General Chemistry Final Exam Review Guide

Hello everyone, here is the video for general chemistry 1 final exam review. In first question, we would like to convert the Fahrenheit unit to degree of Celsius. The formula for this conversion is the temperature in Celsius is equals to 5 9ths times Fahrenheit minus 32. So, 5 9 times the Fahrenheit temperature is 244.2 minus 32. Here the answer is 212.2 and then multiplying to 5 9 the answer is equal to 117.88 Celsius or we can just round it to 117. 0.9 so the answer is c in question 2 we would like to note the mass of a cube of tungsten with the length of 1.10 inch we have the density it is in gram per cubic centimeter so we need to convert the inch to centimeter so we can have the cubic centimeter as a volume so the density is equal to mass over volume. So we can rewrite this equation. Mass is equal to density times volume. The density is 19.3 and volume it is the length of cube to the power of 3. So I need to calculate the length of cube in centimeters. So we have 1.10 inch. and our conversion factor for inch is 2.54 so one inch is equals to 2.54 centimeter the answer is 2.794 so instead of a in this equation i should use this number 2.794 the length of one side of cube to the power of 3 then the answer is equal to 420.95 gram or we can round it to 421 gram the answer is a here in question 3 we would like to convert this distance in kilometer to a centimeter We don't have the direct conversion of kilometer to centimeter but we can convert both of them to meter so we can write it in this form 384 this is in kilometer and we know kilo is means thousand so one kilometer it's thousand meter so kilo and kilo they cancel out then we know centi is equals to 10 to negative 2 So 1 centimeter is equal to 10 to negative 2 meter. So instead of centimeter, I just wrote 10 to negative 2. And final answer is 3.84 times 10 to negative 10 meter. So the answer is D. In question 4, we have the mass and percentage of 3 isotope and we would like to calculate the average atomic mass. We know we supposed to have multiplication of mass under fractional abundance. Fractional abundance is percentage divided by 100. So for first one we have the mass is 72.3942 times instead of 32.60 we write 0.3260 we divided by 100 same thing for second one and the third one for second one we have 72.8968 times 0.4990 and the third one we have 75.1104 times its abundance is 0.1750 The answer for first one is 23.6005. The answer for second one is 36.3755 and the answer for third one is 13.1443 addition of this three is our final answer which is equals to 73.1203 and the answer is b for question five We would like to know what is the name of this ionic compound. PO3, 3 minus, it's phosphide. So A is not the answer, B is not, and C is not. We only need to know if the titanium is 1 or 2. And it's very obvious here that titanium it is 2 because We have 2 for phosphide. We know the 2 is for titanium and 3 is for phosphide. So our ions are titanium 2 and phosphide 3. So the answer is D. In question 6, we would like to know what is the formula for hypobaramos acid. Hypo for oxyacid of halogens is always referring to 1 oxygen. So our answer is HBrO. Question 7. We would like to know how many protons, neutrons, and electrons do we have in indium? 3 plus atomic number is 39 and mass number is 113. So atomic number it give us the number of protons. So the first one is number of protons. and second one is number of neutrons number of neutrons is mass number minus atomic number it equals 74 so number of proton is 39 number of neutrons are 74 and number of electrons this is a cation it has 3 plus charge so it means the atom lost 3 electrons So instead of having 39 electrons from atomic number, we had to have 3 electrons less. So we should have 36 electrons. So the answer is E. For question 8, we would like to know which of these choices have the greatest number of oxygen atom. It's easier to calculate the number of moles for oxygen atom. for first one we have 98 gram nitric acid so i can convert this 98 gram nitric acid to the mole of nitric acid one mole nitric acid is equals to 63.01 gram nitric acid and nitric acid has three oxygen so one mole nitric acid should have three mole oxygen so we can write one mole HNO3 three mole oxygen atom the answer is 4.67 mol of oxygen this is for nitric acid for second one we have 130 gram sulfuric acid we have the same type of calculation the molar mass for sulfuric acid is 98.08 then because sulfuric acid has 4 oxygen We need to multiply the number of moles by 4 to get the mole for oxygen. So 4 moles of oxygen is equal to 1 mole sulfuric acid. And the answer here is 5.30. For the third one, we already have the mole for CO2. But we want to find the mole of oxygen atom. So 2.8 moles of carbon dioxide should multiply by 2. because each carbon dioxide has two oxygen atom the answer is 5.6 so now among this theory C it has the highest number of oxygen atom so to make it easier I just erase the rest of them so C has 5.6 mol of oxygen D we have 5.2 mol of calcium carbonate but because calcium carbonate has 3 oxygen we need to multiply it by 3 so the answer is 15.6 mol of oxygen so now we know the answer is not also C. For last one we have number of molecules. So 9.820 times 10 to 24 of molecule for CO. To convert it to the mole we need to divide it by Avogadro's number. And because carbon monoxide and oxygen they have same number of mole we don't need to convert it to oxygen anymore. The answer is 16.3 mole of CO or mole of oxygen. so the answer is E has the highest number of moles for oxygen atom in question 9 we would like to calculate the empirical formula for this compound we can always convert this percentage to gram we can just simply assume we have 100 gram sample so we have 29.44 gram calcium we have 47.01 gram oxygen and we have 23.55 gram sulfur then we can convert them to their mole one mole calcium is 40.078 so the answer is 0.7346 mole calcium for oxygen one mole oxygen is 15.999 so the answer is 2.938 mole oxygen and for the third one one mole sulfur is 32.065 gram and the answer is 0.7344 mol of sulfur now we have the mol of each elements we need to find their simplest ratio we need to divide it by the smallest value for mol which is the value for sulfur so if i dividing these three values bar the value for mol of sulfur 0.73 4 4 and 0.7344 the value for calcium is almost one the value for sulfur for sure is one and the value for oxygen 2.99.38 divided by 0.7344 is exactly four so in our empirical formula we have one calcium one sulfur and we have four oxygen so it is calcium sulfate. C-A-S-O-4. So the answer is C. In question 10 we would like to balance this equation and find out what is the coefficient for water. Here in our product we have calcium phosphate. So we have 3 calcium and we have 2 phosphate. To balance number of calcium we should have a 3 as a coefficient of calcium hydroxide and because we have two phosphate we need to have two phosphoric acid by having these two coefficients calcium and phosphate they are balanced then we can easily balance number of hydrogen as well here we have two times three six hydrogen and three times two six hydrogen as well so in total we have 12 hydrogen And here we have only two hydrogen. So the coefficient of water should be six. So if I have six here for water, then the equation is balanced. So the answer is B. In question 11, we have an stoichiometry question. We would like to know how many grams of calcium chloride do we need if we want to produce 6.20 grams silver. coloride always for the stoichiometric question the first step is to make sure the equation is balanced it is not balanced because here we have two coloring and here we have only one coloring so i need to balance the equation i put two here for silver color right right now number of silver is not balanced so i need to put a two here as well Right now everything is balanced calcium silver Nitrate and Chloride. All of them they are balanced. Now I can use this equation to find what is the value for Calcium Chloride. So I have 6.20 gram AgCl. I need to convert gram to mole of AgCl. So 1 mole AgCl is 143.32 gram. Then the next step I can convert mol of AgCl to mol of calcium chloride. Regarding the equation 2 mol AgCl is equals to 1 mol calcium chloride. And at the end I can convert mol of calcium chloride to its gram by using its molar mass. the molar mass for calcium chloride is 110.98 gram the answer is equals to 2.4 gram so C is the answer in next question we have again a stoichiometric question but we have the value for both of nitrogen and oxygen so we have the value for both of reactant so it is a limiting reactant question the equation is balanced so we don't need to be worried about it i have to do the calculation for finding the amount of product two times one time based on 25 gram nitrogen and one times based on 47 gram o2 so first i'm going to use n2 gm of N2 converts to mol of N2 1 mol N2 is 28.01 gm N2 it's given in the question then regarding the question we know 2 mol N2 is equal to 2 mol dinitrogen pentoxide and at the end we can convert mol of N2O5 to its gram the molar mass for the nitrogen pentoxide is 108.01 the answer for this one is 96.4 gram and let's find out what is the answer for calculation based on oxygen i can convert gram of oxygen to its mole so 1 mole o2 it's 32 regarding the question so 32 gram O2 times 10 in the equation 5 mol O2 is equal to 2 mol of dinitrogen pentoxide so I can write 5 mol O2 2 mol N2O5 and again the last conversion factor is the same I can convert mol of N2O5 to its gram by using 108.01 its molar mass the answer for the second calculation is 63.5 always in limiting reactant questions the lower value is our actual answer so 63.5 is our answer so oxygen it was our limiting reactant in question 13 We have a reaction of 28 gram ammonia with excess oxygen and it forms 33.8 gram water. And we would like to find the percent yield. So 33.8 gram of water is the actual value because this is what forms in the reaction. We can also calculate the theoretical one by using a stoichiometry. So 28 gram ammonia I convert it to Mol so 17.031 gram ammonia is 1 mol ammonia. Then regarding the question 4 mol ammonia is equal to 6 mol water. Then at the end I can convert mol of water to gram of water. 