average kinetic energy so 3 P.M equals why am I still ready I'm just going to say 3 p.m. has the um more average kinetic energy than 7 a.m. the bright line Spectra observed in the spectroscope for three elements and a mixture of two of these elements are represented in the diagram below the first question says describe in terms of both electrons and energy State how the light represented by the spectral lines is produced uh well I'll explain how it's produced so basically what happens is when you have electrons in uh when it's the electrons are neutral in a ground state um they're pretty stable right um but what happens is they can move up to a more extended state or a higher principal energy level and when they do that um energy is absorbed um however they don't stay there for very long because when they're there they're very unstable so they end up coming back down and when they come down from a higher energy level to a lower energy level that's when energy is released and that is what produces the light so the configuration the variance in the uh valence electrons and all that is what also determines how these bright line Spectras are created so uh there we have it so for number two explain why the Spectrum produced by a one G sample of element Z would have the same spectral lines uh at the same wavelengths as the Spectrum produced by a 2 G sample of of element Z same element so have the same you know like chemical properties and the what it called spectral lines whatever they that is independent of the mass so it doesn't really matter how much mass you have because the electron configuration for the element is the same thing all right moving on state evidence from the bright right line Spectra that indicates element a is not present in the mixture well we can see here element a does it line up with the mixture well the first line doesn't and this one kind of hard to tell but it's like slightly off this one it does line up with one um this one kind of hard to tell um but this one here definitely does not line up um with the mixture and so because the know the lines match up we can say that it is not present in the mixture all right moving on illuminated exit signs are used in public buildings such as schools if the word exit is green the sign may contain the radio isotope TR tritium tritium hydrogen 3 the tritium is a gas sealed in G glass tubes the emissions from the decay of the tritium gas as cause a coating on the inside of the tubes to Glow state in terms of neutrons how an atom of I was going to say hydrogen 3 differs from the atom of hydrogen 1 um so Isotopes right so how many protons does hydrogen have one right has one proton um so all isotopes of a element will have the same protons because protons is what determines you know what element it is in its properties um so if the the only thing if the only thing that stays the same is the proton number what has to change well it's neutron number right because if its mass changes then neutrons plus protons is the mass and so for hydrogen 3 you have a total mass of three and so 3 - 1 is 2 and for hydrogen you have a total mass of one and you have one proton and so it has zero neutrons all righty moving on uh a student compares to models of the atom these models are listed in the table below in order in order of development from top to bottom all righty so the first question says State oops State one way in which the bore model agrees with the thumbs model so the Board model is this one right here and the thet model let's look at their conclusions atoms have small negatively charged particles that seems awfully familiar I mean yeah similar to a electron right they describe each other and so the answer here will be that they both describe a electron or small negative negatively charged particles all right for question six it says using the conclusion from the Rutherford model identify the charged subatomic particle that is located in the nucleus so this one is a bit more interesting so Rutherford model the observations is most alpha particles pass straight through gold foil but a few are deflected so what happens is gold uh sorry alpha particles are positive right so when you shoot these alpha particles uh or at least when he shot these alpha particles through the gold foil um most of them just went straight through right so one of his conclusions was that the atom was made of mostly empty space however of few were deflected now why well because alpha particles are positive he concluded that the nucleus was also positive right because uh like charges repel and opposite charges attract so if they were opposite if they were the same charge they would repel and that's why the alpha particle came back um and we know that what subatomic particle is positive boom protons all right number seven State one conclusion about the internal structure of the atom that resulted from the gold foil experiment well I kind of touched upon that um the atom is mostly empty space um State the model that first included electrons as subatomic particles so these are in chronological order so let's just look down at their conclusions so Doon model anoms are hard indivisible spheres of different sizes uh doesn't really talk about electrons Thompson model atoms have oh yeah it's this one because small negatively charged particles I mean are you kidding me man yes it is this one th a model all right last three questions the element Boron a trace in Earth's crust is found in Foods produced from Plants boron has only two naturally occurring stable isotopes Boron 10 and Boron 11 say in terms of subatomic particles one difference between the nucleus of carbon 11 of a carbon 11 atom and the nucleus of a boron 11 atom so carbon has six protons and boron has five neutrons and they both have a mass of 11 so their neutron number will differ and we can see here uh carbon will have five neutrons carbon 11 will have five neutrons and Boron 11 will have six neutrons number 10 write a isotopic notation of the heavier isotope of the element Boron it has to include atomic number mass number and symbol of isotope so here here is how you would write it so B standing for Boron and then on the top number here you have the mass so that is the mass number and on the bottom you want to write the atomic number um and that just represents the number of protons number 11 compare the abundance of the two naturally occurring isotopes of boron so for this question you'll need your reference table and if you look at Boron the uh atomic mass that's stated is 10.81 so if you know that there are only two naturally occurring stable isotopes Boron 10 and Boron 11 um what this means is that Boron 11 will have to be more naturally occurring because the atomic mass is the average of all the uh naturally occurring Isotopes so it's a weighted average I think so because of that the abundance of boron 11 is more naturally occurring than Boron 10 all right so the first question says Identify the noble gas that has atoms with the same electron configuration as the positive ion represented in diagram one when both the atoms and the ion are in the ground state so the positive ion in diagram one is this potassium ion um so what happens is we can look at uh potassium's electron configuration it's a metal so it tends to lose electrons and its electron configuration as the atom is 2881 um so what happens is it wants to become stable right so if it's 2881 it's going to transfer away one of its electrons um to bromine and then so we're left with 288 so now what we want to do is find a corresponding noble gas that has the same electron configuration as this ion which is the potassium ion so if we look on our periodic table what has the electron configuration of 288 well that is going to be argon so our answer is argon all right let's look at question two so it says draw a Le electron dot diagram for a molecule of br2 so uh bromine and bromine together is just going to have a calent bond because it's two non-metals right and so how I like to approach these questions is add up the total number of valence electrons we're work we're working with so for Bromine we have seven valence electrons so we have uh two bromine atoms together so that's 14 total valence electrons we're working with and so how you would draw it is with this Co valent bond in the center right so you would draw this Bond here and so we know that this represents two right they're sharing the two electrons interchangeably with each other and so since we know there's 14 in total we can subtract two and we end up getting 12 and so we if we do 12 / two that's six and that's an even number and therefore we know that we can add the six to each of the atoms individually and those electrons don't have to be shared but it uses all all the valence electrons and so we don't have to add a double bond or triple bond we know that we've drawn the Lewis electron dot diagram correctly so number three says explain in terms of distribution of charge why a molecule of the substance represented in diagram 3 is non-polar well here's here's something to note so you see how it is a symmetrical structure right you have carbon in the middle and then on each of its sides it has hydrogen and so that indicates that it is a nonpolar molecule because the distribution of charge is equal all right number four determine the number of electrons in the bonds between the nitrogen atom and the three hydrogen atoms represented in diagram 2 so we can just count right so between there's three hydrogens and one nitrogen um and then between each of the nitrogen and hydrogen there is two electrons so 2 * 3 is 6 and so our answer for number four is six number five describe in terms of valence electrons how the chemical bonds form in the substance representing uh sorry how the chemical bonds form in the substance represented in diagram one um so how this works is as we talked about before because potassium the potassium atom is a metal right it wants to become stable and it'll lose a electron in order to do so bromine on the other hand is a non-metal and so it wants to gain electrons because its electron configuration is 2887 right so if it gains electron it'll become stable and so what happens is that electron from potassium is transferred over to bromine so there is our answer number six explained in terms of element classification why k2o is a ionic compound so what is a ionic compound so a ionic compound can consists of a metal and non-metal and potassium we know is a metal and oxygen is a non-metal so because you