Practice Test A4 Stoichiometry Name Per

Size: px

Start display at page:

Download "Practice Test A4 Stoichiometry Name Per"

Giles Wilcox
5 years ago
Views:

1 Practice Test A4 Stoichiometry Name Per Remember, this is practice - Do NOT cheat yourself of finding out what you are capable of doing. Be sure you follow the testing conditions outlined below. DO NOT USE A CALCULATOR. AP Chem does not allow the use of a calculator for the MC part of the exam, so it is time to start practicing without one. You may use ONLY a periodic table. While you should practice working as fast as possible, it is more important at this point in the course, that you practice without a calculator, even if it slows you down. Look for the easy math common factors and rough estimation do not do long division to try to get exact values. Remember it is a MC test, use the answers Mark which questions you would like to go over when we get to school in September. Send me an with the prob # s 1. Balance the following equation: NH3 + O2 NO2 + H2O The balanced equation shows that 1.00 mole of NH3 requires mole(s) of O2 a b c d e Write a balanced equation for the combustion of acetaldehyde, CH3CHO. When properly balanced, the equation indicates that mole(s) of O2 are required for each mole of CH3CHO. a. 1 b. 2 c. 2.5 d. 3 e What is the total mass of products formed when 16 grams of CH4 is burned with excess oxygen? a. 32 g b. 36 g c. 44 g d. 62 g e. 80 g 4. Write a balanced equation for the combustion of propane, C3H8. When balanced, the equation indicates that moles of O2 are required for each mole of C3H8. a. 1.5 b. 3 c. 3.5 d. 5 e Balance the following equation with the SMALLEST WHOLE NUMBER COEFFICIENTS possible. Select the number that is the sum of the coefficients in the balanced equation: a. 3 b. 5 c. 6 d. 7 e. 8 KClO3 KCl + O2 6. Calculate the mass of hydrogen formed when 27 g of aluminum reacts with excess hydrochloric acid according to the balanced equation below. a. 1.5 g b. 2.0 g c. 3.0 g d. 6.0 g e. 12 g 2Al + 6HCl 2 AlCl3 + 3 H2 7. How many grams of nitric acid, HNO3, can be prepared from the reaction of 138 g of NO2 with 54.0 g H2O according to the equation below? a. 92 b. 108 c. 126 d. 189 e NO2 + H2O 2HNO3 + NO 2Ca3(PO4)2 + 10C + 6SiO2 P4 + 6CaSiO3 + 10CO 8. Elemental phosphorus can be produced by the reduction of phosphate minerals in an electric furnace. What mass of carbon would be required to produce 0.2 mol of P4 in the presence of 1 mol of calcium phosphate and 3 mol of silicon dioxide? a. 5 g b. 12 g c. 24 g d. 60 g e. 120 g

2 Practice Test A4 Stoichiometry (page 2" of " 2) 9. The reaction of 7.8 g benzene, C6H6, with excess HNO3 resulted in 0.90 g of H2O. What is the percentage yield? Molar Mass (g/mol): C6H6=78 HNO3=63 C6H5NO2=123 H2O=18 a. 100% b. 90% c. 50% d. 12% e. 2% C6H6 + HNO3 C6H5NO2 + H2O 10. How many grams of H2O will be formed when 32.0 g H2 is allowed to react with 16.0 g O2 according to a g b g c g d g e g 2 H2 + O2 2 H2O 11. When 2.00 g of H2 reacts with 32.0 g of O2 in an explosion, the final gas mixture will contain: a. H2, H2O, and O2 b. H2 and H2O only c. O2 and H2O only d. H2 and O2 only e. H2O only g of metal carbonate, containing an unknown metal, M, were heated to give the metal oxide and 4.4 g CO2. MCO3(s) + heat MO(s) + CO2(g) What is the identity of the metal M? a. Mg b. Cd c. Ca d. Ba e. Cr 14. When a 16.8-gram sample of an unknown mineral was dissolved in acid, 4.4-grams of CO2 were generated. If the rock contained no carbonate other than MgCO3, what was the percent of MgCO3 by mass in the limestone? Molar mass (g/mol): MgCO3 = 84 and CO2 = 44 a. 33% b. 50% c. 67% d. 80% e. 100% N2(g) + 2O2(g) N2O4(g) 15. The above reaction takes place in a closed flask. The initial amount of N2(g) is 8 mole, and that of O2(g) is 12 mole. There is no N2O4(g) initially present. The experiment is carried out at constant temperature. What is the total amount of mole of all substances in the container when the amount of N2O4(g) reaches 6 mole? a. 0 mole b. 2 mole c. 6 mole d. 8 mole e. 20 mole 16. The diagram above represents solid carbon dioxide, better known as dry ice. Which diagram below best represents Some of the dry ice after it sublimes into a gas? a. b. 13. What mass of Al is produced when mole of Al2S3 is completely reduced with excess H2? a. 2.7 g b g c g d g e. 108 g c. d.

