General Chemistry 1 CHM201 Unit 2 Practice Test

Transcription

1 General Chemistry 1 CHM201 Unit 2 Practice Test 1. Which statement about the combustion of propane (C 3H 8) is not correct? C 3H 8 5O 2 3CO 2 4H 2O a. For every propane molecule consumed, three molecules of carbon dioxide are produced. b. For every two molecules of propane consumed, eight molecules of water are produced. c. For every five molecules of oxygen consumed, four molecules of water are produced. d. For every three molecules of carbon dioxide produced, five molecules of oxygen are consumed. e. For every carbon dioxide molecule produced, four molecules of water are also produced. 2. A gallon of water has a mass of 3.79 kg. How many moles of water (18.02 g/mol) is this? a mol d. 68,300 mol b mol e. 386 mol c mol 3. An analisys of the percent composition may be used to determine a compound s a. molecular formula. b. empirical formula. c. empirical and molecular formulas. d. exact number of molecules. e. none of the above. 4. Identify the list below that has these chloride compounds arranged in order of increasing molar mass. a. CCl 4 PCl 3 SCl 2 BCl 3 b. SCl 2 BCl 3 PCl 3 CCl 4 c. BCl 3 CCl 4 PCl 3 SCl 2 d. SCl 2 BCl 3 CCl 4 PCl 3 e. BCl 3 SCl 2 PCl 3 CCl 4 5. Fe 2O 3(s) and powdered aluminum can react with great output of heat to form molten iron and Al 2O 3. When the following reaction equation is balanced, what are the stoichiometric coefficients in order: Fe 2O 3 + Al Fe + Al 2O 3? a. 1, 1, 1, 1 d. 2, 1, 1, 2 b. 2, 2, 2, 2 e. 1, 1, 2, 2 c. 1, 2, 2, 1 6. Nitrogen monoxide undergoes combustion to produce nitrogen dioxide. Which relationship regarding the quantities of reactants and products associated with this reaction is not correct? 2NO O 2 2NO 2 a. 2x g x g 2x g b. 2 molecules 1 molecule 2 molecules c g 32.0 g 92.0 g d. (2)(30.0 g) 32.0 g (2)(46.0 g) e g 8.0 g 23.0 g 7. The combustion of ethanol (CH 3CH 2OH, 46.1 g/mol) results in the formation of water and carbon dioxide according to the following equation: CH 3CH 2OH + O 2 H 2O + CO 2. How many grams of carbon dioxide are produced when 46.1 g of ethanol burns? a g d g b g e g c g

2 8. Fool s Gold is the mineral pyrite (FeS 2). What is the mass percent of sulfur in pyrite? a % d % b % e % c % 9. Combustion analysis of an organic compound to determine the percentages of carbon, hydrogen, and oxygen in the compound depends on which of the following assumptions? a. The moles of water produced equal the moles of hydrogen in the sample. b. All the oxygen in the products comes from the added oxygen gas. c. The mass of the carbon dioxide produced equals the mass of the carbon in the sample. d. The compound burns completely to form carbon dioxide and water. e. The mass of oxygen in the sample is the difference between the mass of the sample and the sum of the masses of carbon dioxide and water produced. 10. One form of asbestos called chrysotile is considered to be a human carcinogen. Mass analysis reveals that it has the empirical formula Mg 3Si 2H 4O 9. If the molar mass is 831 g/mol, which molecular formula is correct? a. MgSiO 4(OH) 2 d. Mg 6Si 4O 10(OH) 8 b. Mg 2Si 4O 8(OH) 4 e. Mg 9Si 6O 15(OH) 12 c. Mg 3Si 2O 5(OH) A reaction vessel contains equal masses of iron and oxygen. How much FeO could theoretically be produced? a. an amount equal to the sum of the masses of iron and oxygen b. an amount greater than the sum of the masses of iron and oxygen c. an amount less than the sum of the masses of iron and oxygen d. It depends on the reaction conditions. e. There is not enough information given to answer this question. 12. A mass of g of phosphoric acid was produced from the reaction of g of P 4O 10 with g water. What was the percent yield for this reaction? P 4O 10 + H 2O H 3PO 4 a % d % b % e % c % 13. Concentrated sulfuric acid contains 4 g of water for every 100 g of solution. The solvent is a. water. d. the same as the solution. b. sulfuric acid. e. the same as the solute in this case. c. concentrated. 14. If 120 g of NaOH were used to prepare 500 ml of solution, what would the concentration be? a. 1 M d. 4 M b. 2 M e. 6 M c. 3 M 15. How many moles of hydrogen peroxide are present in a one pint (473 ml) of a 1.5 M solution? a mol d. 7.1 x 10 2 mol b. 3.2 mol e mol c. 1.5 mol