18.015 gram. The answer for the value of water is 44.4 gram this is our theoretical and we know the percent yield is actual value over theoretical value times 100 that's the percent yield of reaction which is equal to 76.1 percent so the answer is d in next question we would like to know which of these compound is soluble in water so we can take a look to the chart for solubility the group one compounds they are always soluble so when we have sodium potassium lithium they are always soluble in water so c is the answer we can double check based on the solubility chart and we will see all of the rest of these choices are insoluble in water in question 15 we would like to know which of this solution doesn't precipitate when they're mixed together we need to switch the cation and anion to find out what is the product and double check them in solubility chart so here iron and potassium they should change so our final product should be fe cl2 and potassium nitrate chloride is only insoluble for silver mercury and lead so iron chloride it is soluble and nitrate is always soluble so a doesn't form any solid so since both of iron chloride and potassium nitrate our product are soluble we don't have any precipitation reaction so the answer is a we can also write the product for the rest of choices and we will find that in b pi2 it's solid in c copper hydroxide is insoluble in d calcium carbonate is insoluble and in e nickel phosphate so the answer is a for question 16 we have a reaction between hcl and magnesium hydroxide you would like to know what is the net ionic equation so here is the equation hcl it's soluble in water reacts with Magnesium hydroxide but regarding the solubility chart magnesium hydroxide is not soluble in water. Then H and OH they form water. Water is liquid and Cl and magnesium they form magnesium chloride. Formula for magnesium chloride is MgCl2 and MgCl2 is aqueous. So if I want to write the ionic equation HCl is converged to H plus and Cl minus. Magnesium hydroxide is not soluble so I'm not supposed to separate the cation and anion. Water is not ionic compound so we write it in this form. And MgCl2 it's an ionic compound is aqueous so we can write it in their ionic form. this equation it wasn't balanced so actually we should have two HCl and then we should have two water to make the equation balanced so here instead of Cl and H I should write two H plus and two Cl minus and here I should have two water two Cl minus On both sides, they cancel out. It's an spectator ion. And the net ionic equation is 2H+, magnesium hydroxide, 2 water and Mg cation. So the answer is D. In question 17, we have a solution stoichiometry question. We would like to calculate how many grams of aluminum hydroxide do we need? to react completely with 95 milliliter of 2.10 molar nitric acid. So we have 95 milliliter of nitric acid. We can convert volume to mole by using concentration. 2.1 molar it means 2.1 mole in one liter. That's the definition of molarity. mol per liter but because we have milliliter here instead of one liter i would like to write thousand milliliter so milliliter and milliliter they cancel out right now i have mol of nitric acid then i can easily convert it to mol of aluminum three mol nitric acid is equals to one mol aluminum hydroxide then i can convert mol of aluminum hydroxide to gram aluminum hydroxide one mol is equals to 78 gram so the answer is equals to 5.18 7 gram aluminum hydroxide or we can round it up to 5.19 the answer is A. In question 18 we have a dilution question. We would like to calculate the molarity of diluted solution if we add 35 milliliter water to this solution. So V1 is equals to 62.4 milliliter and m1 is equals to 2.75 molar v2 it is equal as to the 35 milliliter plus the original volume of solution so 62.4 plus 35 which is equal to 97.4 milliliter and we want to calculate M2. M1 V1 M2 V2 or we can rewrite it in this form. M2 is equals to M1 V1 over V2. M1 is equals to 2.75. V1 is 62.4 and V2 is 97. The answer is equal to 1.76 molar. So the answer is C. In next question, we would like to calculate the oxidation number of chlorine atom. Oxidation number for hydrogen is 1. For chlorine, we don't know. And for oxygen, It's negative 2. So 4 times negative 2 because we have 4 oxygen. This is the molecule. So in molecule, the total of oxidation number should be equals to 0. So we can say 1 plus Cl minus 8 is equals 0. Or we can write Cl minus 7 is equals to 0. So chlorine is equals to positive. The answer is A. In question 20, we would like to calculate the energy of one photon for a radio wave with frequency of 4.3 times 10 to positive 9 hertz. So the energy for one photon is equal to the Planck's constant times frequency. Planck's constant is equal to 6.626. times 10 to negative 34 and the frequency is 4.3 times 10 to positive 9. the answer is equals to 2.8 times 10 to negative 24. b is the answer in question 21 we would like to know which of this transition in Bohr atomic model has the lowest energy of emissions. So in emission, electron they move from higher level to the lower level. So A is not emissions because electron moving to the higher level. But B, C, D, and E, all of them electron moving to the lower level and we have emission. So this is level 1, level 2, level 3, 4, 5. so we have three to one we have three to two we have four to three and we have five to four so three to one it's two level difference and it has the higher energy but three to two four to three and five to four all of them electron going down to one lower level But we know the higher energy level in Bohr atomic model they have the less distance in their energy. So the distance between 5 to 4 is lower than 4 to 3. And the differences between 4 to 3 is less than 3 and 2. So the lowest energy is between 5 to 4 in this question. So and it has the lowest energy and lowest frequency and longest wavelength. In question 22, we would like to know how many electrons in one atom can have this quantum number. L equals 2 is belong to the D orbital and when it's on level 4, we just write 4D. So N equals 4, L equals 2, it means 4D orbital and we know D orbital, they can have maximum 10 electrons. So the answer is D. In question 23, we would like to know which of these quantum numbers is not allowed. In A, we have n equals 2, l equals 2. It is wrong because l value is always 0, 1 up to n minus 1. So when n is equals 2, 2 minus 1, the maximum possibility for l value is 1. So it's impossible to have L equals 2 in second layer. So this is not allowed. And the answer is A. In question 24, we would like to know which of these diagrams for electron configuration is violating the Pauli exclusion principle. For this principle, we cannot have two electrons with same spin in one orbital. So C. We have two electrons with same spin in one orbital. And this is violation of Pauli exclusion principle. So C is the answer. In A, we have two electrons in one orbital and one electron in another one. Instead of having three single electrons. So A is violating of Hans rule, not Pauli rule. Same thing for B. because when we put single electron regarding the Hans rule they should have same spin not opposite spin and D it is correct it doesn't violate any of Pauli or Afba or Hans rule in question 25 we would like to find the electron configuration of nickel with atomic number 28 nickel 28 its electron configuration is argon 18 after argon we have 4s orbital with 2 electron then we have 3d orbital and we need to have 8 electron then 18 2 and 8 we have 28 electrons when we have nickel cation it means nickel should lose 2 electron and electron should be removed from the highest level highest level is 4s not 3d so two electrons should be removed from 4s and as a result electron configuration of nickel cation is 3d 8. 