3 Practice Test A4 Stoichiometry (page 3" of " 3) 17. When the following chemical equation is balanced using the lowest possible whole numbers, what is the sum of the stoichiometric coefficients? a. 6 b. 8 c. 10 d. 20 SiCl4 + H2O Si(OH)4 + HCl The following three questions refer to the cylinders and molecular models as well as the information below. 21. When the following chemical equation is balanced using the lowest possible whole numbers, what is the sum of the stoichiometric coefficients? H3PO4 + Mg(OH)2 Mg3(PO4)2 + H2O a. 4 b. 6 c. 11 d Which molecular diagram shown below represents a homogeneous mixture? +? Hydrogen and oxygen react together to produce H2O. The number of molecules shown is relative to the number of moles present in each 1 liter cylinder, both of which are at the same temperature and pressure. The two gases are combined into a single cylinder, and sparked to produce water molecules. The reaction runs to completion. The amount of gas after the reaction is not shown. 18. The pressure in the two cylinders before the reaction a. is greater in the left because the molecules are larger. b. is greater in the left because molecules are heavier. c. the same in both cylinders because there are the same number of moles of gas. d. the same in both cylinders because the molecules in both cylinders are moving at the same speed. 19. Which of the following molecules best represent oxygen compared to hydrogen? a. b. c. since both molecules are diatomic, both can equally represent oxygen gas. d. There is no way of knowing which model would best represent oxygen. 20. Which gas(es) will be in the container after the reaction? a. just H2O b. H2O and O2 c. H2O and H2 d. H2O, O2, and H2 23. During the synthesis of a solid organic compound, the following data are collected / calculated. Which shows the correct setup for the calculation of percentage yield? a. b. c. mass of filter paper mass of filter paper and dry compound theoretical yield of dry compound d g 4.3 g 3.7 g

4 Practice Test A4 Stoichiometry (page 4" of " 4) 24. A reaction between 246 g of C6H5NO2, (MM 123 g mol 1 ) and an excess of CH3Cl, (MM 50.5 g mol 1 ) in the presence of a catalyst yields 54.8 g of a compound with the formula C7H7NO2, (MM 137 g mol 1 ). Calculate the approximate percent yield. a. 5.0% b. 20.0% c. 50% d. 75.0% C6H5NO2 + CH3Cl C7H7NO2 + HCl Al + 3O2 2 Al2O3 What mass of oxygen is required to react with 9.0 g of aluminum according to the reaction above? a g b g c. 8.0 g d. 27 g 26. C2H4 + HNO3 C2H3NO2 + H2O The reaction of 5.6 g of ethene with excess nitric acid resulted in 1.2 g of water. What is the percentage yield? Molar masses: C2H4 = 28, HNO3 =63, C2H3NO2 =73, H2O =18 a. 12 % b. 33 % c. 50 % d. 67 % 27. N2 + 3Cl2 2NCl3 When 1.40 g of N2 reacts with 3.55 g of Cl2, the final gas mixture will contain Molar masses: N2 = 28, Cl2 =71, NCl3 =120.4 a. N2, Cl2, and NCl3 b. N2 and NCl3 only c. Cl2 and NCl3 only d. NCl3 only 28. When a 25 g sample of an unknown mineral was dissolved in acid, 4.4 g of CO2 were generated. If the mineral contained no carbonate other than MgCO3, what was the percent of MgCO3 by mass in the mineral? Molar masses: CO2 = 44, MgCO3 =84.3 a. 33 % b. 50 % c. 67 % d. 80 % N2(g) + 2O2(g) N2O4(g) 29. The above reaction takes place in a closed, rigid vessel. The starting moles of N2(g) is 1.0, and that of O2(g) is 1.5 atm. There is no N2O4(g) initially present. The experiment is carried out at constant volume and temperature. What is the total number of moles in the container when the amount of moles of N2O4(g) reaches 0.75 mol? a mol b. 1.0 mol c mol d. 2.5 mol 30. A compound whose empirical formula is C2H4O has a molar mass that lies between 110 and 150 mol -1. What is the molecular formula of the compound? a. C2H4O b. C4H8O2 c. C6H12O3 d. C6H12O2 31. Find the empirical formula for a compound only one element of which is a metal. The compound s percentage composition by mass is 40.0% metal, 12.0% C, and 48% O. a. CaCO3 b. Na2CO3 c. NaHCO3 d. Al2(CO3)3 32. The density of copper decreases as temperature increases (as does the density of most substances). Which of the following will be true upon changing the temperature of a sample of copper from room temperature to 95ºC? a. The copper sample will become lighter. b. The copper sample will become heavier. c. The copper sample will expand. d. The copper sample will contract. 33. A student decomposed 20.0 g of calcium carbonate and was able to recover 6.0 calcium. Their approximate percent yield of calcium would be a. 30 % b. 40 % c. 60 % d. 75 %

