CHEMISTRY 111 MIDTERM

OCTOBER 12, 2000 NAME ____________________________________

SHOW YOUR WORK for CALCULATIONS and GIVE UNITS! WATCH SIGNIFICANT FIGURES.

1. MATCHING: Place the letter corresponding to the best answer in front of each of the following

definitions. (12 pts)

_C__ A substance consisting of two or more elements chemically combined in fixed proportions.

_H__ A mixture in which components separate into physically distinct regions of differing properties.

_F__ Fundamental particles that carry unit negative charge and are found outside the nuclei of all atoms.

_K__ An atom or group of atoms that has a net positive or negative charge.

_L__ One of two atoms of the same element that differ slightly in their masses.

_A__ A hydrogen-containing compound that, under appropriate conditions, can produce H⁺ ions.

_M__ The reactant that is completely consumed in a reaction.

_Y__ The solution component that is dissolved in the other.

_R__ A substance which, when dissolved in water, gives a solution that does not conduct electricity.

_P__ The reaction of an acid with a base. In aqueous solutions, the products are a salt and water.

_W_ Gain of electrons; decrease in oxidation number.

_U__ Substance which gains (accepts) electrons in a redox reaction.

MULTIPLE CHOICE: Place the letter(s) corresponding to the best answer(s) on the line preceding the question. (2 pts each = 38)

_A__2. A silver bracelet tarnishes is an example of a

A. chemical change B. physical change C. neither type of change

_B__3. An inexperienced laboratory student is trying to weigh a known 10.000 g mass. His first set of

measurements is 10.028, 10.034, 10.036. These measurements show

A. good accuracy and good precision B. poor accuracy and good precision

C. good accuracy and poor precision D. poor accuracy and poor precision

Average deviation of the measurements is approximately 1/10^th the error deviation

_B__4. The number 604.0 expressed in exponential notation with the correct number of significant digits is:

A. 6.04x10² B. 6.040x10² C. 6.04x10³ D. 6.040x10³

_B__5. To the correct number of significant figures, the result of the calculation

(830.07 x 0.00286) / 0.1030) = (0.00286 has only 3 significant figures)

A. 23 B. 23.0 C. 23.05 D. 23.049

_D__ 6. The metric prefix pico corresponds to exponential number

A. 10⁹ B. 10¹² C. 10^-9 D. 10^-12 E. none of these

_D__ 7. Calcium carbonate can be decomposed by heating it strongly to form two pure substances. Using

only this information, it can be said with certainty that

A. one of the products is an element

B. both products are elements

C. neither product can be an element

D. calcium carbonate cannot be an element since it can be broken down by heating

_B__8. The law that explains why the analysis of the sodium sulfate, Na₂SO₄, in London and in

Chicago gives the same value for the percent sodium is

A. law of conservation of mass B. law of definite proportions (constant composition)

C. law of multiple proportions D. none of these laws apply

_B__9. 14.0 grams of the element nitrogen occurs naturally as N₂ not N.

A. contains 14.0 N atoms B. contains 6.02 x 10²³ N atoms

C. contains 6.02 x 10²³ N₂ molecules D. contains 1 mole of elemental nitrogen

_B___10. The molecular formula for a hydrocarbon with an empirical formula of C₂H₅O and a

molecular mass of approximately 90 would be empirical formula mass = 45; thus 2

_C__11. In the reaction of 2.00 mol CH₄ with an excess of O₂, 1.70 mol H₂O is formed in the reaction

CH₄ + 2 O₂ -> CO₂ + 2 H₂O

1. the theoretical yield is 1.70 mol H₂O theoretical yield is 2 x 2.00 = 4.00 moles
2. the theoretical yield is 2.00 mol H₂O

_A__ 12. Helium boils at about 4 K. On the Fahrenheit scale this temperature is: 4K - 273^oC = - 269^oC

A. -452 B. -269 C. 39 D. 277 E. none of these

_B__ 13. The one(s) of the following which is (are) (a) molecular compound(s)

A. CaCl₂ B. ClF₃ C. K₂CO₃ D. NaNO₃ all contain metals except ClF₃

_D__ 14. The one(s) of the following which is (are) (an) insoluble compound(s) is (are)

A. Ba(OH)₂ B. CaCl₂ C. KClO₃ D. PbSO₄ most chlorides and chlorates are soluble including two given; Ba(OH)₂ is soluble except at high ion concentrations and is normally considered soluble.