4s0 we don't write it so the answer is a in question 26 we would like to know what is the order of the radius for this species here calcium 2 plus its atomic number is 20. Potassium is 19. Argon is 18. Chlorine minus is 17. And sulfide is 16. All of them they have 18 electrons. So they are isoelectronic. And when we have isoelectronic more atomic number we have a smaller radius we have so among of these calcium has the lowest radius because we have more proton and the attraction of photon on this 18 electron is larger than other ions or atoms so calcium should be a smallest one then potassium then argon then so the answer is a in question 27 we would like to know which one has the highest electronegativity. Highest electronegativity it is on the top right of periodic table and lowest electronegativity it is on the left and down of periodic table. So when we want the highest electron negativity we should take a look to these elements and see which of them is on the most right and top in the periodic table. Chlorine it is on that position comparing to the other elements. So Chlorine has the highest electron negativity. In question 28 we would like to find which of these ionic compound has the highest melting point. highest melting point or highest boiling point it means highest lattice energy because when we have more lattice energy ions they stick together stronger and their melting of them or boiling of them it need more energy or higher temperature so we need to compare them we know lattice energy depending to the size and charge bigger cation has lower lattice energy so we have sodium cation we have potassium cation and we have cesium cation among these three sodium has the smallest radius they are on the same group and sodium should have the higher lattice energy in addition of sodium we have magnesium magnesium is smaller than sodium also its charge is double so we always looking for higher Q over radius so magnesium charge is double comparing to sodium and its radius is a little bit smaller than sodium and their anion are the same so magnesium oxide should have the higher lattice energy comparing to sodium oxide and sodium oxide has higher lattice energy than potassium compounds and cesium compounds so the answer is C in next question we would like to know how many valence electrons do we have in bicarbonate ion hydrogen has one valence electron carbon has four valence electron oxygen has six but we have three oxygen so three times six in addition of these electrons we have a negative charge so we have one extra electron so three times six is eighteen 19, 20 and 4 is 24 valence electron. So B is the answer. In question 30 we would like to know which one is the best Lewis structure for COCl2. Carbon has 4 valence electrons, oxygen 6 and chlorine has 7. So in total we have 24 electrons. carbon is our central atom then we write three bond with chlorine and oxygen then if we put six electron on each chlorine and oxygen to make them octet we can see that carbon has only six electrons so here we have two possibilities we may share the electron of oxygen to make a double bond and make carbon octet or we may share the electron of one of these, Chlorine, to make carbon octet. If we share the electron of Oxygen, we have double bond for Oxygen then E is our Lewis structure and if we have the double bond for Chlorine then A is our Lewis structure. then we need to evaluate which of these Lewis structure is better in both of them all atoms they are octet but we have to check their formal charges if I break these bonds and count number of electrons remaining on chlorine, Chlorine has only six electrons it means we have a positive charge on chlorine because chlorine is supposed to have seven electrons Also here oxygen if we break its bond, oxygen ending up with 7 electrons and we know oxygen supposed to have 6 electrons. So oxygen has a negative charge. But here, chlorine, oxygen, chlorine and carbon all of these atoms they have zero formal charge. So the best Lewis structure is this one. with double bond on oxygen so the answer is D. In question 31 we would like to find the formal charge for nitrogen atom in Nitrite, anion. NO2 minus. Nitrogen has 5 electrons. Oxygen has 6. And we have 1 extra electron for valence shell. So we have 18 electrons. Then we write bond with oxygen. Then I make both of oxygen octet. Right now I have 16 electrons so I should put two more electrons on central atom. Nitrogen still has six electrons so one of these oxygen should share its electron and make a double bond. So this is the Lewis structure for nitrate anion. If I break these bonds this oxygen has seven electrons so its formal charge is negative. If I break these bonds... this oxygen has six electron and its formal charge is zero and nitrogen has five electron and nitrogen is supposed to have five electrons so its formal charge is also zero so the answer is b in question 32 we would like to know what is the electronic arrangement of selenium atom in sebr3 plus Selenium it's belong to group 6. We have 3 Boramine for group 7 and because we have positive charge we lost 1 Electron. So the number of Electron is equals to 26 Electron. So Selenium then Boramine, Boramine and Boramine. If I put lone pairs on Boramines to make them octet. Then right now I have 24 electrons so I need to put two more electrons on selenium. As you can see here selenium has four domains. Four domains it means tetrahedral. So the electronic arrangement of atom is tetrahedral. But if we want to talk about the geometry of atom because one of these domain is lone pair. Instead of tetrahedral, we should say trigonal pyramidal. So we need to pay attention if the question asks about the geometry of molecule or geometry of electron or electronic arrangement of atom. In question 33, we would like to know what is the bond angle of HPH bond. To finding the bond angle, we need to know the geometry of molecule or ion in this case. We can use the shortcut formula or we can just draw the Lewis structure. The Lewis structure for this is in this form. Phosphorus has five electrons, each hydrogen has one electron and one for negative charge. In total we have eight electrons. So two, four, six and eight. Electronic arrangement of molecule is tetrahedral. We have four domains for phosphorus atom. Tetrahedral means the bond angle should be 109.5. But because of the effect of this lone pair on phosphorus atom, they are repelling the bond and push the bond closer. And the bond angle, it will be less than what we expect in the actual tetrahedral. So the answer is A. In question 34, we would like to know which of these molecules has a zero dipole moment. Zero dipole moment, it means which molecule is non-polar. To identify the polarity of molecule, we should derive their Lewis structure. The Lewis structure for NF3 is like ammonia. For IF3... It's T-shape, BF3. As you can see, boron has a symmetry and it doesn't have any lone pair. And this structure can cancel all the polarity vectors. So the answer is BF3. But here we have lone pairs and this lone pair cause the polarity in the molecule, all vectors. they cannot cancel each other. but i'm going to write the Lewis structure for D and E as well PBR tree is like NF tree we have phosphorus and halogens again we have lone pair and BRCL tree is like IF tree it's a T shape So another easier method to identify faster is comparing the atoms. Nitrogen and phosphorus they are on the same group. Fluorine and Boramine also. So their structure should be exactly like them. So if one of them is non-polar the another one should be non-polar. So we can have a faster way to identify which of these choices are unanswered. Same thing for B and E. IF3 and BRCl3. Central atom is group 17 and F and Chlorine also they are group 17. So these two structures they should be exactly the same. So none of them they can be the answer. So answer in this question is C. In question 35 we would like to know how many hybridized orbital do we have in carbonate ion. So CO3 to minus this is its Lewis structure. Carbonate and ion. and we can see that carbon has three domains. Three domains means three orbitals. So it means 1s and 2p. So the answer is sp2. Question 36. We would like to find the bond order for NO-. Nitrogen is on group 5. So we have 5 valence electrons. Oxygen 6 valence electron and we have 1 negative charge. So we have 12 valence electron. N2 triple bond has 10 electron and O2 double bond has 12 electron. So the bond order of NO minus it should be exactly like O2 because both of them they have 12 electron. Bond order should be 2. In question 37, we would like to know the number of sigma and pi bond in nitrate ion, NO3-. If we draw the Lewis structure for nitrate anion, we have two single bond and one double bond between nitrogen and oxygen. Each single bond is sigma and and one of the bond for double bond is sigma as well so in total we have three sigma and the other bond in double bond is pi bond so double bond is one sigma and one pi so the answer is three sigma and one pi d is the answer in question 38 we would like to calculate the delta e the delta e of system is equal to heat and work it says the system absorb 1700 joules so when it's absorbed it gain energy so the sign is positive and it says the system does 5460 joule work on surrounding