5 " Practice Test A4 Stoichiometry (page 5" of " 5) 34. The density of aluminum is 2.7 g/ml and the density of ethanol at room temperature is g/ml. Calculate the length of one side of the perfect cube of aluminum that is placed in the amount of ethanol in the left cylinder and is shown in the right cylinder. a. 1.8 cm b. 2.0 cm c. 2.7 cm d. 4.0 cm e. 8 cm 37. An unnamed element with an atomic number of 130 is vaporized and injected into a mass spectrometer. The results are shown below. Use the data to calculate the average atomic mass of this element. 20 ml a. 320 amu b. 321 amu c. 322 amu d. 323 amu e. 324 amu 10 ml 35. When a hydrate of CuSO4 *? H2O weighing g is heated until all the water is removed. The anhydrate weighs g. What is the value of? in the hydrated salt? C3H8 + O2 CO2 + H2O 38. When the equation for the reaction represented above is balanced and all coefficients are reduced to the lowest whole-number terms, the coefficient for O2(g) is a. 1 b. 2 c. 3 d. 5 e. 6 a. 2 b. 3 c. 4 d. 5 2 KClO3 2 KCl + 3 O2 36. What is the percentage yield of O2 if 12.3 g of KClO3 (123 g mol -1 ) is decomposed to produce 3.2 g of O2 (32 g mol -1 ) according to the equation above? a. 100 % b. 67 % c. 50 % d. 33 % e. 10 %

6 Practice Test A4 Stoichiometry (page 6" of " 6) Mass Spectrum for Chlorine 42. Estimate the approximate mass of a mole of natural sample of chlorine gas? a. 35 b. 70 c. 71 d How many neutrons does the least abundant chlorine isotope have? a. 17 b. 18 c. 19 d. 20 Use the mass spectrum shown below to answer the following five questions. The mass spectrum of a naturrally occuring sample of chlorine is shown. Detailed analysis shows that the two stable isotopes of chlorine have masses of amu and amu. 39. Which are the mass numbers of the two isotopes of chlorine? a and b. 34 and 36 c. 35 and 37 d. 17 and Estimate the approximate percent abundance of the lighter isotope. a. 10 b. 25 c. 50 d How many types of molecules with different masses exist in a sample of chlorine gas if the sample exists entirely as diatomic molecules a. 1 b. 2 c. 3 d. 4