3 16. If 100. ml of 3.0 M solution were diluted to 250 ml, what would the concentration be? a M d. 12 M b M e. 120 M c. 1.2 M 17. Which one of the following compounds is a strong electrolyte? a. methane, CH 4 d. hydrofluoric acid, HF b. methanol, CH 3OH e. potassium chloride, KCl c. ammonia, NH What is the molar concentration of sodium ions in a 0.3 M sodium phosphate (Na 3PO 4, 164 g/mol) solution? Sodium phosphate is used as a cleaning agent, food additive, and stain remover. a. 0.1 M d. 0.9 M b. 0.3 M e. 1.0 M c. 0.6 M 19. In the following reaction, H 2O CH 3COOH(aq) H 2O(l) CH 3COO - (aq) H 3O + (aq) a. acts as an acid. b. acts as a base. c. acts neither as an acid nor as a base. d. serves as both an acid and as a base. e. causes a precipitate to form. 20. Which one of the following reaction equations is the net ionic equation for the reaction of hydrochloric acid, HCl, with lithium hydroxide, LiOH? All species are in aqueous solution. a. HCl LiOH LiCl H 2O b. H + Cl Li + OH Li + Cl H 2O c. H + Cl Li + + OH LiCl H 2O d. Li + Cl LiCl e. H + OH H 2O 21. Ammonia (NH 3) is a weak base that reacts with a strong acid to form the ammonium ion, NH 4+. If 5.00 ml of a solution of an ammonia cleaner is titrated directly with 42.6 ml of M HCl, what is the concentration of the NH 3 in solution? (Assume that the ammonia is the only solute that reacts with the acid.) a M d M b M e M c M 22. Which of the following ionic compounds is insoluble in water? a. BaCl 2 d. Ba(NO 3) 2 b. BaSO 4 e. MgCl 2 c. NaOH

4 23. Write the complete ionic equation for the reaction that takes place between a solution of lead(ii) nitrate and sodium sulfate. a. Pb(NO 3) 2(aq) Na 2SO 4(aq) PbSO 4(s) 2NaNO 3(aq) b. Pb(NO 3) 2(aq) 2Na + (aq) 2 SO 4 (aq) PbSO 4(s) 2Na + (aq) 2NO 3 (aq) c. Pb 2+ (aq) 2 SO 4 (aq) PbSO 4(s) d. Pb 2+ (aq) 2NO 3 (aq) 2Na + (aq) 2 SO 4 (aq) PbSO 4(s) 2Na + (aq) 2NO 3 (aq) e. 2NO 3 (aq) 2Na + (aq) 2NaNO 3(aq) 24. What mass of calcium hydroxide (molar mass = 74.1 g/mol) is produced when 15 ml of a M solution of calcium nitrate is mixed with 25 ml of a M solution of potassium hydroxide? Ca(NO ) 2(aq) + KOH(aq) Ca(OH) 2(s) + 2KNO (aq) a g d g b g e g c g 25. Sodium metal easily loses an electron to form the sodium ion. In this process, sodium is a. reduced. d. complexed. b. oxidized. e. both ionized and oxidized. c. ionized. 26. What is the oxidation number of S in H 2SO 4? a. +2 d. 6 b. 4 e. +6 c When the oxidation reduction reaction shown here is balanced, how many electrons are transferred for each atom of copper that reacts? Ag + (aq) Cu(s) Ag(s) Cu 2+ (aq) a. 1 d. 4 b. 2 e. 0 c. 3