have a metal and a non-metal that creates a ionic compound all right number seven so there are six elements in group 14 periodic table one of these elements has the symbol uuq which is temporary which is a temporary systematic symbol this element is now known as whatever that is so it says State the expected number of valence electrons in an atom of the element of that thing in the ground state all right so group 14 we know all the atom uh all the elements in a group have the same number of valence electrons because that's what determines their certain types of properties right their ordered in groups by their valence electrons and so if you look at a group 14 um valence electrons we can see carbon has four silicon has four and so we can assume that this uuq also has four valence electrons number eight explain in terms of electron shells why each successive element in group 14 has a larger atomic radius as the elements are considered in order of increasing atomic number all right so group 14 we can see that carbon has electron configuration of 24 and then the successive element right below it is silicon with the electron configuration of 284 all right so it has a additional electron shell and that increases its atomic radius all right number nine identify a element in group 14 that is classified as a metaloid so if you just look at group 14 down the down the row um you'll just have to notice metalloids are on that staircase thing right so one of those metalloids that happens to be in group 14 is silicon number 10 explain in terms of electrons why the radius of a potassium atom is larger than the radius of a potassium ion in the ground state so we kind of talked about this but we know that potassium is wants to lose that electron right so originally it's 2 8 81 so what determines the radius that atomic radius is determined by the number of uh electron shells right so its atom has four shells whereas its ion has three shells because after it loses that one electron that valence electron to become stable it has three shells and so because it has three shells it has a smaller uh radius compared to its original atom number 11 says explaining terms of atomic structure why group 18 elements on the per periodic table rarely form compounds so compounds are formed in order to either transfer electrons or share electrons in the case of non-metals right nonmetal and a nonmetal but we know that group 18 elements are noble gases and we can just look on the periodic table their electron configurations they all have eight in their valence electrons so they've already reached stability and therefore they rarely form compounds number 12 it says silverplated utensils are popular before stainless steel became wiely used to make uh eating utensils silver tarnishes when it comes in contact with hydrogen sulfide H2S which is found in the air and in some foods however stainless steel does not tarnish when it comes in contact with hydrogen sulfide draw a leis do diagram for the compound that uh Silvers um so another thing here is just what are we dealing with well hydrogen is a nonmetal uh and sulfur is also a non-metal and so we know we have we're going to have uh calent bonds so I'm just going to actually just draw this out on the side so we start off with the sulfur and we have two hydrogens right so we can just make these coent bonds and we have have finished the hydrogens right because the hydrogens only need one electron um and so they can share interchangeably with the sulfur uh atom so how many total electrons do we have so sulfur has six valence electrons whereas each of the hydrogens has one so that's just plus two so we have to have a total of eight valence electrons here we know that each of these represent two right so that's 2 * 2 four so we still need four um electrons on this Le dot diagram and it works out nicely because we can just put these two on the outside for sulfur and we know sulfur wants to reach eight right so it can reach stability right 2 4 6 8 and therefore it is stable and so that is the Lis electron dot diagram for H2S first question says the volume of one mole of hydrogen bromide at STP is 22.4 L the grand formula mass of hydrogen bromide is 80 .9 G per mole what is the density of hydrogen bromide at STP so this is kind of a weird question um but what we need to do is use the density formula so the density formula is just density equals mass over volume so we can just plug our numbers in so our Mass they give us is 80.9 G per mole and we have one mole of hydrogen bromine so that is 80.9 g over the volume which they give us as uh 22.4 l so we see the math there um and I have it under this so we can see that it is 3.6 is uh gr per liter as the density all right let's move on to number two identify the type of bonding in solid potassium so what is potassium potassium is a metal right so if there's B Bing between two metals that indicates that it's a metallic bond all right let's go on to number three so it says um in 1864 the Solway process was developed to make soda ash one step in the process is represented by the balanced equation below uh in the space draw a Lewis Dot Diagram for the reactant containing nitrogen in the equation so if we look at our reactants um we can see that over here the one that has nitrogen as uh in its equation would be NH3 right here and so if we were to draw the Lewis Dot Diagram we'd have to find the number of total valence electrons so nitrogen has five um and then each of the hydrogens has one so that's eight total so if we draw this out we have nitrogen and then our hydrogens on the outside and we can see that the hydrogens are taken care of but we have eight total electrons um and each of these represent two electrons so 2 4 six and then we need two on the outside and so it looks something like that going to reveal this and boom yeah I just drew it reverse but it's the same thing write the chemical formula for one compound in the equation that contains both ionic bonds and coent Bond so let's look over here um and we have one right here let switch colors so this one right here na3 so let's first identify the ionic compound so we know na sodium is a metal so if we combine metal with whatever this is I see a bunch of non-metals that is going to form a ionic compound and then the calent compounds is between each of these non-metals so CO3 here has uh calent bonds and so that is one of the possible answer choices moving on to number five um ozone o03 is produced from oxygen O2 by electrical discharge during thunderstorms the unbalanced equation below represents the reaction that forms ozone explain in terms of electron configuration why a oxygen molecule so this is a oxygen molecule um is more stable than just a oxygen atom so oxygen atom its electron configuration is 26 so it's unstable um and then for our oxygen molecule that is where you have uh something like this so you have your two oxygens it has 12 total valence electrons it needs to share so like this we add two bonds because the reason I know to add two bonds is because if I have 12 and I have one Bond right right that's two 10 so I can't share 10 equally um in terms of pairs between these two right because 10 ID two is five and I can only have pairs of two so if I have two instead so a double bond i' have two minus sorry 10 minus 2 is 8 and now this works out because eight oops 8 divided two between each of these atoms is four and so four is an even number I can express in these uh electrons so I just add two on each side boom and that gives me a total of four pairs and so we can see here that because this is stable right the uh they're sharing electrons in the middle and therefore they able to reach stability versus just the oxygen atom which is 26 which is unstable all right number six identify the type of bonding between the atoms in a oxygen molecule well you can see right here actually it works out this thing calent bonds um so it is calent bonding boom and it's also nonpolar because it's the same atom uh seven draw a loose electron dot diagram for a molecule of chlorine so let's move over here so we look at chlorine right we're going to do the same strategy we did before with these we look at chlorine how many total valence electrons is seven each so seven + 7 is 14 let's add our single Bond right does that work out so then I would have 12 um so if I do this then I would have 12 12 divided by two because there's two atoms right here gives me six can six be expressed by my valence electrons yes it can and so I can just add these on the outside and that should give me 14 total and therefore was a question just draw the diagram that should be the diagram right there so boom and there it is all right question eight so explain in terms of molecular polarity why hydrogen chloride is more soluble than hydrogen and water under the same conditions of temperature and pressure all right so first let's take a look at hydrogen chloride so what exactly is hydrogen chloride written out as a chemical formula so you have hydrogen and then chloride so I just want to make sure this is balanced so hydrogen is plus one and chlorine is minus one so this is already balanced all right so first off what type of molecule is it well this is a polar molecule right because they're electro negativity differences um indicate that and it's not the same molecule either so the distribution of their charges um is not equal so knowing this why is it more soluble than just hydrogen and water well if we had hydrogen alone that is not just polar right and we know that water so H2O is polar so if these two are polar and like dissolves like so polar things dissolve in polar things and non-polar things dissolve in non-polar things that's why oil uh doesn't you know mix with water very well now we know that because they're both polar and like dissolves like hydrogen chloride um is more soluble than water number nine explain in terms of electro negativity difference why the bond in HCL is more polar than the bond in hii so for this take out your reference table and look up their electro negativity so hydrogen's electr negativity is 2.2 um and then chlorine electron negativity is 3.2 so that's a difference of one um so let's keep that in mind so now we want to look at iodines electro negativity uh iodine electro negativity is 2.7 so 2.7 minus 2.2 is 0.5 so we know that uh HCL has a greater electronegative difference now what does this mean when we're talking about polarity well the greater the electr negativity difference the more polar it is because