it means when system works it lose energy so the sign is negative 54 60. the answer is equal to negative 3760 joule so the answer is e in next question we add a 60 gram aluminum cube its temperature is 25 into 130 gram hot water at 90 Celsius and we want to calculate what is the final temperature when aluminum and water they're reaching to same temperature when they have equilibrium to solving this colorimetry question we can say that q of water the amount of energy that water lose transfer it to aluminum is equal to the q of aluminum but because one of them losing one of them gaining we have a negative sign then we can write m of water c of water delta t of water delta T of water is equals to is equals to T of final minus T of initial initial is 90 Celsius we can write the same thing for aluminum so negative mass of aluminum C of aluminum and T of final minus T of initial initial is 25 so here Mass of water is 130. C of water is 4.184. Mass of aluminum is 60 gram. And C of aluminum is 0.90. Then if I'm multiplying these numbers to the parentheses, then I have 543.92 T of final. then multiplication of these two number to 90 minus 48952.8 this is for left side for right side then multiplication of these two number to t of final i have negative 54 tf plus 13 50. multiplication of these two number to 25. then we need to solve this equation for final temperature if i add 54 tf to both sides then on the right side negative 54 tf cancel out and if i add 4 8 9 5 2 8 to both sides then these two they cancel out so i can write the equation in this for 489528 plus 1350 and on the left side 543.92 plus 54 tf it's equals to 597.92 tf and the answer for these two is equals to 5302.8 If I divide in both sides by 597.92, then I have the answer. The answer is equal to 84. So that's the final temperature of the mixture of water and aluminum. In question 40, we have the heat stoichiometry. You would like to find how much heat is Release when we have formation of 62.5 carbon monoxide in this reaction. So we have the mass of carbon dioxide we can convert it to heat. 62.5 grams CO2. We can convert gram to mole. 1 mole CO2 is equal to 44.01 gram. then grams of carbon dioxide cancel out now regarding the equation 3 mol of carbon dioxide is equal to this value of heat so i can write 3 mol CO2 negative 20 43 kilo joule then mol of carbon dioxide they cancel out the answer is equal to 967.1 kilojoule so a is the answer in question 41 you would like to calculate the delta h of this reaction by using this three equation it is a Hess's law question so we need to take a look to our equation and find the pieces from this tree given informations in our equation we have Hydrazine N2H4 on the left side and its coefficient is 1. Here hydrazine it is on the right side and its coefficient is 2. So we need to flip the equation. So I need to write 2N2H4 plus 2H2 it's 4 ammonia. Then we flip the equation delta H also. changing the sign is changing so instead of 44 it should be negative 44 but we don't need two mole of hydrazine we need one mole So we actually need to divide the whole equation by 2. By doing this, we have 1 here, we have 1 here for hydrogen, and we have 2 for ammonia. Then the delta H of reaction also should be divided by 2. So instead of 44, it is negative 22. So we're done with N2H4. The next one is methanol, CH3OH. It is on the right side but it should be on the left side. So I need to flip this equation as well. But the coefficients are fine. CH3OH. Then on the right side CH2O plus H2. And because I flipped this equation, the delta H should change from positive to negative. so it's negative 81 and for the third part we have N2 and 3H2 on the right side here we have both of them on the left side so i need to flip this equation as well so i can write 2 ammonia then on the right side N2 and 3H2 and the sign for delta H should change from negative to positive then if i add this three equation ammonias are cancel out hydrogen and hydrogen they cancel out and what is remaining is n2h4 methanol formaldehyde on the right side and n2 and 3h2 then the delta h of this reaction should be addition of these three numbers positive 58 negative 81 and negative 22. the answer is equals to negative 45. so c is the answer in question 42 we would like to know which of these choices has the zero value for delta h of formation delta h of formation is zero only for elements under a standard condition. A, B, C, and D they are compound. So the only element in this question is D, N2. So the answer is D. In question 43 we would like to calculate delta H from the bond energy. We know the equation is the sigma for bond energy of reactants minus the sigma for bond energy of products. And we have all the bond energy for our reactants and product. In our reactants, we have CH4. CH4 has 4 CH bond. So we need to multiply the value of CH by 4. So we have 4 times the value for CH is 413. Then O2 is double bond and we have 2 molecule of O2. O2 so for O2 also we need to multiply the value by 2 495 so this is the value for reactants then let's calculate the product this is the Lewis structure for CO2 so we have 2 C double bond O so 2 times 799 and for water we have 2 OH bond for each water molecule and because we have two water molecule in total we have 4 OH bond so 4 times the value for OH it's 460 T the value for the reactant is 2642 and the value for product is 3450 So 2642 minus 3450 is equals to negative 808 kilojoule per mile. So the answer is E. In question 44, we would like to know which of these laws stated that the volume of gas is directly proportional to its absolute temperature. The answer is Charles'Law. Question 45 we have volume temperature and pressure of neon gas and we would like to calculate the volume another temperature and pressure so the formula for this type of question is combined gas law P1V1 T1 P2V2 T2 P1 it is 1.65 ATM V1 volume is 26 0.99 T1 is not 77 Celsius because we should use Kelvin in this type of question. So 273 plus 77 which is 350. We don't know V2 but P2 is 2.35 times V2 and again for second temperature final temperature. we should convert it to the kelvin 273 plus 355 it is 628 now we can solve this question for v2 so 628 is here and 350 is here if we cross multiplying v2 is equal to multiplication of this three number dividing by these two number So V2 is equal to 1.65 times 26.99 times 628 divided by 350 times 2.35. And the answer is equal to 34.00 liter. B is the answer. In question 46... We want to calculate the volume of O2 gas and we have its pressure and temperature. So we should use ideal gas law PV nRT. Or we can rewrite it in this form. V is equals to nRT over P. N is 2.1 mol. R value when we have ATM and leader it is 0.0821. then for temperature we should convert it to the kelvin so 273 plus 110 it is 383 kelvin divided by pressure which is 1.45 the final answer is equals to 45.5 liter so d is the answer in question for this 7 we would like to calculate the volume of so2 at STP condition when we have reaction of 42.60 iron sulfide with enough amount of oxygen so we have a gas stoichiometry question so we have 42.60 gram of iron sulfide i can convert gram of iron sulfide to its mole so one mole FES is equal to 87.91 gram FES. Then I can convert mol of FES to mol of SO2. 4 mol FES is equal to 4 mol SO2. 4 and 4 they cancel out and mol of iron sulfide they cancel out. We know at STP condition, 1 mol of any gas is equal to 22.4 liter. So the answer is 10.85 liter. E is the answer. In question 48, we would like to calculate the density of xenon gas at STP condition. The formula... for density is pm drt or if i rewrite the equation by dividing both side by rt i can have density is equals to pm over rt pressure at stp is 1 atm the molar mass for xenon it is 131.29 r value is 0.0821 and temperature it is zero Celsius we need to convert it to Kelvin so is 273 Kelvin then the answer is equals to 5.86 gram per liter this is the answer in question 49 we would like to know which of these following gas has the lower rate of effusion R1 over R2 is equal to m2 over m1 so this is the garahams law and we can see that heavier molecule they have lower rate of effusion we need to find out which of these compound or elements have the heavier particle n2 nitrogen mass is 14 so n2 is 28 oxygen o2 is 32. Chlorine, each chlorine is 35.45. So the molar mass or molecular mass of chlorine is 70.91. For F2, almost 38. And for carbon dioxide, it is 44. So the heaviest gas here is chlorine. So it should have the lower rate of effusion and the last question question 50 we have a mixture of three gases 3.2 mole o2 2.1 mole co2 1.7 mole helium and the total pressure is 9.5 atm and we would like to calculate partial pressure of helium so partial pressure of helium is equal to the mole fraction of helium times the total pressure. Mole fraction for helium is equal to mole of helium 1.7 over the total mole 3.2 for O2, 2.1 for CO2 and 1.7 for helium times the total pressure is 9.580 m. Here the total is equals to 7 so 1.7 divided by 7 times 9.5 then the answer is 2.3 atm so d is the answer thank you for watching this video and good luck with your final exam