7 " Practice Test A4 Stoichiometry ANSWERS 1. d It might be easiest to balance the equation with mostly whole numbers: 2 NH3 + ⁷ ₂O2 2NO2 + 3H2O. The question asks about the amount of oxygen reacting with ONE mole of ammonia, thus cut the ⁷ ₂ (3.5) of oxygen in half to c Balance: CH3CHO + ⁵ ₂ O2 2CO2 + 2H2O Note: If you are having trouble balancing equations, you MUST act fast on the first day of school and get in for some extra help. 3. e Balance: CH4 + 2O2 CO2 + 2H2O Then do some stoichiometry using easy math 16 g of methane (MM = 16) is 1 mole and 1 mole of methane will produce 1 mole of CO2 = 44 g, and 2 moles of H2O which is 36 g for a total of 80 g 4. d Balance: C3H8 + 5O2 3CO2 + 4H2O 5. d Balance: 2KClO3 2KCl + 3O2 6. c In multiple choice questions without a calculator, you must look for the easy math You will be most successful at this if you put all the numbers in the dimensional analysis on the page and look for common factors you can cancel out. 1mol 27gAl 27g 3H 2 2Al 2g = 3 g H2 7. c First you must realize this is a limiting reactant problem. You can tell this since you are given quantities for both reactants. 1mol Convert both values to moles: 138gNO 2 and Clearly the NO2 limits 46g = 3molNO 2 54gH 2 O 1mol 18g = 3molH O 2 since the balanced equation tells us that 3 moles are required for every one mole of water, thus use the limiting reactant to 2HNO determine the amount of acid that can be produced. 3molNO 3 63g 2 3NO 2 = 126gHNO 3 8. c In this problem you want to produce 0.2 mol of P4, thus you must figure the minimum amount of C to produce that. The quantities given for calcium phosphate and silicon dioxide are just distractors. Although it is important that you confirm that you have enough of these two reactants to produce the 0.2 mol of P4 that the problem asks for. 10C 12g " 0.2molP 4 and finish the mass calculation: " 1P 4 = 2molC 2molC = 24gC It is worth confirming that you have enough of each of the other reactants as given in the problem. 2Ca So calculate: " 0.2molP 3 (PO 4 ) 2 4 is enough as 1 mol was given 1P 4 = 0.4molCa (PO ) required SiO and " 0.2molP 2 4 is also enough as 3 mol was given. 1P 4 = 1.2molSiO required 2 1mol 9. c Look for easy approximations: " 7.8gC 6 H 6 and continue " 78g = 0.1molC H 1H molC 2 H 2 O 18g 6 1C 6 H 6 = 1.8gH 2 O 0.9gH then percent yield also will be easy values: " 2 O 1.8gH 2 O 100 = 50%H O c Again, this is a limiting reactant problem, 32 g of H2 is 16 moles and is far in excess of the 0.5 moles of O2 available. Thus the O2 limits the reaction. Set up the dimensional analysis and look for easy factors. 1mol 16gO 2 32g 2H 2 O 1O 2 18g = 18gH 2 O 11. c First you should recognize that you need to write a balanced equation: 2H2 + O2 2H2O Next, realize that this is a limiting reactant problem 2 g of H2 is 1 mole and 32 g of O2 is 1 mole. The 1 mole of H2 will limit the reaction, thus we know that after the reaction stops, O2 will remain in the container with the H2O produced.