um essentially what po po polarity of a molecule uh determines within the bond is that uh the sharing of that charge right the sharing of electrons is going to be more unequal for more polar molecules versus a non-polar one so there we have it Greater difference means it's more polar number 10 the density of hydrogen at STP is 0.0899 g g per liter Express this density to two significant figures all right so if we look at this sigfigs right we're going to start with a decimal so there's a decimal so we're going to start on the left side so if it's zero 0 right so right now number of SigFig sigfigs is z zero um but now we have our first term 8 but we see this is 8 99 so we can actually round this to just n so how many sigfigs is this well we just move from the left to the right until we reach the first term so this is one SigFig so we need another term here and since we're rounding it we can just call the zero oops and once we reach the first nonzero number everything after that counts as a SigFig so this is two SigFig and so our answer is 0.090 grams per liter as the density to two s figs all right number 11 State evidence that indicates NH3 has stronger intermolecular forces than cf4 so if it has stronger intermolecular forces right so intermolecular forces if this is super high that means it's harder to break apart the bonds and stuff and so it's boiling point should be very high and its melting point should be very high as well so we just look over here it's the evidence that NH3 has stronger intermolecular forces than cf4 so let's just look at their boiling points and melting points is NH 3's boiling point higher um - 33.3 is indeed higher than 77.7 is indeed greater than 183.0 um and boom smiley face number 12 explain in terms of molecular structure or distribution of charge so we have a option here why a molecule of methane which is this NH 4 is non-polar well if you were to draw this um out it'll look like this and we can just talk about the distribution of charge right it's a symmetrical distribution of charge symmetrical molecule here and therefore it is nonpolar equal distribution of charge number 13 draw the electron uh dot diagram structure of calcium chloride so calcium chloride let's write this out calcium and then chloride and then once you write this out you want to make sure it's balc so calcium's charge is + two chlorine charge is minus one and so if this is + one I mean sorry plus two and this is minus one we'll need two chlorines to balance it out so now we just need to draw the electron do structure for this and something to not here is calcium is a metal and chlorine is a nonmetal so metal plus nonmetal ionic bonding so how you would draw this is like this so first we want to express the metal it's a bad bracket so Express the metal calcium and then you're just going to draw the brackets and then the plus two on the outside to show that it lost two electrons and transferred those two electrons to chlorine so how I like to do is just draw each of the chlorine atoms CU so let's just draw the first one and then the one that gains electrons so in our case the two chlorine atoms you will draw in each of those balance electrons and then outside the brackets we can just write minus one to show it gained one and then do the same thing for this one there are a couple ways you can do it as well you can put like a two on the outside of the first chlorine atom to show that there's two of those but that works two let's see y boom um they're drawing electrons when they gain that was what I was talking about when you would draw in the electrons the full shell um for the atom that gains those electrons so yeah that does it the first one says show numerical set for calculating the percent composition of mass of silicon and silicon uh O2 too uh so pretty pretty straightforward so the equation for percent comp is just uh let me p over here it is mass of part over mass of whole time 100 so for this question we can just have the mass of the part which is the Silicon um over the mass of the entire thing of silicon O2 * 100 and so that is the numerical setup you don't have to write your answer but you can you won't get dock points for that obviously um but yeah pretty straightforward so let's move on number two determine the the total mass of ethanol produced when 270 G of glucose react completely to form ethanol and 132 G of carbon dioxide so this one I'm actually going to do out over here on the side U because it is like more traditionally a mole question so the first I'd like to do is just say what we start off with so we have um what happens when 270 G of glucose react completely form ethanol and 132 G of carbon oxide so 270 G of that glucose now I like to put this over one all right that's our first bracket here and now we're trying to find the total mass of ethanol and ethanol is this thing over here on the products um and so in order to get to that we need to First convert our glucose into moles so we have 270 G of glucose so we want to find how many moles that is so the grand form mass for glucose is 180 g all right and then we can just put one on top for one mole you'll see the reason I'm doing this is because the grams on both sides cancel so this is how you know you're on the right track and so you're left with moles and so this uh part of the equation here would tell you how many moles it translates to but the next thing you want to do is just find the ratio so we can see that there is one mole of glucose which produces two moles of the ethanol right so now we just want to convert this so one mole then we can just put this over that ratio over one Mo so now we're saying that for every glucose mole we produce over here what we're going to get is 2 moles of the ethanol and so now we know how many moles of ethanol we have so we have to convert those moles of ethanol into G because that is the total Mass so the grand F mass of ethanol is 46 G we put that on top because one mole will cancel with the units so you'll see here moles cancel with moles moles cancel with moles and now we're left with our final answer of 138 G and we can see right here so I made sure to spend time on that because that's more traditionally the um harder mole concepts of just multi-step layered questions like that right number three determine the marity of the solution so marity of the solution the equation let me just pull that out marity marity is moles of solute over lers of solution all right so how many uh moles do we have here it says in the equation that this 2.5 L aquous solution contains 1.25 moles so we have 1.25 moles on on top and then the second part of the equation is just the liter of solution and so we just have 2.5 L on the bottom so 1.25 / 2.5 gives us 0.5 M all right number four write the empirical formula for absorbit acid it says observ acid is also just vitamin C um it has a molecular formula of C6 h8 so I write this out C6 h8 06 so the empirical formula is just the most condensed simplified version here so let's find a common factor between all these we can divide each of these by two right simplifying 6 divide by two uh up three and then 8 divide by two for the hydrogens gives us four and oxygen is the same thing and so now these don't have a common factor anymore so that is the most condensed version which is our empirical formula number five show a numerical setup for calculating the percent composition of mass of oxygen in absorbing acid so very similar question to number one so we just find the total mass of the oxygen which is 96 G over absorb acid which is this formula over here so you can add that up but I will tell you that it is 176 G um so like as you uh number six determine the number of moles of vitamin C in the orange that contains 0.071 G of vitamin C um so we have 0.0 uh 71 G um and we're trying to find how many moles that is well we know that the gram for mass is 176 moles per uh G per mole right so we can just set 0.071 over 176 to find our number of moles so you'll see here I did a little differently here but you see it's using the same similar setep to number two all right number seven uh in the space below calculate the gr for the mass of glycine your response must include both a numerical setup and calculator result so glycine is this formula up here that organic compound um so pretty straightforward here we're just going to literally just add I'm going to find like nitrogen on the reference table find its uh Mass uh hydrogen carbon Etc um so it should look something like this and you should get 75 gram as your final numerical answer all right number eight so we have hydrogen peroxide is a water soluble compound the concentration of aquous hydrogen peroxide solution that is 3% let's underline that 3% by mass is used as an antiseptic and that's the equation for it calculate the total mass of H2O2 and 20 G of a aquous H2O solution that is used as a anti septic you must include both the numerical setup and calcul results so this is to find the mass here because they give us the percent composition of by mass which is 3% they stated we can just use the same equation we did before with um percent composition of mass equals mass of part over mass of um whole * 100 um so we have the percent composition which is 3% and then what we're trying to find is the total mass of H H22 um so that would be our part because our whole is just the entire solution which is the aquous H22 solution where the uh the septic has been dissolved in so we can just call this X and then 20 G on the bottom as our whole and then times 100 and now we just solve for x and so X ends up being zero oops that's a bad X so X ends up being 0 6 G and we can see here check yes sir all right number nine identify the type of chemical reaction represented by the balanced equation so we have a compound here right and it breaks apart into two smaller parts and so that is a clear indicator of decomposition Bo all right number 10 calculate the percent error of the student's experimental result uh need the correct numerical setup and calculate result all right so the student measured had an accepted value of 20.9% so that's what he was supposed to get um they did an experiment and they got it to be 21.4% so the equation for percent error is just measured minus accepted over accepted time 100 so like I said before it's a very plug heavy on plugging and chugging with equations um but you should get positive 2.39% all number 11 determine the number uh the total number of moles of oxygen that completely react with 8 moles of c26 all right