So we can rewrite this equation. Mass is equal to density times volume. The density is 19.3 and volume it is the length of cube to the power of 3. So I need to calculate the length of cube in centimeters. So we have 1.10 inch. and our conversion factor for inch is 2.54 so one inch is equals to 2.54 centimeter the answer is 2.794 so instead of a in this equation i should use this number 2.794 the length of one side of cube to the power of 3 then the answer is equal to 420.95 gram or we can round it to 421 gram the answer is a here in question 3 we would like to convert this distance in kilometer to a centimeter We don't have the direct conversion of kilometer to centimeter but we can convert both of them to meter so we can write it in this form 384 this is in kilometer and we know kilo is means thousand so one kilometer it's thousand meter so kilo and kilo they cancel out then we know centi is equals to 10 to negative 2 So 1 centimeter is equal to 10 to negative 2 meter.

So instead of centimeter, I just wrote 10 to negative 2. And final answer is 3.84 times 10 to negative 10 meter. So the answer is D. In question 4, we have the mass and percentage of 3 isotope and we would like to calculate the average atomic mass. We know we supposed to have multiplication of mass under fractional abundance.

Fractional abundance is percentage divided by 100. So for first one we have the mass is 72.3942 times instead of 32.60 we write 0.3260 we divided by 100 same thing for second one and the third one for second one we have 72.8968 times 0.4990 and the third one we have 75.1104 times its abundance is 0.1750 The answer for first one is 23.6005. The answer for second one is 36.3755 and the answer for third one is 13.1443 addition of this three is our final answer which is equals to 73.1203 and the answer is b for question five We would like to know what is the name of this ionic compound. PO3, 3 minus, it's phosphide. So A is not the answer, B is not, and C is not. We only need to know if the titanium is 1 or 2. And it's very obvious here that titanium it is 2 because We have 2 for phosphide.

We know the 2 is for titanium and 3 is for phosphide. So our ions are titanium 2 and phosphide 3. So the answer is D. In question 6, we would like to know what is the formula for hypobaramos acid. Hypo for oxyacid of halogens is always referring to 1 oxygen. So our answer is HBrO.

Question 7. We would like to know how many protons, neutrons, and electrons do we have in indium? 3 plus atomic number is 39 and mass number is 113. So atomic number it give us the number of protons. So the first one is number of protons.

and second one is number of neutrons number of neutrons is mass number minus atomic number it equals 74 so number of proton is 39 number of neutrons are 74 and number of electrons this is a cation it has 3 plus charge so it means the atom lost 3 electrons So instead of having 39 electrons from atomic number, we had to have 3 electrons less. So we should have 36 electrons. So the answer is E. For question 8, we would like to know which of these choices have the greatest number of oxygen atom.

It's easier to calculate the number of moles for oxygen atom. for first one we have 98 gram nitric acid so i can convert this 98 gram nitric acid to the mole of nitric acid one mole nitric acid is equals to 63.01 gram nitric acid and nitric acid has three oxygen so one mole nitric acid should have three mole oxygen so we can write one mole HNO3 three mole oxygen atom the answer is 4.67 mol of oxygen this is for nitric acid for second one we have 130 gram sulfuric acid we have the same type of calculation the molar mass for sulfuric acid is 98.08 then because sulfuric acid has 4 oxygen We need to multiply the number of moles by 4 to get the mole for oxygen. So 4 moles of oxygen is equal to 1 mole sulfuric acid. And the answer here is 5.30. For the third one, we already have the mole for CO2.

But we want to find the mole of oxygen atom. So 2.8 moles of carbon dioxide should multiply by 2. because each carbon dioxide has two oxygen atom the answer is 5.6 so now among this theory C it has the highest number of oxygen atom so to make it easier I just erase the rest of them so C has 5.6 mol of oxygen D we have 5.2 mol of calcium carbonate but because calcium carbonate has 3 oxygen we need to multiply it by 3 so the answer is 15.6 mol of oxygen so now we know the answer is not also C. For last one we have number of molecules.

So 9.820 times 10 to 24 of molecule for CO. To convert it to the mole we need to divide it by Avogadro's number. And because carbon monoxide and oxygen they have same number of mole we don't need to convert it to oxygen anymore. The answer is 16.3 mole of CO or mole of oxygen. so the answer is E has the highest number of moles for oxygen atom in question 9 we would like to calculate the empirical formula for this compound we can always convert this percentage to gram we can just simply assume we have 100 gram sample so we have 29.44 gram calcium we have 47.01 gram oxygen and we have 23.55 gram sulfur then we can convert them to their mole one mole calcium is 40.078 so the answer is 0.7346 mole calcium for oxygen one mole oxygen is 15.999 so the answer is 2.938 mole oxygen and for the third one one mole sulfur is 32.065 gram and the answer is 0.7344 mol of sulfur now we have the mol of each elements we need to find their simplest ratio we need to divide it by the smallest value for mol which is the value for sulfur so if i dividing these three values bar the value for mol of sulfur 0.73 4 4 and 0.7344 the value for calcium is almost one the value for sulfur for sure is one and the value for oxygen 2.99.38 divided by 0.7344 is exactly four so in our empirical formula we have one calcium one sulfur and we have four oxygen so it is calcium sulfate.

C-A-S-O-4. So the answer is C. In question 10 we would like to balance this equation and find out what is the coefficient for water. Here in our product we have calcium phosphate. So we have 3 calcium and we have 2 phosphate.

To balance number of calcium we should have a 3 as a coefficient of calcium hydroxide and because we have two phosphate we need to have two phosphoric acid by having these two coefficients calcium and phosphate they are balanced then we can easily balance number of hydrogen as well here we have two times three six hydrogen and three times two six hydrogen as well so in total we have 12 hydrogen And here we have only two hydrogen. So the coefficient of water should be six. So if I have six here for water, then the equation is balanced. So the answer is B. In question 11, we have an stoichiometry question.

We would like to know how many grams of calcium chloride do we need if we want to produce 6.20 grams silver. coloride always for the stoichiometric question the first step is to make sure the equation is balanced it is not balanced because here we have two coloring and here we have only one coloring so i need to balance the equation i put two here for silver color right right now number of silver is not balanced so i need to put a two here as well Right now everything is balanced calcium silver Nitrate and Chloride. All of them they are balanced.

Now I can use this equation to find what is the value for Calcium Chloride. So I have 6.20 gram AgCl. I need to convert gram to mole of AgCl. So 1 mole AgCl is 143.32 gram. Then the next step I can convert mol of AgCl to mol of calcium chloride.

Regarding the equation 2 mol AgCl is equals to 1 mol calcium chloride. And at the end I can convert mol of calcium chloride to its gram by using its molar mass. the molar mass for calcium chloride is 110.98 gram the answer is equals to 2.4 gram so C is the answer in next question we have again a stoichiometric question but we have the value for both of nitrogen and oxygen so we have the value for both of reactant so it is a limiting reactant question the equation is balanced so we don't need to be worried about it i have to do the calculation for finding the amount of product two times one time based on 25 gram nitrogen and one times based on 47 gram o2 so first i'm going to use n2 gm of N2 converts to mol of N2 1 mol N2 is 28.01 gm N2 it's given in the question then regarding the question we know 2 mol N2 is equal to 2 mol dinitrogen pentoxide and at the end we can convert mol of N2O5 to its gram the molar mass for the nitrogen pentoxide is 108.01 the answer for this one is 96.4 gram and let's find out what is the answer for calculation based on oxygen i can convert gram of oxygen to its mole so 1 mole o2 it's 32 regarding the question so 32 gram O2 times 10 in the equation 5 mol O2 is equal to 2 mol of dinitrogen pentoxide so I can write 5 mol O2 2 mol N2O5 and again the last conversion factor is the same I can convert mol of N2O5 to its gram by using 108.01 its molar mass the answer for the second calculation is 63.5 always in limiting reactant questions the lower value is our actual answer so 63.5 is our answer so oxygen it was our limiting reactant in question 13 We have a reaction of 28 gram ammonia with excess oxygen and it forms 33.8 gram water. And we would like to find the percent yield. So 33.8 gram of water is the actual value because this is what forms in the reaction.