8 Practice Test A4 Stoichiometry (page 8" of " 8) 12. e This problem requires a bit more clever thought and understanding. Work the problem backwards starting with the 1mol information that the carbon dioxide product tells you. Since " 4.4gCO 2, the balanced equation tells us 44g = 0.1molCO 2 that 0.1 mol of CO2 must come from 0.1 mole of MCO3. Further, we know that 0.1 mole of MCO3 contains 0.1 mole of M and also 0.1 mole of CO3. Since CO3 has a mass of 60 g/mol, we can calculate that the 0.1 mole of the CO3 has a mass of 6 g and the remainder of the compound (whose mass we were told was 11.2 g), the M part, must weigh 5.2 g. But remember 5.2g there is 0.1 mole of M in this compound, thus " and so you are looking for an element that has a molar 0. = 52g mass of 52, which is Cr. 13. c You could consider the chemical reaction, (Al2S3 2 Al + 3S) or you could simply realize that since the Al would be written as a monatomic atom in the balanced equation and since there are 2 Al s in the compound, there must be a 2:1 ratio 2Al between the Al2S3 and Al : 0.5molAl 2 S 3 1Al 2 S 3 = 1moleAl 27g = 27gAl 14. b As in problem 13, it is an important concept to realize that metal carbonates decompose or react with acid to produce the same amount of moles of carbon dioxide as the moles of the original compound. All metal carbonates will react with acid to produce carbon dioxide and a metal oxide. This reaction is specifically: MgCO3 + H + MgO + CO2. The original 16.8 g sample contains both MgCO3 and some other inert substances that do not react with acid to produce any gas. 1mol " 4.4gMgCO 3 thus " 44g = 0.1molMgCO 1molMgCO 3 84g 8.4gMgCO 3 1molCO 2 = 8.4gMgCO gsample 100 = 50% 15. d This problem is a simple stoichiometry problem that you can certainly do without a calculator. When the reaction reaches a quantity of 6 mole of product (remember the problem states that there was no product, N2O4 to start with), use the coefficients in the balanced equation to determine that 12 mole of O2 and 6 mole of N2 must have reacted to produce the 6 mole of product. and since the reaction was started with only 12 mole of O2, there will be none left 2molO 6molN 2 O 2 4 1molN 2 O 4 = 12molO 2 and since the reaction was started with 8 mole of N2, there will be 2 mole of N2 left. 1molN 6molN 2 O 2 4 1molN 2 O 4 = 6molN 2 Thus 2 mole N2 left over, no O2 left over, and 6 mole of N2O4 produced, will mean a total of 8 mole of substances left in the flask. 16. a When a substance sublimes, it converts from solid directly to gas. At atmospheric pressure, CO2 does not melt, hence its name, dry ice. When the solid CO2 sublimes and turns to gas, the substance still remains CO2 molecules. 17. c SiCl4 + 4 H2O Si(OH)4 + 4HCl 18. c The pressure in the two containers must be the same because the moles of gas cause the pressure. Avogadro s Law tells us that equal volumes of gases with the same number of moles of molecules, at the same temperature, will exert the same pressure. 19. a Both oxygen and hydrogen are diatomic, however, oxygen is a larger atom than hydrogen, thus the larger atom molecule best represents oxygen, O2. You should know about the size of oxygen compared to hydrogen from their locations on the periodic table. You learned this in your periodicity unit in first year chemistry. 20. b Anytime you are doing a stoichiometry problem you need a balanced equation; 2H2 + O2 2H2O Also it is important to recognize that when you are given equal moles of oxygen and hydrogen molecules. You should determine which one is the limiting reactant. Clearly the hydrogen limits as the 6 moles of H2 would require only 3 moles of O2 and we have more than that. This means that the excess reactant, O2 will also be in the container along with the product H2O after the reaction is complete. Using the balanced equation, 6 H2 would require 3 O2 and produce 6 water with 3 O2 left over. 21. d 2 H3PO4 + 3 Mg(OH)2 Mg3(PO4)2 + 6 H2O 22. c Since the diagram in (c) shows two different molecules, integrated together, unlike diagram (b), we can assume (c) is a homogeneous mixture. (a) would be a pure substance made of molecules, (d) would be a pure substance made of elemental atoms, and (b) would represent a heterogeneous mixture with two distinct layers.

9 " Practice Test A4 Stoichiometry (page 9" of " 9) 23. a % yield is the same calculation as a grade on a quiz; " experimental( whatyoudidget) theoretical( whatyoushouldget) b 246 g of C6H5NO2 is 2 mol of which would produce 2 mole of C7H7NO2 which would weigh 274 g thus the yield is which is a close enough approximation to " ~ = 1 =~ 20% 5 ~ c 9g 1mol 27g = 1 3 mol 3O 2 4Al = 1 4 molo 2 32gmol 1 = 8gO 2 required 26. b Again, be on the lookout for factors, which will cause math you can do without a calculator. 5.6 looks like 56, which is 2 28, thus 5.6 g is 0.2 mole of ethene, and the stoichiometry is 1:1, and is of course 3.6 then 1.2 is ⅓ of 3.6 = 33% The formal dimensional analysis is shown here. 5.6g 1mol and 28g 1mol 18g 1.2g = 3.6gH 2 O 3.6g 100 = 33% 27. b Hopefully you can immediately recognize this as a limiting reactant problem. First, determine which gas limits. 1.4g 1mol > 28g = 0.05molN g 1mol 1 71g = 0.05molCl 2 3 since after normalizing, Cl2 is smaller, thus the Cl2 limits. Thus the final mixture will have both product and the excess gas, N2, present. 28. a There is no better time than now, to learn that 1 mole of a metal carbonate, will give off 1 mole of carbon dioxide. The reaction is shown here: MgCO3 + 2H + Mg 2+ + CO2 + H2O 1mol 4.4gCO 2 which must have come from 0.1 mol of MgCO3, and thus 44g = 0.1molCO 84.3g 2 0.1mol = 8.4gMgCO 3 8.4g 25g 100 =~ 33% 29. b You may find it helpful to set up a RICE box, which is simply tracking stoichiometry in a table. The I row stands for initial conditions, the C rowstands for change or where you apply the stoichiometry, and the E row is the moles of the gases at the end of the reaction. Add the moles together to get the total moles. R N2 + 2 O2 N2O4 I C E c Remember that the empirical formula is the lowest whole number ratio. The molecular formula will be the same, or some whole number multiple of the empirical formula, thus d is not even an option. Add the molar mass of the empirical formula, 44 g/mol, then divide that into the molar mass given.i ll use 125 and see what s close " 125 thus I need to push a 3 44 = 2.8 ~ 3 through the empirical formula to end up with C6H12O3 12g(%) 1mol 12g = 1C 31. b If you take a look at the carbon and oxygen percentages " you can see that you need the rest of the 49g(%) 1mol 16g = 3O compound to weigh a total of 40 g Na is close (46 g) but the Ca (40 g/mol) is closer. The percetages of each element in the d option turn out to be 23% Al, 15% C, and 61% O but those # s are hard to get without a calculator. 32. c Heating or cooling will not change the mass of a substance (unless some evaporates, which is not likely to happen for a solid) but heating will increase the volume (of most substances). Constant mass with a larger volume will decrease the density. 33. d Since the molar mass of CaCO3 is 100 g and calcium would be 40% of that, thus the decomposition of CaCO3 should yield " 20 4 but the student only recovered 6 g, thus the student s percent yield is " 10 (40%) = 8g 6g 100 = 75%yield 8g