so if we had eight moles of C2 H6 on this side in this equation it shows two moles all right and so we're trying to find the number of moles of oxygen so for two moles so 2 moles of C2 H6 this reacts with 7 moles of o2 so what happens if we bring this up to 8 moles of C2 H6 well that's a factor of four right so this increases by four right here and so we can pretty safely assume that this will also increase by 4 so we're just 7 * 4 that is 28 so we are ending up with 28 moles of o2 and boom all right number 12 what is the R infl mass of uh CA co4 um and then that weird simple to H2O um so this symbol right here that little dot I'm not really sure what to call it but essentially what you do here when calculating the grand F mass is that is the indicator of a plus so it's sort of like combining these two compounds together but you can see like the two would apply um like you have two full H2O molecules and so what you would do is just calculate CA co4 together and then add that to two * um the gr folium mass of H2O which I'm pretty sure is 18 so we would get something like this in our final answer would be 172.2 G all right number 13 given the compound C4 h108 calculate the molar mass of the compound all right so molar mass uh pretty sure for just calculate the grand flum mass of four carbons 10 hydrogens and eight oxygens and then combine them together so we should get 186 G calculate the number of moles and 17.7 G of the compound so if we have 17.7 G we can just set this over uh that's s put this over 186 right because that is the molar mass which is also the number of GRS per per mole and so we should end up getting boom 0.095 moles what is the empirical formula so what is the most condensed version of C4 h108 well let's see let's see what we can do here what is the common factor between 4 10 and 8 well that would be two so we get C2 H5 H5 and then o4 and since five is odd there's no common factor between these guys so that is our empirical formula and the first one says a 2.5 L aquous solution contains 1.25 moles of a dissolved sodium chloride and the equation is below compare the freezing point of this solution to the freezing point of a solution that contains 0.75 moles of sodium chloride per 2.5 L Sol of solution so it's the same total amount of solution except the only thing that's changing is the number of moles and so the concentration is changed um so to sort of illustrate the thinking here let's see you had 5 G of water and 1 liter of water so obviously the more water you have it's going to be hard to freeze or at least it's going to take a longer amount of time um because there's more surfaced area the amount of heat loss changes as well um but if you think about this question the more concentration you have it takes away from the existing water right in the solution that's trying to freeze and so what happens is it creates what is called a freezing point depression so it's going to go down to the extremes the freezing point is going to be lower the greater concentration you have so boom there we have it uh total of 1.4 moles of sodium nitrate is dissolved in enough water to make 2 L of a aquous solution it's 1.4 moles of sodium nitrate is dissolved enough water the gram full of mass of sodium nitrate is 85 gr per mole and it wants us to determine the marity of the solution so the marity equation on the reference table is marity is equivalent to moles of solute over lers of solution so how many moles of the solute do we have well we have 1 point4 moles right here because it's dissolved in enough water to make 2 L of equ solution so our total lers of solution is 2 L and so ultimately this 85 G per mole that's not really we don't really need that piece of information to solve the problem but we can just up the equation and we get 0. 7 molar all right so number seven what is the mass of K3 that must dissolve in 100 G of water to form a saturated solution at 50° C so for this one go to table G on your reference table and we know that it has to be dissolved in 100 G of water um which is nice because the reference table is already in scale with solubility per 100 G of H2O um it needs to form a saturated solution at 50° C all right so go to 50° C and you're just going to line it up right so you go from 50° C all the way up to kn3 and then just find where that intersects with the Curve of ko3 and we can see it's in between 80 and 90 um so let say about 84 G and so that is the total amount of K3 that's is going to be dissolved in that 100 G of water to form a saturated solution at 50° C all right number eight explain in terms of distribution of particles why the solution is a homogeneous mixture so what we have here is is HCl gas and it is is dissolved and looks like 200 g of water at 20° C and so what happens here is that the uh HCL becomes H+ ions and is CL ions and so these ions are now in that water and so when you have a gas that dissolves in a liquid like water it's going to be homogeneous because the particles or those ions will be equally distributed right so it's a uniform distribution and that is what makes it homogeneous all right number nine based on reference table G identify in terms of saturation the type of solution made by the scientist so the scientists uh made a solution that has 44 G of hydrogen chloride gas and 200 g of water at 20° C so the first thing notice here is that they give us this in terms of 200 g of water so because the reference table is scale down to 100 G of water we need to scale everything else to that proportion so if we think about this it can dissolve 44 G in 200 g of water right so 44 G in 200 g of water so what this means is that well the first thing you have to realize is that the more water you have the more solute that you're going to be able to dissolve um so because of this if we scale this down to 100 G of H2O what do you think will happen to to the amount of um gr of hydrogen gas that will dissolve well it's going to be half as well because from 200 100 it's 1 half so this will go from 44 G to 22 G and so now we can just line this up right 22 G and 20° C 20° C and 22 G we can line up and we can see that it is under the curve for for the HCL uh line and HCL is one of those examples that you know it's a gas because the curve goes down so as you increase temperature it gets less soluble um so as a matter of fact even if you didn't scale this down even if you were at 44 G at 20° C you'd still be under that curve and so it is unsaturated so see here boom all right so number 10 identify the two ions present in the solute so what is a solute looks like it is 3 mol of dissolved nh4 CL all right so nh4 and cl um so pretty simple here we just want to identify the charges on each of these so nh4 you can find the charge on table e nh4 is ammonium which has a positive charge so our first one is just nh4 plus and then our second one will be CL and I'm assuming this is a balanced equation so CL should be negative so that is our two ions all right determine the marity of the solution so very similar to a question we did up there marity is just moles of solute over moles of solution and so we can just have the moles of solute which in our case is 3 moles over the uh lers of solution which is 2 L so 3 over 2 is 1.5 and there we have it few pieces of dry ice are at -78° C they're placed in a flask that contains Air at 21° C the flask is sealed uh with a uninflated balloon as the balloon inflates the dry ice disappears and there's no liquid in the flask compare the entropy of the CO2 molecules in that dry ice to the entropy of the CO2 molecules in the inflat balloon so first off what is entropy so entropy is just the measure of a disorder of a system all right so if we think about it if we go from CO2 molecules in dry ice which is a solid to CO2 molecules in inflated balloon where they're going to be gas right what happens to the disorder right well the structure of a solid is much more structurally fixed right versus gas molecules which go all over the place and so it's not hard to see which one's more disordered here it's going to be the gas and so the entropy of those the molecules in dry ice will be less than that of the ones in inflated balloon number two describe how the Catalyst zy speeds up this reaction so we've got this reaction here and it's a pretty simple question that's just asking oops that's just asking what happens if we add a catalyst so what does a catalyst actually do in a reaction well it doesn't really do anything to the start of your reactants right the potential energy of your rea reactants or your final product right you're going to end up with the same thing so what does it actually do well a catalyst just speeds up your reaction and so when you have a reaction Let's see we have a chart here you're going to start off with let's say just your potential energy of your reactants it goes through with activation energy the initial activation energy and then it's going to drop back down and like this so this would be like exothermic or if you had something that was endothermic it would look like this go like this boom and it comes back down and so Catalyst at the end of the day is just what dictates this thing right here all that so that would be lowered and so ultimately it would be speeding up the reaction so it decreases that activation energy all right number three state the effect on the concentration of the CL ion when there is a decrease in concentration of the O minus ion so o minus is right here so it wants us to um it wants us to State what happens to this when this goes down so let's think about it if this goes down because according to La chat's principle what needs to happen is this needs to be balanced right so if this side if the left side goes down what happens is the right side has more and so the right side will start producing um and shift this equilibrium position this way to create more and maintain that equilibrium and so if the products on the right side are being used that means clo will decrease so ultimately there's also a decrease uh for clo ions number four explain why the container must be closed to maintain equilibrium so equilibrium is just the uh forward and reverse reactions of each of these reactions and so it needs to be closed so first off gas doesn't escape right like cl2 gas you wouldn't want that escaping and that would just mess up the entire reaction and so also in addition so like matter doesn't just leave the system um so that is why it has to be at equilibrium and that's why it has to be closed so number five