We can also calculate the theoretical one by using a stoichiometry. So 28 gram ammonia I convert it to Mol so 17.031 gram ammonia is 1 mol ammonia. Then regarding the question 4 mol ammonia is equal to 6 mol water. Then at the end I can convert mol of water to gram of water. 18.015 gram.

The answer for the value of water is 44.4 gram this is our theoretical and we know the percent yield is actual value over theoretical value times 100 that's the percent yield of reaction which is equal to 76.1 percent so the answer is d in next question we would like to know which of these compound is soluble in water so we can take a look to the chart for solubility the group one compounds they are always soluble so when we have sodium potassium lithium they are always soluble in water so c is the answer we can double check based on the solubility chart and we will see all of the rest of these choices are insoluble in water in question 15 we would like to know which of this solution doesn't precipitate when they're mixed together we need to switch the cation and anion to find out what is the product and double check them in solubility chart so here iron and potassium they should change so our final product should be fe cl2 and potassium nitrate chloride is only insoluble for silver mercury and lead so iron chloride it is soluble and nitrate is always soluble so a doesn't form any solid so since both of iron chloride and potassium nitrate our product are soluble we don't have any precipitation reaction so the answer is a we can also write the product for the rest of choices and we will find that in b pi2 it's solid in c copper hydroxide is insoluble in d calcium carbonate is insoluble and in e nickel phosphate so the answer is a for question 16 we have a reaction between hcl and magnesium hydroxide you would like to know what is the net ionic equation so here is the equation hcl it's soluble in water reacts with Magnesium hydroxide but regarding the solubility chart magnesium hydroxide is not soluble in water. Then H and OH they form water. Water is liquid and Cl and magnesium they form magnesium chloride.

Formula for magnesium chloride is MgCl2 and MgCl2 is aqueous. So if I want to write the ionic equation HCl is converged to H plus and Cl minus. Magnesium hydroxide is not soluble so I'm not supposed to separate the cation and anion. Water is not ionic compound so we write it in this form.

And MgCl2 it's an ionic compound is aqueous so we can write it in their ionic form. this equation it wasn't balanced so actually we should have two HCl and then we should have two water to make the equation balanced so here instead of Cl and H I should write two H plus and two Cl minus and here I should have two water two Cl minus On both sides, they cancel out. It's an spectator ion. And the net ionic equation is 2H+, magnesium hydroxide, 2 water and Mg cation.

So the answer is D. In question 17, we have a solution stoichiometry question. We would like to calculate how many grams of aluminum hydroxide do we need?

to react completely with 95 milliliter of 2.10 molar nitric acid. So we have 95 milliliter of nitric acid. We can convert volume to mole by using concentration.

2.1 molar it means 2.1 mole in one liter. That's the definition of molarity. mol per liter but because we have milliliter here instead of one liter i would like to write thousand milliliter so milliliter and milliliter they cancel out right now i have mol of nitric acid then i can easily convert it to mol of aluminum three mol nitric acid is equals to one mol aluminum hydroxide then i can convert mol of aluminum hydroxide to gram aluminum hydroxide one mol is equals to 78 gram so the answer is equals to 5.18 7 gram aluminum hydroxide or we can round it up to 5.19 the answer is A. In question 18 we have a dilution question. We would like to calculate the molarity of diluted solution if we add 35 milliliter water to this solution.

So V1 is equals to 62.4 milliliter and m1 is equals to 2.75 molar v2 it is equal as to the 35 milliliter plus the original volume of solution so 62.4 plus 35 which is equal to 97.4 milliliter and we want to calculate M2. M1 V1 M2 V2 or we can rewrite it in this form. M2 is equals to M1 V1 over V2.

M1 is equals to 2.75. V1 is 62.4 and V2 is 97. The answer is equal to 1.76 molar. So the answer is C.

In next question, we would like to calculate the oxidation number of chlorine atom. Oxidation number for hydrogen is 1. For chlorine, we don't know. And for oxygen, It's negative 2. So 4 times negative 2 because we have 4 oxygen. This is the molecule.

So in molecule, the total of oxidation number should be equals to 0. So we can say 1 plus Cl minus 8 is equals 0. Or we can write Cl minus 7 is equals to 0. So chlorine is equals to positive. The answer is A. In question 20, we would like to calculate the energy of one photon for a radio wave with frequency of 4.3 times 10 to positive 9 hertz.

So the energy for one photon is equal to the Planck's constant times frequency. Planck's constant is equal to 6.626. times 10 to negative 34 and the frequency is 4.3 times 10 to positive 9. the answer is equals to 2.8 times 10 to negative 24. b is the answer in question 21 we would like to know which of this transition in Bohr atomic model has the lowest energy of emissions. So in emission, electron they move from higher level to the lower level.

So A is not emissions because electron moving to the higher level. But B, C, D, and E, all of them electron moving to the lower level and we have emission. So this is level 1, level 2, level 3, 4, 5. so we have three to one we have three to two we have four to three and we have five to four so three to one it's two level difference and it has the higher energy but three to two four to three and five to four all of them electron going down to one lower level But we know the higher energy level in Bohr atomic model they have the less distance in their energy.

So the distance between 5 to 4 is lower than 4 to 3. And the differences between 4 to 3 is less than 3 and 2. So the lowest energy is between 5 to 4 in this question. So and it has the lowest energy and lowest frequency and longest wavelength. In question 22, we would like to know how many electrons in one atom can have this quantum number. L equals 2 is belong to the D orbital and when it's on level 4, we just write 4D. So N equals 4, L equals 2, it means 4D orbital and we know D orbital, they can have maximum 10 electrons.

So the answer is D. In question 23, we would like to know which of these quantum numbers is not allowed. In A, we have n equals 2, l equals 2. It is wrong because l value is always 0, 1 up to n minus 1. So when n is equals 2, 2 minus 1, the maximum possibility for l value is 1. So it's impossible to have L equals 2 in second layer.

So this is not allowed. And the answer is A. In question 24, we would like to know which of these diagrams for electron configuration is violating the Pauli exclusion principle. For this principle, we cannot have two electrons with same spin in one orbital.

So C. We have two electrons with same spin in one orbital. And this is violation of Pauli exclusion principle. So C is the answer. In A, we have two electrons in one orbital and one electron in another one. Instead of having three single electrons.

So A is violating of Hans rule, not Pauli rule. Same thing for B. because when we put single electron regarding the Hans rule they should have same spin not opposite spin and D it is correct it doesn't violate any of Pauli or Afba or Hans rule in question 25 we would like to find the electron configuration of nickel with atomic number 28 nickel 28 its electron configuration is argon 18 after argon we have 4s orbital with 2 electron then we have 3d orbital and we need to have 8 electron then 18 2 and 8 we have 28 electrons when we have nickel cation it means nickel should lose 2 electron and electron should be removed from the highest level highest level is 4s not 3d so two electrons should be removed from 4s and as a result electron configuration of nickel cation is 3d 8. 4s0 we don't write it so the answer is a in question 26 we would like to know what is the order of the radius for this species here calcium 2 plus its atomic number is 20. Potassium is 19. Argon is 18. Chlorine minus is 17. And sulfide is 16. All of them they have 18 electrons.