_B__ 15. The one(s) of the following which is(are) (a) strong acid(s) is (are)

A. Ba(OH)₂ B. HClO₃ C. HF D. H₂SO₃ Ba(OH)₂ is a strong base. HClO₃ has 2 more O's than H's. Strong binary acids include only HCl, HBr, and HI

_B__ 16. When KClO₃ is converted to KCl, the Cl is corresponds to Cl⁺⁵ going to Cl^-1 or a gain of e^-

A. oxidized B. reduced C. both D. neither

_E__ 17. The oxidation number for Bi in the compound NaBiO₃ is (+1 + Bi + 3(-2) = 0; Bi = 6-1

A. +1 B. +2 C. +3 D. +4 E. +5 F. none of these

_B__ 18. Of the following metals, the most easily oxidized is (corresponds to most active metal = Ca)

A. Ag B. Ca C. Cu D. Fe E. Zn

_B__ 19. The reaction of Co with MnCl₂ to form Mn and CoCl₂ is expected to occur Mn is more active than Co so that it is metal that should be found in compound; thus expect reactants

1. in the forward direction B. in the reverse direction C. to not occur at all

_B__ 20. The chemical formula for the compound that is formed between Li and N is

A. LiN B. Li₃N C. LiN₃ D. none of these Li⁺¹ + N^-3 -> Li₃N

21. Determine the number of protons, neutrons, and electrons in the following atoms and ions. (6 pts)

22. Correct if necessary, the names for the following compounds. More than one error may occur: write correct if correct. (6 pts)

1. IBr₃, iodine tribromine binary non-metals so prefixes and ide ending

Iodine tribromide

2. CoF₃, cobalt trifluoride binary variable (oxidation state) metal + nonmetal ide ending

Cobalt(III) fluoride

3. Cu₂SO₃, copper(II) sulfate ternary variable (oxidation state) metal + 1 less O than ate anion

22. Correct if necessary, the names for the following compounds. More than one error may occur: write

correct if correct. (6 pts)

1. hypophosphorous acid. H₂PO₃ hypo-ous indicates 2 fewer O's than phosphoric acid = H₃PO₄

22. You are given a bottle that contains 2.36 ml of a yellow liquid. The mass of the bottle and the liquid is 5.26 g. . The empty bottle weighs 3.01 grams. What is the density of the liquid? (4 pts)

D = M/V = (5.26 - 3.01) g / 2.36 mL = 2.25 g/ 2.36 mL = 0.953₄ g/mL

 

Density = |_0.953 g/mL_

 

23. Determine the empirical formula for a compound with the following composition: 10.4% carbon, 27.8% sulfur and 61.7 % chlorine. Consider 100 g sample so % becomes g (6 pts)

10.4 g/ 12.01 g/mol = 0.8659 mol C

27.8 g/ 32.66 g/mol = 0.8670 mol S

61.7g / 35.45 g/mol = 1.7405 mol Cl; dividing by smallest moles = 0.8659 gives

1.00 C

1.001 S

2.01 Cl so that empirical formula is CSCl₂

Empirical formula = |_CSCl₂___

 

24. What is the molarity of the solution prepared by dissolving 1.50 g of (NH₄)₂SO₄ in enough water to make 250 ml of solution? (6 pts)

Molar Mass = 2(14.008) + 8(1.008) + 32.066 + 4(16.00) = 132.146 g/mol; Molarity = mol solute/L soln

1.50 g (NH4)2SO4 x	1 mol x	1000 mL	= 0.04540 M
250 mL	132.146 g	L

M = |_0.0454 M _

25. Calculate the mass of AgCl that will precipitate if a 45.0 mL portion of a 0.190 M silver nitrate solution is added to 26.0 mL of a 0.250 M ferric chloride solution: 3 AgNO₃(aq) + FeCl₃(aq) --> 3 AgCl(s) + Fe(NO₃)₃(aq). Molar Mass AgCl = 35.45 + 107.88 = 143.33 g / mol (6 pts)

mass = |_1.23 g __

28. Complete and balance the following metathesis reactions and then write the ionic equation, net ionic equation, and the driving force for the reaction, if any. (10)

a) ___Na₂S + __Mn (NO₃)₂ --> 2 Na⁺¹NO₃^-1 + Mn⁺²S^-2(s)

Ionic

equation: 2 Na⁺ + S^2- + Mn²⁺ + 2 NO₃^- à 2 Na⁺ + 2 NO₃^- + MnS(s)

Net ionic

equation: Mn²⁺ + S^2- à MnS(s)

Driving Force: formation of the precipitate MnS

b) ___Ba(OH)₂ + _2_ HNO₂ --> Ba²⁺(NO₂^-)₂ + 2 H⁺OH^- (HOH is H₂O, a nonelectrolyte)

Ionic

equation: Ba²⁺ + 2 OH^- + 2 HNO₂ à Ba²⁺ + 2 NO₂^- + 2 H₂O

Net ionic

equation: OH^- + HNO₂ à H₂O + NO₂^- (note the final equation should be reduced from 2's above)

Driving force: formation of the nonelectrolyte H₂O

29. Balance the following redox reactions, adding H⁺, OH^-, and H₂O as necessary. (10 pts)

2 e^- (to balance charge) + 2 H⁺ + 2 NaOCl à Cl₂ + 2 NaOH (to balance Na)

3[2 e^- + 2 H⁺ + 2 NaOCl à Cl₂ + 2 NaOH] = 6e^- + 6 H⁺ + 6 NaOCl à 3 Cl₂ + 6 NaOH

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Species oxidized: loses electrons: NH₃

Oxidizing agent: is reduced, gains electrons: NaOCl

8OH^- + ___ S^2- + _4_ Cl₂ --basic-> ___ SO₄^2- + _8_ Cl^- + 4 H₂O

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Species reduced: gains electrons = Cl₂

Reducing agent: is oxidized = loses electrons = S^2-