compare the rate of the four reaction to the rate of the reverse reaction so this one because it's an equilibrium it's just equal and reversible uh number eight state two methods to increase the rate of reaction and explain terms of particle Behavior how each method increases the reaction rate so two simple ones here are first temperature all right so what is temperature do so if you increase the temperature you're going to increase the speed of the uh of the molecules right and so because what happens with that the particles will then collide with each other uh much more frequently because they're moving much faster right there's an increased likelihood of them colliding with each other another one you could use is something like I don't know like surface area um so surface area means that there's more uh area for them to collide with each other versus uh so let's say add a drawing like here here you could hit it at any of these points right any of these points versus if you had very small increased smaller molecules but increased surface area you can hit them from all these other angles and so that would increase the chances for a collision and so let's see what you could also increase the concentration which would mean you would just have more particles in general which would ultimately lead to having more collisions so if you have more collisions that's what ultimately leads to having a greater reaction rate number nine determine the amount of heat released by the production of one mole of SO3 so here we have two moles of um so yeah we have 2 moles of s SO3 and this produces 392 uh well it releases 2 392 uh k of energy so what happens if we only have one mole of SO3 well uh this 30 392 would just be half to make up for that and so we get a 196 number 12 explain in terms of collision 3 by using one g of powder zinc instead of one G strip of zinc would have an increased rate of reaction so if we have powder zinc that means the surface area is increased right because if we have a strip of zinc it look something like this so yeah you could hit it from all these other points um but what happens is that if you have smaller molecules like powdered zinc for example look at all this extra surface area where you can hit it from all these other angles and therefore increasing the number of reactions um and thus increasing the rate of reaction all right number 13 explaining in terms of l l chaters principle why the equilibrium shifts to the right to relieve the stress when the pressure on the system is increased at constant temperature okay so this is something you need to know what happens when you increase the pressure in the system and you have gases all right so if you increase the pressure the system will move to the side with less moles all right so that's just something you need to know and so in our case if we have two moles on this side and one mole on this side if we increase the pressure it's going to go in the forward reaction or it's going to go to the right side so boom shft shifts to the side less moles when the pressure increases um so I'll save you from reading all this uh number 14 the potential energy diagram represents reaction one this one right here you notice there's a lot of stuff you definitely want to read everything but sometimes you can get away with not reading the little passage background to give you uh so represents reaction one without a catalyst on the same diagram draw a dash line to indicate how potential energy changes when the reaction is catalized in the converter so a catalyst speeds up the reaction right and so what happens is like we talked about before it decreases that activation energy and so the potential energy of the reactants is going to be the same thing except when it comes to uh this activation energy it's not going to reach as high so just go boom so it's going to follow the same pattern but when it comes back down it's going to have the same Trend it's going to finish with the same potential energy of the products here so curve down somewhat like this um I couldn't really cover that for first question says using the data determine the concentration of HCL which is the acid all right so the data here is in the passage so let's actually read that so a 25ml sample of HCL is neutralized by 15 mL of 0.2 M NaOH which is our base so we can just use the titration equation which is M of the acid so the mity of acid oops I wanted to write the volume of that acid is equivalent to the marity of the base and the marity sorry the volume of the base so we already know the marity of the base which is 0.2 and we know the volume of the base which is 15 mL and we know the volume of the acid which is 25 M so all we have to do is just solve for the concentration which is just the marity of the acid so accept the equation and we find that the marity is 0.12 M oops based on the data the calcul marity of HCL should be expressed to what number of significant digits so or figures so for multiplication how sigfigs work is you're going to find the number of sigfigs in each of the values that you calculate and then you have to round to the one that is the least number of Sig fixs so in our case we have 25.1 uh sorry 25.0 which has three si figs and then we have 15.0 which is also has three six figs and then we have 0.20 which has two Sig fix right because when you're counting with a decimal from left to right you don't negate the zero and then once you reach two that's just one and then two Sig fix so we have to round to two sigfigs here and so that is our answer for two number three complete the equation for the neutralization reaction that occurs by writing a formula for each product so what is the neutralization reaction so a neutralization reaction is where acid and a base come together to create water and a salt and so water we know is just H2O right so the easy way to solve these problems is to just take out H2O so we have the h from the HCL and then the other H and the oxygen from the NaOH so we can just cross this out and make our first product which is H2O and so now we just need to determine what the salt is and we just have the remaining products so we have Na and Cl so we just combine those two and they're already balanced by the way so we just create na as our salt and that is our answer number four identify the laboratory process described in this passage so this passage basically um well we kind of talked about this we use the titration equation and it talked about a neutralization reaction um so can you guess yep titration all right number five say how many times greater the hydronium ion concentration in HCL uh is than the hydronium ion concentration in whatever that is so we're comparing a pH of 2 to a pH of three and we know that hydronium ions increase as you become more acidic right and we know that a pH of 2 is more acidic than a pH of 3 um by a factor of 10 because it scales right so if we had uh difference between two and four in terms of pH that would be a difference of 100 but in our case a difference between two and three is just 10 so the hydronium ion concentration also has a factor difference of 10 number six complete the table by writing the color of thol blue in the na uh aquous solution and in the NaOH aquous solution so for this one just go to your reference table and let see if I can find it go to table M and we're just going to compare the ph and what they would be um in terms of the color change so for NAC we have a ph of 7 and then the color in thymol blue thol blue ranges from 8 to 9.6 which gives you yellow and blue respectively and so because the pH of s is under uh eight we're going to have a color change of yellow for NAC boom and then for NaOH we have a pH of 12 12 is above 9.6 so we're going to get blue all right let's move on so number seven determine the concentration of Koh and this is another titration question so we just set up our equation with marity of the acid times the volume of the acid equals marity of the base times volume of the base so here we have we're just trying to find the marity of our base here so we just set the equation like so we find the marity is 0.17 M number eight compare the number of moles of H+ aquous ions to the number of moles of O minus ions in the titration mixture when the HCL acreas is exactly neutralized by k um so if it's neutralized so the first thing I think would be useful here is what exactly is titration so titration is where like let's say you have this is not really accurate drawingle let's say you have like a beaker here and you have you have I don't know like a base I don't know what base this is let's just call this oops let's call this Koh because that's the example they gave us so let's say you had Koh in here and then you like it's like mixed in weird solution and then up here you have had uh your acid right so you had HCL so what you would do is you would incrementally so you add like drops of this HCL right you would add drops of this into the Koh um until it turns color so you have some sort of indicator in there that shows you when it turns slightly acidic and so when it just turns slightly acidic remember it starts as a base then the difference in between how much HCL you use or how much acid you you use uh will tell you a lot about either the marity of the solutions or in our case the uh number of moles and so if it just turns acidic you're trying to get it to not turn acidic to be honest you're just trying to neutralize it you're trying to strike a balance at a pH of s so if you strike a balance at a pH of s that is where you have a balance actually of H+ ions and oh minus ions and so in our case it would be oops I keep clicking the equal sign but they would be equal I feel like this titration thing right here this was kind of irrelevant for this question but whatever number nine um carbonated beverages are made by forcing carbon dioxide into a beverage solution when a bottle of one kind of carbonated beverage is first open it has a pH of three so when it's left open for several hours hydronium concentration is decreases to one Oops why did it go okay decreases to 1 1,000th of the original concentration determine the new pH of the beverage so the hydronium ion we talked about this before hydron ions uh stick with acidic Solutions right and so if it decreases so if this hydronium ion decreases where we know it gets more basic so the the pH will increase and it starts at a pH of three right and since we know that each pH that goes up is a factor of 10 and we have 1 1000th um so if we go to four