So they are isoelectronic. And when we have isoelectronic more atomic number we have a smaller radius we have so among of these calcium has the lowest radius because we have more proton and the attraction of photon on this 18 electron is larger than other ions or atoms so calcium should be a smallest one then potassium then argon then so the answer is a in question 27 we would like to know which one has the highest electronegativity. Highest electronegativity it is on the top right of periodic table and lowest electronegativity it is on the left and down of periodic table.

So when we want the highest electron negativity we should take a look to these elements and see which of them is on the most right and top in the periodic table. Chlorine it is on that position comparing to the other elements. So Chlorine has the highest electron negativity. In question 28 we would like to find which of these ionic compound has the highest melting point.

highest melting point or highest boiling point it means highest lattice energy because when we have more lattice energy ions they stick together stronger and their melting of them or boiling of them it need more energy or higher temperature so we need to compare them we know lattice energy depending to the size and charge bigger cation has lower lattice energy so we have sodium cation we have potassium cation and we have cesium cation among these three sodium has the smallest radius they are on the same group and sodium should have the higher lattice energy in addition of sodium we have magnesium magnesium is smaller than sodium also its charge is double so we always looking for higher Q over radius so magnesium charge is double comparing to sodium and its radius is a little bit smaller than sodium and their anion are the same so magnesium oxide should have the higher lattice energy comparing to sodium oxide and sodium oxide has higher lattice energy than potassium compounds and cesium compounds so the answer is C in next question we would like to know how many valence electrons do we have in bicarbonate ion hydrogen has one valence electron carbon has four valence electron oxygen has six but we have three oxygen so three times six in addition of these electrons we have a negative charge so we have one extra electron so three times six is eighteen 19, 20 and 4 is 24 valence electron. So B is the answer. In question 30 we would like to know which one is the best Lewis structure for COCl2. Carbon has 4 valence electrons, oxygen 6 and chlorine has 7. So in total we have 24 electrons.

carbon is our central atom then we write three bond with chlorine and oxygen then if we put six electron on each chlorine and oxygen to make them octet we can see that carbon has only six electrons so here we have two possibilities we may share the electron of oxygen to make a double bond and make carbon octet or we may share the electron of one of these, Chlorine, to make carbon octet. If we share the electron of Oxygen, we have double bond for Oxygen then E is our Lewis structure and if we have the double bond for Chlorine then A is our Lewis structure. then we need to evaluate which of these Lewis structure is better in both of them all atoms they are octet but we have to check their formal charges if I break these bonds and count number of electrons remaining on chlorine, Chlorine has only six electrons it means we have a positive charge on chlorine because chlorine is supposed to have seven electrons Also here oxygen if we break its bond, oxygen ending up with 7 electrons and we know oxygen supposed to have 6 electrons. So oxygen has a negative charge. But here, chlorine, oxygen, chlorine and carbon all of these atoms they have zero formal charge.

So the best Lewis structure is this one. with double bond on oxygen so the answer is D. In question 31 we would like to find the formal charge for nitrogen atom in Nitrite, anion. NO2 minus.

Nitrogen has 5 electrons. Oxygen has 6. And we have 1 extra electron for valence shell. So we have 18 electrons.

Then we write bond with oxygen. Then I make both of oxygen octet. Right now I have 16 electrons so I should put two more electrons on central atom. Nitrogen still has six electrons so one of these oxygen should share its electron and make a double bond. So this is the Lewis structure for nitrate anion.

If I break these bonds this oxygen has seven electrons so its formal charge is negative. If I break these bonds... this oxygen has six electron and its formal charge is zero and nitrogen has five electron and nitrogen is supposed to have five electrons so its formal charge is also zero so the answer is b in question 32 we would like to know what is the electronic arrangement of selenium atom in sebr3 plus Selenium it's belong to group 6. We have 3 Boramine for group 7 and because we have positive charge we lost 1 Electron. So the number of Electron is equals to 26 Electron.

So Selenium then Boramine, Boramine and Boramine. If I put lone pairs on Boramines to make them octet. Then right now I have 24 electrons so I need to put two more electrons on selenium. As you can see here selenium has four domains. Four domains it means tetrahedral.

So the electronic arrangement of atom is tetrahedral. But if we want to talk about the geometry of atom because one of these domain is lone pair. Instead of tetrahedral, we should say trigonal pyramidal.

So we need to pay attention if the question asks about the geometry of molecule or geometry of electron or electronic arrangement of atom. In question 33, we would like to know what is the bond angle of HPH bond. To finding the bond angle, we need to know the geometry of molecule or ion in this case. We can use the shortcut formula or we can just draw the Lewis structure.

The Lewis structure for this is in this form. Phosphorus has five electrons, each hydrogen has one electron and one for negative charge. In total we have eight electrons. So two, four, six and eight. Electronic arrangement of molecule is tetrahedral.

We have four domains for phosphorus atom. Tetrahedral means the bond angle should be 109.5. But because of the effect of this lone pair on phosphorus atom, they are repelling the bond and push the bond closer. And the bond angle, it will be less than what we expect in the actual tetrahedral.

So the answer is A. In question 34, we would like to know which of these molecules has a zero dipole moment. Zero dipole moment, it means which molecule is non-polar. To identify the polarity of molecule, we should derive their Lewis structure.

The Lewis structure for NF3 is like ammonia. For IF3... It's T-shape, BF3.

As you can see, boron has a symmetry and it doesn't have any lone pair. And this structure can cancel all the polarity vectors. So the answer is BF3. But here we have lone pairs and this lone pair cause the polarity in the molecule, all vectors.

they cannot cancel each other. but i'm going to write the Lewis structure for D and E as well PBR tree is like NF tree we have phosphorus and halogens again we have lone pair and BRCL tree is like IF tree it's a T shape So another easier method to identify faster is comparing the atoms. Nitrogen and phosphorus they are on the same group. Fluorine and Boramine also.

So their structure should be exactly like them. So if one of them is non-polar the another one should be non-polar. So we can have a faster way to identify which of these choices are unanswered.

Same thing for B and E. IF3 and BRCl3. Central atom is group 17 and F and Chlorine also they are group 17. So these two structures they should be exactly the same. So none of them they can be the answer.

So answer in this question is C. In question 35 we would like to know how many hybridized orbital do we have in carbonate ion. So CO3 to minus this is its Lewis structure. Carbonate and ion. and we can see that carbon has three domains.

Three domains means three orbitals. So it means 1s and 2p. So the answer is sp2.

Question 36. We would like to find the bond order for NO-. Nitrogen is on group 5. So we have 5 valence electrons. Oxygen 6 valence electron and we have 1 negative charge. So we have 12 valence electron.

N2 triple bond has 10 electron and O2 double bond has 12 electron. So the bond order of NO minus it should be exactly like O2 because both of them they have 12 electron. Bond order should be 2. In question 37, we would like to know the number of sigma and pi bond in nitrate ion, NO3-. If we draw the Lewis structure for nitrate anion, we have two single bond and one double bond between nitrogen and oxygen.