5 CHM201 Unit 2 Practice Test Answer Section MULTIPLE CHOICE 1. ANS: A ; The balanced chemical equation is give. We are asked to determine the fixed ratios of molecules of products and reactants from the coefficients and identify the incorrect ratio among the choices. Selection a. proposes a ratio of propane to carbon dioxide of 1:3. That is the ratio in the balanced equation. Correct statement. Selection b. proposes a ratio of propane to oxygen of 2:8, which is, of course, 1:4. That is the ratio in the balanced equation. Correct statement. Selection c. proposes a ratio of oxygen to water of 5:4. That is the ratio in the balanced equation. Correct statement. Selection d. proposes a ratio of carbon dioxide to oxygen of 3:5. That is the ratio in the balanced equation. Correct statement. Selection e. proposes a ratio of carbon dioxide to water of 1:4. The ratio in the balanced equation is 3:4. Incorrect statement. The ratio suggested in selection e. does dot match the coefficients in the balanced equation. PTS: 1 DIF: Easy REF: 3.1 OBJ: Analyze a chemical reaction equation in terms of reactant and product molecules and their states. MSC: Understanding 2. ANS: B We are given a mass and a molar mass of water, H 2O. ~4000 divided by ~20 = ~200. The answer is reasonable. PTS: 1 DIF: Easy REF: 3.2 OBJ: Convert between mass and moles of particles. 3. ANS: B a. molecular formula.- NO. You also need the molecular mass to find this. b. empirical formula. - YES. This is all that is needed to establish the molar, and, therefore, atomic ratios.

6 c. empirical and molecular formulas - NO. See a. d. exact number of molecules - NO. Besides, who cares? e. none of the above - NO. There is a correct answer above. PTS: 1 DIF: Easy REF: 3.7 OBJ: Distinguish between empirical formula and molecular formula. MSC: Understanding 4. ANS: B We are given four molecular formulas. The molar mass is the molecular mass expressed in grams, so we need to calculate all the molecular masses and compare them. (close approximations are sufficient) CCl 4 x + x = + = 154 g/mol PCl 3 x + x = + = g/mol SCl 2 x + x = + = 99 g/mol BCl 3 x + x = + = g/mol SCl 2 BCl 3 PCl 3 CCl 4 The number of chlorine atoms in each molecule indicates the smallest and largest, while a comparison of phosphorous and boron quickly orders the other two. PTS: 1 DIF: Medium REF: 3.2 OBJ: Calculate the molar mass given a molecular formula. MSC: Understanding 5. ANS: C We are given an unbalanced equation: Fe 2O 3 + Al Fe + Al 2O 3 It may be helpful to create a table, Balance each element, Element Reactant atoms Product atoms Fe 2 1 O 3 3 Al 1 2 Element Reactant atoms Product atoms Fe 2 1x2=2 O 3 3 Al 1x2=2 2

7 Insert the indicated coefficients, Fe 2O 3 + 2Al 2Fe + Al 2O 3 : After inserting coefficients, the number of atoms of each element on each side of the equation are equal. PTS: 1 DIF: Easy REF: 3.3 OBJ: Given reactants and products, write a balanced reaction equation. 6. ANS: A We are given a balanced equation, 2NO O 2 2NO 2 In order to evaluate all of the statements, we will have to know the molar masses of the reactants and products, NO g/mol g/mol = 30.0g/mol O 2-2 x 16.0g/mol = 32.0g/mo NO g/mol + (2 x 16.0g/mol) = 46.0g/mo a. 2x g x g 2x g - INCORRECT. The stoichiometric relationship is based on moles, NOT grams. b. 2 molecules 1 molecule 2 molecules - CORRECT. This is one interpretation of a balanced chemical equation. c g 32.0 g 92.0 g - CORRECT - Using the molar masses above, this may be stated, 2 mol 1 mol 2 mol, which is also a proper interpretation of the balanced chemical equation. d. (2)(30.0 g) 32.0 g (2)(46.0 g) - CORRECT. Identical to c. e g 8.0 g 23.0 g - CORRECT. Using the molar masses above, this may be stated, mol mol mol, which maintains the ratios of the balanced chemical equation. Thin about it: The relationships expressed in balanced chemical equation are ratios of atoms, molecules and formula units, NOT masses. PTS: 1 DIF: Difficult REF: 3.2 OBJ: Explain a chemical reaction equation in terms of molecules, moles, and masses of reactants and products. MSC: Understanding 7. ANS: A We are given a chemical equation, a mass of one reactant and are asked to determine the mass of a product. We will need to see if the equation is balanced, and, since it is not, we must balance it, CH 3CH 2OH + 3O 2 3H 2O + 2CO 2. (You already know how to do this.) We must also know the molar masses of the reactant and product in question,