five and that's six so from 3 to 4 it' be 1/10th hydronium like ion concentration four to 5 would be 1/ 100th and 5 to 6 is just 1 1,000 actually this is inaccurate it's 3 3 to 4 is 1110th 3 to 5 is 1/ 100th and 3 to 6 is 1 1,000th so in our case the new pH would be six all right number 10 state in terms of the pH scale why this beverage is classified as acidic well it's under seven seven is neutral anything under it is acidic anything above it is basic so yeah pH St number 11 identify the negative ion found in most of magnesia so milk of magnesia it's a mixture of magnesium hydroxide which is um so wait I should actually read this so in liquid water a equilibrium exists between H2O molecules H+ ions oh ions okay and if someone is experiencing acid and digestion after drinking tomato juice they can ingest milk of magnesium to reduce acidity so I'm assuming it's basic okay so tomato juice has a pH of four milk and magnesia has a pH value of 10 okay that makes more sense okay so because has a pH value of 10 it's above seven right so seven is seven is neutral so 10 is basic so negative ion there is just going to be the ion that's associated with bases which is oh minus because we know that hydronium or H+ ions are associated with acidic Solutions so in our case it would be o minus all right number 12 compare the hydrogen ion concentration in tomato juice to the hydrogen ion concentration in milk of magnesium so tomato juice is obviously more acidic um so because we're talking about hydrogen ion concentration hydrogen ions are more concentrated in acids than bases so boom Tom juice has a greater hydrogen ion concentration says compare the intermolecular forces of the two substances at STP so the two substances here is ethane uh which has a boiling point of 89° C at STP and then ethanol which has a they say has a much higher boiling point than ethane at STP so intermolecular forces are the just the bonds in between molecules in these substances and so um if you have a much higher boiling point what that means means is it's going to take a lot more energy to break those bonds right so if it takes more energy to break those bonds it means the bonds are stronger and so the intermolecular forces of ethanol are stronger there you go all right number two a liquid boils when the vapor pressure of the liquid equals the atmospheric pressure on the surface of the liquid based on table H what is the boiling point of ethanol at STP uh so for this question we just take out table H and then we're going to go down down to uh STP and see where it lines up so we can see on the dotted line where it says 101.3 kPa that line intersects at a point with ethanol right so then we can go down here and see what temperature that lines up with and we can see it's around uh 80° C so our answer is 80° cus number three show a numerical setup for calculating the volume of the gas in cylind B and STP all right so it looks like we have cylinder a is a movable piston and is hydrogen gas and cylinder B is exact same thing except it has methane gas um and then it looks like between the two gases the pressure is the same the volume is the same and the uh temperatures are also the same so if we want to find if we're just looking at cylinder B right so these condition the pressure conditions right here and the temperature conditions are currently not STP so for STP the temperature is 173 Kelvin and then the pressure is just one atmosphere all right so in order to solve for the volume because right now we don't know what the volume is at STP we will have to use the gas laws equation and the gas laws equation is just pressure of one volume of one over tempure of one so these are just the initial conditions so we know all three of these variables and then we're just going to set this equal to the pressure of two in our case is one atmosphere over the temperature of two we know is would be 273 Kelvin for STV conditions and then we're just going to solve for the final uh volume here and so if you plug in the numbers you should get that so we only need a numerical setup we don't actually have to solve for V number four state a change in temperature and a change in pressure that will cause the gas in cylinder a to behave more like ideal gas so uh you're just going to have to know this off the top of your head so ideal gases they want to have high temperature and lower pressure now real gases in contrast have lower temperature or closer together and therefore have high pressure um but ideal gases you want high temperature and low pressure number five compare the total number of gas molecules in cylinder a to the total number of gas molecules in cylinder B so this is going to have to you're going to really have to read in between the lines here so if we look at the volumes here we have 1.25 L of gas here and 1.25 L of gas here and since pressure and temperature are constant constant between the hydrogen gas and the methane gas because the volume is the same um you're also going to have the same uh number of of gas molecules right so the amount of gas molecules is going to be constant between the two as well so it is the same all right number six compare the total number of gas particles in the sample under the initial conditions to the total number of gas particles in the sample under the final conditions um so let's read up here it says a sample of helium gas is in a closed system with a movable piston volum of the gas has changed um y okay so basically you have a closed system and has a movable piston and there's some you know just gas particles in here and what they did is they so they increased both the temperature right temperature went from 200 Kelvin to 300 Kelvin and they also increased the pressure from two atmospheric uh to 7 atmosphere so we don't know what happened to the volume but right now the first question here is asking about the total number of gas particles so because this is a closed system it doesn't really matter what we do to the temperature or the pressure the total number of particles because it's a Clos system no matter no matter can escape and so the total number of gas particles will be the same as well number seven convert convert the final temperature of the helium gas sample to degrees C so pretty straight forward this equation is in your reference table so Calvin is equivalent to Celsius plus 273 and so in our case we know the Calin so we want the final temperature into degrees so the final temperature is 300 right so we just write 300 and it's just some simple under route we can just subtract 273 to find our Celsius so that should be uh 27 boom uh number eight in the space below show a correct numerical set up for calculating the final volume of the helium gas sample so this question right here uses the gas laws again uh so it's the same equation that we talked about up here now I'm pretty sure it's the same missing variable too let me just check yeah because we're trying to find the volume uh of this one as well so for this one we would just plug in uh so for P1 that would just be two for V V1 right here that would just be 500 and then for the initial temperature just 200 final temperature would would be 300 uh and then final pressure would be 7even and then we can just solve for the final volume and so you should get that all right number nine when the airbag inflates the nitrogen gas is at a pressure of 1.3 atmospheres a temperature of 301 Kelvin and has a volume of 40 L calculate the volume of the nitrogen gas at STP your response must include both a correct numerical setup and calculated volume so this question is another gas loss question all right so let's just set this up this time um so we know that it starts with a pressure of 1.3 atmospheres has a temperature of 301 Kelvin uh and 40 l so this out like this boom and then it's going to calculate the volume of nitrogen at STP so STP conditions you have to pretty much know this it's on the reference table by the way U so STP conditions 273 Kelvin one atmosphere and then we're just solving for that final volume so that final volume I've calculated right here but you need both a correct numerical setup which is this thing over here and also the calculated volume so I don't I I don't put units but you probably should put units on the actual regions but yeah there we go 47.2 L all right number 10 balance the equation for the decomposition of Na n three using the smallest whole number code efficient so for this one uh here's what I like to do SL out three goes to na process boom boom boom all right here we have a voltaic cell which has zinc on the left side the anode and then it has uh iron on the right side as the cathode and it says the cell and the unbalanced ionic equation representing the reaction that occurs in the cell r as follows shown below number one says explain in terms of zinc atoms and zinc ions why the mass of zinc of the zinc electrode decreases as the cell operates uh so this is the electrode right here so what happens is because this is the anode um zinc is going to be losing electrons right so part of this is going to just like fly off as electrons come off and so the mass actually decreases and you can see this in the equation right you have zinc solid going to zinc plus two but this plus two means that it's lost two electrons and so the mass of zinc uh the mass of the zinc electrode decreases as it operates because it is oxidized all right number two identify the subatomic particle that flows through the wire as the cell operates so this is the wire right here that connects the uh anode and cathode So What flows through here are the electrons now the salt bridge thing um that's where the actual ions will flow and that just uh keeps a balance between the two electrodes but for number two that is going to be electrons all right moving on number three write a balanced half reaction equation for the reduction of the iron ions and iron 3 oxide to iron atoms all right so for this one we're going to start off with our iron three iron plus three because it wants us to show the reduction um for iron 3 oxide to iron atoms so we're going to have a addition of three electrons and that is all going to go to iron one iron atom so the reason for this is because in these Redux reactions there's going to be a conservation of mass and conservation of charge right so the charge on this side plus three on this side is going to be zero and so we have that -3 from those three electrons and so this cancels out and gives up zero and then we have one mole of Fe one Mo of that