CHM1045 Practice Test 3 v.1 - Answers Name Fall 2013 & 2011 (Ch. 5, 6, 7, & part 11) Revised April 10, 2014

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1 CHM1045 Practice Test 3 v.1 - Answers Name Fall 013 & 011 (Ch. 5, 6, 7, & part 11) Revised April 10, 014 Given: Speed of light in a vacuum = 3.00 x 10 8 m/s Planck s constant = 6.66 x J s E (-.18x10 1 J) n final n initial Part A. Multiple Choice Circle the best answer, FILL IN THE BLANK or ANSWER. 1. The first law of thermodynamics is: (1 pt.) A) Mass is neither created nor destroyed. *B) Energy is neither created nor destroyed. C) Pressure is constant. D) Gases obey the gas laws.. Standard Temperature and Pressure (STP) for gases is. ( pts.) A) 5C, 1 atm B) 0C, 760 atm *C) 0C, 760 Torr D) 1C, 0 atm 3. If the temperature of a gas decreases, what will happen to the volume if all other variables are held constant? ( pts.) a) increase *b) decrease c) stays the same 4. Question #3 above illustrates: ( pts.) *a) Charles Law b) Avogadro s Law c) Boyle s Law d) Graham s Law e) Dalton s Law 5. What are the possible values of n and m l for an electron in a 3d orbital? ( pts.) a) n = 3 and m l = b) n = 3 and m l = 1, 0, or +1 c) n = and m l =, 1, 0, +1, or + *d) n = 3 and m l =, 1, 0, +1, or + e) n = 3 and m l = 3,, 1, 0, +1, +, or Which of the following set of quantum numbers are not allowed? ( pts.) a) n=, l=1, m l = 1,0,+1 b) n=3, l=, m l =, 1, 0, +1, + *c) n=3, l=3, m l = 3,, 1, 0, +1, +, or +3 d) n=3, l=1, m l = 1,0,+1

2 7. Kinetic Energy is defined as: ( pts.) = the energy of a moving object 8. The change of state from a gas to a solid is called deposition. (1 pt.) 9. The name of the quantum number n is the principal quantum number. (1 pt.) 10. Give a general description in words of the van der Waals equation (don t need to give the actual equation) and list at least one thing that this equation corrects for. (3 pts.) - Calculates P, V, n, or T of a real gas, includes constants a & b that must be measured in the lab for each gas. - Corrects for the volume of a gas molecule - Corrects for attractions between molecules 11. The temperature of a 5.00 L container of N gas is increased from 0ºC to 50ºC. If the volume is held constant, predict qualitatively how this change affects the following: (a) the average kinetic of the molecules; (b) the average speed of the molecules; (c) the total number of collisions of molecules with the walls per second; (d) the pressure in the container. (8 pts.) (a) The average kinetic energy increases since the temperature increases (energy is being added). (b) The average velocity of the molecules increase because kinetic energy (K.E.) increased and K.E. = (1/) mv, (mass doesn t change so velocity must) (c) the total number of collisions of molecules with the walls increase per second because they are moving faster, so they hit walls more frequently (d) the pressure in the container increases because pressure is created by collisions of molecules with the walls, if there are more collisions per sec then the pressure increases.

3 3 Part B. Calculations. MUST SHOW ALL WORK. Give answer to correct significant figures and units. 1. (a) Is the following reaction exothermic or endothermic? Circle one. ( pts.) (b) How much heat is absorbed or released when 7.55g of hydrogen are produced? (4 pts.) NH 3 (s) N (g) + 3 H (g) ΔH= kj 9.38 kj H 3 moles H kj 1mole H 1mole H 7.55 g g moles H kj moles H 1mole H OR: 115 kj 1mole H 7.55 g g 9.38 kj 3 mole H 115 kj 13. We obtain uranium-35 and U-38 by fluorinating the uranium to form UF 6 (which is a gas) and then taking advantage of the different rates of effusion and diffusion for compounds containing the two isotopes. Calculate the ratio of effusion rates for 38 UF 6 and 35 UF 6. The atomic mass of U-35 is amu and that of U-38 is amu (6 pts.) r r mass of mass of UF UF g/mole g/mole