8 CH 3CH 2OH - given as 46.1g/mol CO 2 - we calculate as 44.0g/mol 1 mol of ethanol should produce 2 moles of carbon dioxide. PTS: 1 DIF: Medium REF: 3.5 OBJ: Use a balanced reaction equation to relate amounts of reactants and products for a given reaction. 8. ANS: D We are given the formula for pyrite, FeS 2. Assume exactly 1 mol of FeS 2, which is g/mol (Calculate molar mass from the formula.). That would mean there are 2 moles of sulfur in each mole of pyrite (from the formula). The value should be close to 50% from the known atomic mass. The last digit depends on the accuracy of the molar masses used in the calculations. PTS: 1 DIF: Easy REF: 3.6 OBJ: Determine the percent composition given a chemical formula. 9. ANS: D a. The moles of water produced equal the moles of hydrogen in the sample. - NO. Each molecule of water produced will require two atoms of hydrogen from the reactant, therefore, the moles of water produced equal the one-half moles of hydrogen in the sample. b. All the oxygen in the products comes from the added oxygen gas. - NO. This statement assumes that the organic compound contains no oxygen, and that is not guaranteed. c.the mass of the carbon dioxide produced equals the mass of the carbon in the sample. - NO. carbon dioxide is not pure carbon.

9 d. The compound burns completely to form carbon dioxide and water.yes. Otherwise, there will be other products we haven t accouted for, or there are reactants left over. e. The mass of oxygen in the sample is the difference between the mass of the sample and the sum of the masses of carbon dioxide and water produced. - NO. It is the difference between the mass of the sample and the sum of the masses of carbon and hydrogen present in the two products. PTS: 1 DIF: Easy REF: 3.8 OBJ: Describe combustion analysis and explain how the information that it provides can be used. MSC: Understanding 10. ANS: E We are given an empirical formula, Mg 3Si 2H 4O 9, and a molar mass, 831 g/mol. We must determine the molecular formular. In order to determine the molecular formula from this information, we must detemine the formula mass of the empirical formula and use that to determine a ratio of molecular mass to formula mass. The empirical formula mass is 277 g/mol. (You know how to determine this.).. Multiply the empirical formula s coefficients by the ratio, and you get Mg 9Si 6H 12O 27, which may als0 be written, Mg 9Si 6O 15(OH) 12. The subscript of magnesium must be tripled from 3 to 9, and e. is the only selection that satisfies this requirement. PTS: 1 DIF: Easy REF: 3.7 OBJ: Determine a molecular formula from the empirical formula (or percent composition) and molar mass of a substance. 11. ANS: C a. an amount equal to the sum of the masses of iron and oxygen - NO. Iron accounts for much more than half of iron oxide (~75%), making it the excess reagent. Some of it would be left over after the maximum amount of FeO formed, so FeO couldn t be responsible for the entire mass after the reaction completed. b. an amount greater than the sum of the masses of iron and oxygen - NO. This would violate conservation of mass. c. an amount less than the sum of the masses of iron and oxygen - YES. The reasoning is shown in a. above. d. It depends on the reaction conditions. - NO. There are no reaction conditions that affect theoritical yield. Conditions might affect actual yield. e. There is not enough information given to answer this question. - NO. See a. and c.. PTS: 1 DIF: Medium REF: 3.9 OBJ: Identify the limiting reactant in a chemical reaction. 12. ANS: D MSC: Understanding

10 We are given masses of both reactants, enough information to determine limiting reactant and theoretical yield, and we are also givrn the actual yield. 1) Balance the equation - P 4O H 2O 4H 3PO 4. 2) Determine the limiting reagent. 3) Determine the theoretical yield. 4) Determine the percent yield. P 4O H 2O 4H 3PO g = mol x 4/1 yield = mol 12.00g = mol x 4/6 yield = mol The reactant that produces the smaller yield is the limiting reactant. P 4O 10 produces mol, which is g. That is the theoritical yield. PTS: 1 DIF: Medium REF: 3.9 OBJ: Given amounts of reactants and the actual yield, calculate the theoretical yield and the percent yield for a chemical reaction. 13. ANS: B a. water. - NO. Water is the smaller amount, therefore, it is the solute by convention. b. sulfuric acid. - YES. Sulfuric acid is the larger amount, therefore, it is the solvent by convention. c. concentrated. - NO. This makes no sense in this context. d. the same as the solution.- NO. The solution is the combination of the solute(s) and the solvent. e. the same as the solute in this case. - NO. The solvent and solute are distinct in all solutions. PTS: 1 DIF: Easy REF: 4.1 OBJ: Identify the solvent and the solute(s) in a solution. MSC: Understanding 14. ANS: E We are given the mass of the solute, NaOH, and the volume of the solution. The molar concentration of a solution is defined as the number of moles (n) of the solute divided by the volume of the solution (in liters). We must compute the molar mass of NaOH and convert the volume to liters.