V and electron's mass is non neg is negligible so there we have it let's number three boom all right number four identify one metal from the passage that is more active than carbon and one metal from the passage that is less active than carbon So In this passage here it says metallic elements are obtained from their ores by reduction some metals such as zinc lead iron and copper can be obtained by heating their oxides with carbon so we can pick from any of these we can pick zinc lead iron copper doesn't matter all right so for this one we can just pick something like zinc and now for the one that is more active we can just look below it says more active metals such as aluminum magnesium and sodium cannot be reduced so let's just pick aluminum here um so yeah there we there we have it identify the component required for electrolysis of Molen uh NAC that is missing from the cell diagram so basically uh this is electrolysis and it can't really operate right now because it's missing something in the middle and what is that something well it is um energy right it needs a power source and so answer for this is a battery all right number six state in terms of energy why the cell is a volic cell all right so it's a volic Cel because first off is there a battery no so there is no battery and so what you want to look for is how energy is created right so these electrodes um the Redux reactions that they go through the half reactions that they go through are producing chemical energy chemical energy and this chemical energy ultimately is going to power this entire system right so it's like a circuit of electrons that flow through and so this chemical energy will go and create electrical energy and so in this case because it goes from chemical energy to electrical energy that's how you know it is a volic cell as opposed to something like electrolysis which requires a battery so there we go number seven determine the number of moles of aluminum needed to completely react with N9 moles of nickel ions so we have two aluminum which reacts with uh sorry two moles of aluminum reacts with 3 moles of nitrogen all right oops so if we have a ratio of so we have 9 moles of nickel and so if initially we had three moles of nickel and this gave us two moles of aluminum that is a 3 or 1.5 ratio right and so we can do the same thing here is just 9 / 1.5 equals that should give us boom six all right number eight write a balanced half reaction for the oxidation that occurs when the switch is closed all right so oxidation means a loss of electrons so let's see which of these electrodes actually loses electron so we can see here that aluminum goes from a neutral charge of zero zero and it goes to looks like plus three and so it is actually losing three electrons and so we know that aluminum the solid aluminum and the electrode is being oxidized and therefore if we were to write this out do this to the side we would start off with our two aluminum right and this is already balanced and so this would just go to it's losing electrons so the electrons would be on the right side and then our products is right here so two aluminum plus three ion and then we have to add our electrons right so if this is six this is a positive six we need a negative six and so we need six electrons so that is one of the answers for eight you could also simplify this so you could say like just 1 Mo of aluminum solid is equivalent to 1 Mo of alum aluminum aluminum ion plus three electrons um either one works number n see the direction of electron flow through the wire when the switch is closed so electron flow flows from the anode to the cathode so in our case the anode is just the aluminum and that's where electrons are lost and so the electrons will just go like this so left to right so boom all right number 10 explain in terms of uh AG atoms and AG ions why the mass of the silver electrod decreases as the cell operates well pretty similar situation here except we have a power source right so this is a electronic uh cell right here so AG is our anode all right so that's going to be oxidized so when we oxidize it it's going to lose electrons and like we talked about before if you lose electrons you're going to be losing mass as well so boom number 11 State the purpose of the power source in the cell well the power source um is really what drives this reaction so the purpose of this power source or in our case I'm assuming it's a battery here is to drive the spontaneous reaction so boom all right number 12 identify the cathode in the cell well we kind of did that because we know we said it starts in the anode with the Argon and then we end on the other other side which in our case would just be nickel right so our nickel solid is going to be the cathod so nickel key all right so the first question says draw a structural formula for the hydrocarbon that is approximately 2% of natural gas so we want to read this thing up here and it says that natural gas and coal are two fuels burn to produce energy natural gas consists of Approximately 80% methane 10% ethane 4% propane and 2% butane so that is the 2% natural gas and that butane is a hydrocarbon that we're trying to draw all right so let's go on the side here so how do we draw butane well first let's identify that prefix right and the prefix is but and so but this uh prefix here tells us a lot so it tells us the number of carbons that are in this hydrocarbon so if you go on your reference table and you go to table P it's says that the but prefix u means that there are four carbons so we know that for our structural formula we're going to have four carbons and now let's look at the ending right so it's butane so it is part of the alcane homologous series and so what this means is that there are single bonds so single bonds and now why is this important because if we know that there are single bonds that helps us determine how many hydrogens we have right so first if we were to draw this let's just write out our four carbons and then connect them we know there's single bonds because it's uh the alcane series and now we want to figure out how many uh hydrogens there are well we can use the formula that is provided on the reference table as well if you go to table Q alanes the general formula is as follows so carbon uh n hydrogen 2 n + 2 so what does this mean so n is just the um number of carbon atoms right so in our case we would have C4 because we have four carbons and then we just plug in four for n here so we have 2 * 4 is 8 + 2 is 10 so you have 10 hydrogens now works out because if we just draw a hydrogen on each of these remaining parts on the carbon we can see that we will have uh 10 hydrogen so that is the structural formula number two write the general formula for the homologous series that includes the components of the natural gas listed in this passage well we already did that right here that is the uh general formula you can use to help you determine how to draw your structural formulas um and for questions like these very helpful so there we have it all right number three identify the class of organic compounds to which ethanol belongs so ethanol it says up here that uh ethanol is C2 H5 o so how exactly do we determine uh what class this compound is in well we want to look for that uh functional group so this functional group here um we can see we just have the regular carbon hydrogen but now we have this o right here so this o tells us a lot because that is the functional group O the functional group for alcohol all right so that indicates that ethanol is a part of the alcohol class now another way you could look at this is just by the ending right so we have ethanol so we have that uh let's bring this over we have that anol ending so where else can you find this anal ending well if you look at the reference table table R we can see that the example for uh alcohol is one propenol so we can see that same ending here and so you can use that comparison let's use the highlighter you can use that comparison there that's actually good but you can see the null a null so that lines up and so we know that is part of the alcohol class all right number four to which class of organic compounds does ethyl uh sterate belong so where is eth so eth sterate is this thing right here so we're just going to look at this structural formula and figure out what class it belongs to so we have a single bonded uh oxygen on this side and then a double bonded uh oxygen on this side so let's look on our reference table reference table are the reference table is your best friend and we can see here that it lines up uh nicely with the Esther all right so we have a Esther the functional group has a double bonded oxygen and then a single bonded oxygen as well so it lines up there and therefore we can determine that this ethyl sterate is a part of the Ester group two hydrocarbons that are isomers of each other are represented by the structural formulas and molecular formulas below explain in terms of structural formula and molecular formula why these hydrocarbons are isomers of each other so isomers all right you're going to have to know this pretty stra forward though so isomers essentially have the same actually fast hon they're going to have the same molecular formula but they're going to have different structural formulas all right so what this means is if we take a look case at hydrocarbon 1 and hydrocarbon 2 we can see that for the first one uh they both have a molecular formula of C5 h8 right so if you count up the total carbons and hydrogens they line up however if you were to to count um the actual naming of these hydrocarbons right so that would be determined by the structural formula that would be different right because here we have uh we have double bonds and here we have triple bonds and the bonds are also in different locations so in our case we can just say it has the same molecular formula in different structur formula all right number seven write the name of the homologous series to which the hydrogen belongs to so in this case we have a looks like a reaction between bromine and hydrocarbon um so this is the bromine and so this is the hydrocarbon in question um so we can see here immediately we have a double bond and that indicates table Q double bonds are a part of the alken uh Series so the answer here is alkenes number eight identify the type of organic reaction so what what's happening here let's erase this so we can see very clearly there is is a addition symbol so here we can just uh very confidently just say this