10 Practice Test A4 Stoichiometry (page 10 " of " 10) 34. b 2 cm In this problem, all that matters is that the block of aluminum displaces 8 ml of liquid. It does not matter that the liquid is alcohol, the volume of alcohol displaced is 8ml, thus the aluminum object is 8 ml. Since 8 ml = 8 cm 3, and because the 3 object is a cube, " 8cm 3 = 2cm on each side. 35. d Remember that determining formulas is about finding the mole ratios; in this case the mole ratio between the 16 g of anhydrate 16g 1mol 160g = 0.1ish and the 10 g of water. Thus 1 hydrate for every 5 waters = CuSO4 5H2O 1mol ~ 10g 18g(let 's call it 20g) = 0.5ish 36. b First add the molar mass of KClO3 to get ~123 and thus you have 0.1 mol of reactant. Given the equation, you can see a 3:2 ratio between O2 and KClO3 which indicates 0.15 mole of O2 should be produced. Thus calculate the mass of O2, resulting in 3.2g 4.8 g. The percent yield is of course " 100 = 67% This is a it messy in your head, so estimate your way into answer (b). 4.8g 100% yield would mean 4.8 should be collected, and 50% yield would mean 1.6 should be collected, thus the collected amound must be between answer (a) and answer (c). 37. b Some graphs will show percent abundance on the y-axis, but other graphs like this one will just show intensity. Thus you must compare the intensity values. The lower peak looks to be about ⅓ of the larger peak. Thus 1 part heavier isotope, 3 parts lighter isotope, giving 75 % light isotope, and 25% heavier isotope. If there were 50% of each of the two isotopes, the average molar mass would have to be right in the middle at 322, but since there is 25 % of the heavier isotope and 75% of the lighter isotope, the molar mass would have to be one step to the left at 321. Perhaps the number line shown below will help. 38. d Simply balance by inspection. 39. c Mass number is the mass rounded to the nearest whole number. You can get these for the Cl isotopes right off the graph, or by rounding the isotope masses given in the paragraph before the graph. 40. b Just inspect the graph and report ~25% Lighter isotope 100% 75% 50% 25% 0% Heavier isotope 0% 25% 50% 75% 100% Molar Mass c Because there are two isotopes, there are three ways you can combine them to make Cl2 molecules , 37+37, and c The easiest way to get this is by looking at the average molar mass on the periodic table ~35.5 and since natural chlorine gas is Cl2, double 35.5.for 71. You could of course calculate the 35.5 from the mass # s and isotopic percantages but that is not necessary. 43. d The mass number of the least abundant isotope is 37 (protons + neutrons), and the atomic number (protons)is 17, thus the nujber of neutrons is = 20 neutrons.

Practice Test Unit A Stoichiometry Name Per

Practice Test Unit A Stoichiometry Name Per This is practice - Do NOT cheat yourself of finding out what you are capable of doing. Be sure you follow the testing conditions outlined below. DO NOT USE A