11 120 g is about 3 mol of NaOH. It is dissolved in a half liter of solution. 3/0.5 = 6. PTS: 1 DIF: Easy REF: 4.2 OBJ: Calculate the molarity of a solution given the number of moles or mass of the solute and the volume of the solution. 15. ANS: E We are given the volume and molarity of a solution, 473 ml = L 1.5 M = 1.5 mol/l We have about a half-liter of solution. 1.5/2=0.75. The answer is reasonable. PTS: 1 DIF: Easy REF: 4.2 OBJ: Determine the amount of solute present (mass, moles, number of particles) in some volume of solution given the concentration of the solution. 16. ANS: C We are given the data for the initial solution and the desired volume of the diluted solution, M initial = 3.0 M V initial = 100. ml M diluted =? V diluted = 250 ml

12 We are diluting by a factor of about one half, and the result is about one half of the initial concentration (3.0 to 1.2), so the answer seems reasonable. PTS: 1 DIF: Easy REF: 4.3 OBJ: Determine the concentration of a solution after it has been diluted. 17. ANS: E a. methane, CH 4H - Methane is a molecular compound. (Nonelectrolyte) b. methanol, CH 3OH - Methane is a molecular compound. (Nonelectrolyte) c. ammonia, NH 3 - Methane is a molecular compound that will cause slight dissociation of water. (Weak electrolyte.) d. hydrofluoric acid, HF - Methane is a polar molecular compound that dissociates only slightly. (Weak electrolyte.) e. potassium chloride, KCl - Potassium chloride is an ionic compound, and it will dissociate. (Strong electrolyte.) PTS: 1 DIF: Easy REF: 4.4 OBJ: Distinguish among strong electrolytes, weak electrolytes, and nonelectrolytes. MSC: Understanding 18. ANS: D We are given the molarity of sodium phosphate, 0.3M, and its formula, Na 3PO 4. We are also given the molar mass of sodium phosphate. The molecular equation, Na 3PO 4 Na + + PO 4 3- shows that each mole yields 3 moles of the ion in solution. The molarity of the compound must be multiplied by three to determine the molarity of sodium ions: 3(0.3 M) = 0.9 M The math is trivial. PTS: 1 DIF: Medium REF: 4.4 OBJ: Calculate concentrations of ions in electrolyte solutions. 19. ANS: B We are given the reaction, CH 3COOH(aq) H 2O(l) CH 3COO - (aq) H 3O + (aq). A look at the products showshydronium ion, H 3O + (aq).

13 H 3O + forms when a water molecule accepts a proton, in this case from acetic acid, CH 3COOH(aq). Since water accepts a proton, it is acting as a base. This is the Br nstead definition of a base. PTS: 1 DIF: Medium REF: 4.5 OBJ: Identify acid-base reactions. MSC: Understanding 20. ANS: E We are given the reactants which imply the products after ions are exchanged. The molecular equation is HCl LiOH LiCl H 2O (hydrogen from HCl and lithium from LiCl) swap places in the molecular interpretation. This will yield the complete ionic equation. Complete ionic equation, H + Cl Li + OH Li + Cl H 2O (Lithium chloride dissociates, water does not) Cancelling spectator ions, H + Cl Li + OH Li + Cl H 2O Leaving the net ionic equation H + OH H 2O The net ionic equation is what can t be cancelled fron the complete ionic equation. PTS: 1 DIF: Easy REF: 4.5 OBJ: Write balanced molecular, overall ionic, and net ionic equations for neutralization reactions. 21. ANS: C We are given the volume of the base, NH 3 (5.00 ml), volume of the acid, HCl, (42.6 ml) and molarity of the acid ( M). The number of moles of titrant must equal the number of moles of analyte, adjusted for the stoichiometry of the neutralization equation. NH 3 + HCl NH 4Cl (balanced)