Each single bond is sigma and and one of the bond for double bond is sigma as well so in total we have three sigma and the other bond in double bond is pi bond so double bond is one sigma and one pi so the answer is three sigma and one pi d is the answer in question 38 we would like to calculate the delta e the delta e of system is equal to heat and work it says the system absorb 1700 joules so when it's absorbed it gain energy so the sign is positive and it says the system does 5460 joule work on surrounding it means when system works it lose energy so the sign is negative 54 60. the answer is equal to negative 3760 joule so the answer is e in next question we add a 60 gram aluminum cube its temperature is 25 into 130 gram hot water at 90 Celsius and we want to calculate what is the final temperature when aluminum and water they're reaching to same temperature when they have equilibrium to solving this colorimetry question we can say that q of water the amount of energy that water lose transfer it to aluminum is equal to the q of aluminum but because one of them losing one of them gaining we have a negative sign then we can write m of water c of water delta t of water delta T of water is equals to is equals to T of final minus T of initial initial is 90 Celsius we can write the same thing for aluminum so negative mass of aluminum C of aluminum and T of final minus T of initial initial is 25 so here Mass of water is 130. C of water is 4.184. Mass of aluminum is 60 gram. And C of aluminum is 0.90. Then if I'm multiplying these numbers to the parentheses, then I have 543.92 T of final. then multiplication of these two number to 90 minus 48952.8 this is for left side for right side then multiplication of these two number to t of final i have negative 54 tf plus 13 50. multiplication of these two number to 25. then we need to solve this equation for final temperature if i add 54 tf to both sides then on the right side negative 54 tf cancel out and if i add 4 8 9 5 2 8 to both sides then these two they cancel out so i can write the equation in this for 489528 plus 1350 and on the left side 543.92 plus 54 tf it's equals to 597.92 tf and the answer for these two is equals to 5302.8 If I divide in both sides by 597.92, then I have the answer.

The answer is equal to 84. So that's the final temperature of the mixture of water and aluminum. In question 40, we have the heat stoichiometry. You would like to find how much heat is Release when we have formation of 62.5 carbon monoxide in this reaction. So we have the mass of carbon dioxide we can convert it to heat.

62.5 grams CO2. We can convert gram to mole. 1 mole CO2 is equal to 44.01 gram. then grams of carbon dioxide cancel out now regarding the equation 3 mol of carbon dioxide is equal to this value of heat so i can write 3 mol CO2 negative 20 43 kilo joule then mol of carbon dioxide they cancel out the answer is equal to 967.1 kilojoule so a is the answer in question 41 you would like to calculate the delta h of this reaction by using this three equation it is a Hess's law question so we need to take a look to our equation and find the pieces from this tree given informations in our equation we have Hydrazine N2H4 on the left side and its coefficient is 1. Here hydrazine it is on the right side and its coefficient is 2. So we need to flip the equation.

So I need to write 2N2H4 plus 2H2 it's 4 ammonia. Then we flip the equation delta H also. changing the sign is changing so instead of 44 it should be negative 44 but we don't need two mole of hydrazine we need one mole So we actually need to divide the whole equation by 2. By doing this, we have 1 here, we have 1 here for hydrogen, and we have 2 for ammonia. Then the delta H of reaction also should be divided by 2. So instead of 44, it is negative 22. So we're done with N2H4. The next one is methanol, CH3OH.

It is on the right side but it should be on the left side. So I need to flip this equation as well. But the coefficients are fine. CH3OH.

Then on the right side CH2O plus H2. And because I flipped this equation, the delta H should change from positive to negative. so it's negative 81 and for the third part we have N2 and 3H2 on the right side here we have both of them on the left side so i need to flip this equation as well so i can write 2 ammonia then on the right side N2 and 3H2 and the sign for delta H should change from negative to positive then if i add this three equation ammonias are cancel out hydrogen and hydrogen they cancel out and what is remaining is n2h4 methanol formaldehyde on the right side and n2 and 3h2 then the delta h of this reaction should be addition of these three numbers positive 58 negative 81 and negative 22. the answer is equals to negative 45. so c is the answer in question 42 we would like to know which of these choices has the zero value for delta h of formation delta h of formation is zero only for elements under a standard condition. A, B, C, and D they are compound. So the only element in this question is D, N2.

So the answer is D. In question 43 we would like to calculate delta H from the bond energy. We know the equation is the sigma for bond energy of reactants minus the sigma for bond energy of products.

And we have all the bond energy for our reactants and product. In our reactants, we have CH4. CH4 has 4 CH bond.

So we need to multiply the value of CH by 4. So we have 4 times the value for CH is 413. Then O2 is double bond and we have 2 molecule of O2. O2 so for O2 also we need to multiply the value by 2 495 so this is the value for reactants then let's calculate the product this is the Lewis structure for CO2 so we have 2 C double bond O so 2 times 799 and for water we have 2 OH bond for each water molecule and because we have two water molecule in total we have 4 OH bond so 4 times the value for OH it's 460 T the value for the reactant is 2642 and the value for product is 3450 So 2642 minus 3450 is equals to negative 808 kilojoule per mile. So the answer is E.

In question 44, we would like to know which of these laws stated that the volume of gas is directly proportional to its absolute temperature. The answer is Charles'Law. Question 45 we have volume temperature and pressure of neon gas and we would like to calculate the volume another temperature and pressure so the formula for this type of question is combined gas law P1V1 T1 P2V2 T2 P1 it is 1.65 ATM V1 volume is 26 0.99 T1 is not 77 Celsius because we should use Kelvin in this type of question. So 273 plus 77 which is 350. We don't know V2 but P2 is 2.35 times V2 and again for second temperature final temperature. we should convert it to the kelvin 273 plus 355 it is 628 now we can solve this question for v2 so 628 is here and 350 is here if we cross multiplying v2 is equal to multiplication of this three number dividing by these two number So V2 is equal to 1.65 times 26.99 times 628 divided by 350 times 2.35.

And the answer is equal to 34.00 liter. B is the answer. In question 46... We want to calculate the volume of O2 gas and we have its pressure and temperature.

So we should use ideal gas law PV nRT. Or we can rewrite it in this form. V is equals to nRT over P.

N is 2.1 mol. R value when we have ATM and leader it is 0.0821. then for temperature we should convert it to the kelvin so 273 plus 110 it is 383 kelvin divided by pressure which is 1.45 the final answer is equals to 45.5 liter so d is the answer in question for this 7 we would like to calculate the volume of so2 at STP condition when we have reaction of 42.60 iron sulfide with enough amount of oxygen so we have a gas stoichiometry question so we have 42.60 gram of iron sulfide i can convert gram of iron sulfide to its mole so one mole FES is equal to 87.91 gram FES.

Then I can convert mol of FES to mol of SO2. 4 mol FES is equal to 4 mol SO2. 4 and 4 they cancel out and mol of iron sulfide they cancel out. We know at STP condition, 1 mol of any gas is equal to 22.4 liter.

So the answer is 10.85 liter. E is the answer. In question 48, we would like to calculate the density of xenon gas at STP condition.

The formula... for density is pm drt or if i rewrite the equation by dividing both side by rt i can have density is equals to pm over rt pressure at stp is 1 atm the molar mass for xenon it is 131.29 r value is 0.0821 and temperature it is zero Celsius we need to convert it to Kelvin so is 273 Kelvin then the answer is equals to 5.86 gram per liter this is the answer in question 49 we would like to know which of these following gas has the lower rate of effusion R1 over R2 is equal to m2 over m1 so this is the garahams law and we can see that heavier molecule they have lower rate of effusion we need to find out which of these compound or elements have the heavier particle n2 nitrogen mass is 14 so n2 is 28 oxygen o2 is 32. Chlorine, each chlorine is 35.45. So the molar mass or molecular mass of chlorine is 70.91.

For F2, almost 38. And for carbon dioxide, it is 44. So the heaviest gas here is chlorine. So it should have the lower rate of effusion and the last question question 50 we have a mixture of three gases 3.2 mole o2 2.1 mole co2 1.7 mole helium and the total pressure is 9.5 atm and we would like to calculate partial pressure of helium so partial pressure of helium is equal to the mole fraction of helium times the total pressure. Mole fraction for helium is equal to mole of helium 1.7 over the total mole 3.2 for O2, 2.1 for CO2 and 1.7 for helium times the total pressure is 9.580 m.

Here the total is equals to 7 so 1.7 divided by 7 times 9.5 then the answer is 2.3 atm so d is the answer thank you for watching this video and good luck with your final exam

Transcript for:General Chemistry Final Exam Review Guide

Transcript for:
General Chemistry Final Exam Review Guide