is addition um so what happens here is you'll see that double bond that was existing before in this hydrocarbon is broken down in order to make room for these bromine uh atoms to add on to the existing hydrocarbon so there we can say addition all right number nine explain in terms of bonding why c3h6 is classified as the unsaturated hydrocarbon so unsaturated hydrocarbons are going to have either double bonds or triple bonds because saturated hydrocarbons have single bonds and so is C3 H6 a part of the alkane series well it's not because the alcane series like we talked about before the uh it fils this formula here so in our case we have three carbons and six hydrogens um it does not follow the formula here because here if we had three carbons we would have eight hydrogens Instead This follows the formula for l l kees so this follows the formula for alkenes and therefore it's going to have a double bond and so it is a unsaturated hydrocarbon all right the last question here number 11 in the space below draw a structural uh formula for a molecule of 22 for trimethyl pentane so when you get these types of questions how should you approach them well first I would identify on the very right side what you actually have as the base the base of the hydrocarbon in our case that is pentane so this is not one of those fancy alcohol Esther Ketone whatever um so we can just write draw pentane right so we have pentane U so pentane uh the prefix pent on the reference table p uh has five carbons so we can just draw five carbons and these are single bonds as well because the pre prefix sorry the suffix at the end is a a the Alan's group has all single bonds all right so now let's approach the next part which is the trimethyl so trimethyl means there's going to be three additional meth methyls that are attached on um each of these carbons now where are these going to be attached right like which carbons are these uh methyl is going to be coming off of and this is where this part comes in wish I could switch color but whatever so 224 what does this mean that means you're going to have um methyls on both sides of the two the second carbon and then one on the fourth carbon so I'll show you what I mean right here so if we were to draw this out we would have 22 so this be the second one so the first carbon right here and so this would be the second carbon so then we could just draw a ch3 which represents methyl like this let's erase this at the bottom and we can have another ch3 so in reality these aren't really 2D they're 3D and can move around in all that but that's just how you want to draw it in this case now for the fours we just count here so this is 1 2 3 four so we know our fourth carbon is right here and let's attach another methyl so now we finished the tri methyl so try means three we have three methyls there and now we just have to complete the rest of the molecule so carbon here we can just add a hydrogen to each side of this um same thing here and same thing here boom boom boom nothing fancy here because if we zoom in on this right here it's a single bond between those uh two carbons so it looks like this oops it's going to look like like this boom boom and then boom boom boom so it says a breeder reactor is one type of nuclear reactor in a breeder reactor uranium 238 is transformed in a series of nuclear reactions into plutonium 239 uh the plutonium 239 can undergo fision as shown in the equation below the X represents a missing product in the equation right so the first question wants us to write a notation for the nucleid represented by the missing product X in this equation all right so for all these fision reactions there's going to be a balance of charge and Mass all right so here we can see on the left side we have a total mass of 240 and then a total charge of 94 and on the right side we have a so far 94 + 1 so we have a mass of 95 right now but ultimately we need to balance the mass so it's 240 on the left side oops so if it's 240 on the left side we want to subtract 95 and that gets us 144 so our mass is going to be 144 so for our final answer we can write like one oops this is not good at all we're just write 144 over here all right now for the bottom side we have 36 on the right side and 94 on the left side so that is a difference of 58 so we can write 58 on the bottom so this 58 right here also represents the protons all right so because we know the number of protons um even though this may be a isotope of a element then we can determine what that element is now case that is going to be C so there we go and so that should be the nucleid um the missing product X in this equation there we go all right number two compare the amount of energy released by one Mo of completely Vision plutonium 239 to the amount of energy released by the complete combustion of one Mo of methane all right so here's the thing right so combustion is a chem chemical reaction or a chemical product process whereas fision is nuclear and the thing is nuclear reactions and processes release a lot more energy than chemical ones do right so like nuclear think like atomic bombs right just a vast amount of energy very unstable and so in our case it would be the fision plutonium so fision does release greater amounts of energy uh number three based on table n identify the Decay mode of plutonium radioisotope produced in the braer reactor so pretty straightforward just go to your reference table on reference table n um so you won't find plutonium there but you want to like read the question right so it go it says it goes from uranium 238 to plutonium 239 so what you want to do is just look for the Decay mode of uranium 238 which in this case is Alpha so there we go that's the alpha symbol all right number four determine the number of neutrons in an atom of the uranium isotope used in the breeder reactor all right so if we write out uranium and I should just use another color for this uranium um it's 238 for its mass and then it has 92 protons or 92 protons well pretty straightforward we just find the difference between mass and the protons and we get the number of neutrons which is 146 and so there we go Bo all right number five complete the nuclear equation for the radioactive decay of tritium by writing a notation for the missing part so tritium is also known as hydrogen 3 all right so hydrogen 3 good to know they list it right here so balance of uh mass in charge so on the right uh sorry on the left side we have a mass of three on the right side it's zero because the electron is not as neg negligible in mass and so we know we have to write a three on the top all right on the bottom we have a one on the left side and a negative one on the right side and so to balance that out we need a two and so what has two protons well that's going to be helium so that is going to be that notation for that part boom number six determine the fraction of a original sample of trium that remains unchanged after 24.6 2 years so this is a halflife uh question all right so first off what is the half life of trium or hydrogen 3 well on the reference table it states that the halflife is 12. 31 years so that's just something to keep in mind here now it says the fraction after 24.6 2 years so this is the total amount of time that has elapsed and this is the total halflife and so each half life 50% of it decays and so we have 24. 62 and how many times does 12.31 go into 24.6 two two times so that is two half lives that is very messy but whatever so two half lives pass all right so let's write this out um we wants the total fraction so we start off as one because that's just one whole and after one half life it decays to 1/2 right and then it decays 1/2 again so what's 1/2 of 1/2 well 1 14 and so that is our fraction and boom I did another way here but um both ways work moving on number seven determine the total amount of time that has elapsed on 12.5 G of the original Cobalt 60 sample at the hospital remains unchanged so what is the original sample here so it says the hospital keeps a 100 G sample of Cobalt 60 Okay cool so we start off with 100 right right it wants us to know the total amount of time that it will elapse when it reaches 12.5 G okay so 100 will Decay to 1 half of 100 is 50 so that's 1 half life and then decays 25 that's two half lives and then finally decays to 12.5 so how many half lives is that let's count so this is one two three Half Lives okay so you have three half lives right so what is the actual halflife like that the time that it takes to go through 1 half lifee for Cobalt 60 well you can go on your reference table and the reference table says let's pull this out let's see it says 5271 all right so five [Music] 5271 I'm not sure what that is but whatever that is together should be the total amount of time that has elapsed so 15.8 is years all right number eight complete the nuclear reaction below for the beta decay of Cobalt 60 by running an isotopic notation for the missing product all right so very similar question like the ones we've done before left side is 60 right side is zero so we need 60 on top okay uh left side is 27 right side is negative negative 1 so we need 28 so what has 28 protons let's check uh looks like nickel all right so nickel has 28 protons and so that should be our notation for that question number nine compare the penetrating power of the two emissions from Cobalt 60 all right so if we look above here I wonder if the highlighter actually works it says that the Cobalt 60 is the artificially produced radio isotope that emits gamma rays and beta particles okay wow that actually looks very realistic on the paper but whatever um but yeah so it emits uh gamma rays and beta particles so gamma rays have no charge right so when you put it through a magnetic field it's just going to go past whereas beta particles are negatively charged and so they will actually get uh deflected and so here gam gamma rays have a much stronger emission pen penetrating power so yeah there we go number 10 State one risk to human tissue associated with the use of radio uh Isotopes to treat cancer um well here's the thing it's it's radioactive material right so I mean if you get inside your body okay it can kill the cancer cells but it also poses a big risks to the healthy cells so yeah uh boom there we go it can be harmful to healthy cells so that does it for the review of unit 12 and that concludes chemistry uh regions at least um so yeah hope you guys learned something if you did make sure you 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