14 Since the coefficients are all unity, and both acid and base are solutions, we determine the number of moles the same way for both analyte an titrant, molarity x volume. Since they must be equal, we use the equation below and solve for the molarity of the titrant. The volume of the titrant is about 10x the volume of the analyte, so the molarity must be about 1/10 the molarity of the analyte. PTS: 1 DIF: Easy REF: 4.6 OBJ: Determine an unknown concentration from titration data. 22. ANS: B a. BaCl 2 - SOLUBLE. Barium is not an exception to chloride solubility. b. BaSO 4 - INSOLUBLE. Barium is an exception to sulfate solubility. c. NaOH - SOLUBLE. All sodium compounds are soluble. d. Ba(NO 3) 2 - SOLUBLE. All nitrate compounds are soluble. e. MgCl 2 - SOLUBLE. Magnesium is not an exception to chloride sollubility. PTS: 1 DIF: Medium REF: 4.7 OBJ: Identify and predict precipitation reactions and insoluble salts based on solubility guidelines. MSC: Remembering 23. ANS: D a. Pb(NO 3) 2(aq) Na 2SO 4(aq) PbSO 4(s) 2NaNO 3(aq) - This is the molecular equation. b. Pb(NO 3) 2(aq) 2Na + (aq) SO 4 2 (aq) PbSO 4(s) 2Na + (aq) 2NO 3 (aq) - Pb(NO 3) 2(aq) dissociates and should be represented as ions. c. Pb 2+ (aq) SO 4 2 (aq) PbSO 4(s) - This is the net ionic equation. d. Pb 2+ (aq) 2NO 3 (aq) 2Na + (aq) SO 4 2 (aq) PbSO 4(s) 2Na + (aq) 2NO 3 (aq) - All dissociated reactants and products are represented as ions. e. 2NO 3 (aq) 2Na + (aq) 2NaNO 3(aq) - NaNO 3(aq) dissociates and should be represented as ions. These are actually the spectator ions. PTS: 1 DIF: Easy REF: 4.7 OBJ: Write balanced molecular, overall ionic, and net ionic equations for precipitation reactions. MSC: Understanding 24. ANS: C We are given the volume (15mL) and molarity (0.220 M) of the reactant, calcium nitrate, and the volume (25mL) and molarity (0.140 M) of the reactant, potassium hydroxide. We are asked to predict the yield of calcium hydroxide that will be produced. We must first find the limiting reactant, by determining the potential production for each of the reactants. We can use n = MV to determine the moles of each reactant.

15 Ca(NO ) 2(aq) + KOH(aq) Ca(OH) 2(s) + 2KNO (aq) M x L = mol x 1/1 yields mol M x L = mol x 1/2 yields mol KOH is the limiting reactant and will yield mol of Ca(OH) 2. With a similar number of moles, the larger coefficient on KOH makes the fact that it is the limiting reactant reasonable. The rest of the math is stoichiometry. PTS: 1 DIF: Difficult REF: 4.7 OBJ: Determine the mass of a precipitate given the amounts of reactants. 25. ANS: E The gain or loss of an electron from an atom transforms it into an ion, and an increase in oxidation number indicates oxidation. It is both ionized and oxidized. PTS: 1 DIF: Easy REF: 4.9 OBJ: Write definitions of the following terms: oxidation, reduction, oxidizing agent, and reducing agent. MSC: Remembering 26. ANS: E Following the guidelines for determining oxidation number in a formula: Hydrogen is +1 except in certain metal hydrides, and this is not a metal hydride. Oxygen is nearly always -2, and this instance is not an exception. There are two hydrogens, 2 x (+1) = +2. There are four oxygens, 4 x (-2) = -8. Sulfuric acid is electrically neutral, i.e., it is not an ion, therefore the sum of all the charges must equal 0. (+2) + (-8) + ox # of S = 0 ox # of S = +6 PTS: 1 DIF: Easy REF: 4.9 OBJ: Define the term oxidation number, and determine the oxidation number of an atom in a compound. 27. ANS: B We are given the unbalanced redox equation showing the ionization of copper.

16 The equation does not need to be balanced to determine the number of electrons are transferred for EACH (my emphasis) atom of copper that reacts. Copper s oxidation number goes from 0 (no charge) as a reactant to +2 as a product. This represents a transfer of 2 electrons. Electrons have a negative charge. Losing two will cause a change from 0 to +2. PTS: 1 DIF: Easy REF: 4.9 OBJ: Determine the number of electrons transferred in a redox reaction.