4 4 14. (a) Convert 738 mmhg into Torr. (b) Convert 738 mmhg into atm. (c) Convert 738 mmhg into in.hg. Show work. (8 pts. total) (a) 1Torr 738 mmhg 1mmHg 738 Torr (b) 1atm 738 mmhg 760 mmhg 0.971atm (c) 1m 100 cm 1in. 738 mmhg 1000 mm 1m.54 cm 9.1in. Hg 15. Zinc metal reacts with hydrochloric acid according to the balanced equation: Zn (s) + HCl (aq) ZnCl (aq) + H (g) When g of Zn (s) is combined with enough HCl to make 50.0 ml of solution in a coffee-cup calorimeter, all of the zinc reacts, raising the temperature of the solution from.5c to 3.7C. Find H for this reaction as written. (Use 50.0 g as the mass of the solution and 4.18 J/g ºC as the specific heat.) (7 pts.) 4.18 J g C, q rxn = 50.8 J q surr = 50.0 g3.7 C.5 C 50.8 J 1mole Zn g Zn mole g q rxn H # of moles 50.8 J mole 1kJ 1000 J kj 1mole Zn

5 5 16. A 48 ml gas sample has a mass of g at a pressure of 745 Torr and a temperature of 8C. What is the molar mass of the gas? (6 pts.) P = 745 Torr 1atm 760 Torr atm, T = 8 C = K, V = 48 ml = 0.48 L PV = nrt n PV RT atm 0.48 L L atm K mole K mole Molar mass = grams moles g mole 44.0 g mole 17. A local radio station broadcasts and an energy of 7.8 x 10 6 J. What is this frequency in MHz? Show work. (4 pts.) 6 E E 7.8x10 J 1Hz 1MHz h, 118 MHz 34 1 h J s x s 1x10 6 Hz 18. Green light has a wavelength of about m. What is the frequency in Hertz? (3 pts.) 6 1x 10 m m m x 7 m c, 8 c 3.00x10 m/s 5.45x x10 m 14 s x10 14 Hz

6 6 19. A gas mixture contains 1.45 g N and 0.65 g O in a 1.75 L container at 19C. Calculate the partial pressure of each component in the gas mixture and calculate the total pressure. (7 pts.) There are several ways to do this problem, all are shown below. T 19 C = 9.15 K, V = 1.75 L 1mole N 1.45 g N mole g 1mole O 0.65 g O mole g n total = mole n RT P total = total V L atm mole K 1.75 L mole K atm n N PN V RT L atm mole K 1.75 L mole K atm n O PO V RT Latm mole K 1.75 L mole K atm PO P P total N atm 0.8 atm P O n O mole P total n total mole atm atm 0.8 atm P N n N mole P total n total mole atm atm atm

7 7 0. How much energy is needed to heat 45.0 g of H O (s) from 5.0C to 0.0C? (4 pts.).09 J q g K 1kJ 45.0 g0.0 C 5.0 C.35 kj 1000 J 1. How much energy in kj is needed to convert g of H O from a liquid at 100ºC to a gas at 100ºC? (4 pts.) (see Given information on page 1) 1mole H O kj g H O g 1mole H O 59.6 kj. CH 3 OH can be synthesized by the reaction: CO (g) + H (g) CH 3 OH (g) What volume of H gas (in liters), at 85C and 746 mmhg partial pressure of hydrogen gas, is required to synthesize 7.7 g of CH 3 OH? MUST SHOW WORK. (7 pts.) 1atm T = 85 C = K, P = 746 mmhg atm 760 mmhg 1mole CH OH mole H 7.7 g CH 3 3OH mole H g 1mole CH3OH PH V n RT H n H RT V P H Latm mol K mole K atm L

8 8 3. The volume of helium gas in a balloon is 5.0 L at 788 Torr. What is the volume of the balloon at 465 Torr if the temperature is constant? (6 pts.) P V P V P V 1 1, T 1 = T, 1 1, P1 V1 P T1 T V T 1 T 788 Torr5.0 L P1 V1 V L P 465 Torr P V 4. Use the change in enthalpy of formations to calculate ΔH of the following reaction. (7 pts.) N O 4 (g) + 4 H (g) N (g) + 4 H O (g) The enthalpy of formations of H O (g) and N O 4 (g) are given on an attached table. ΔH rxn = [ (1 mole)δh f (N (g)) + (4 mole)δh f (H O(g))] [(1 mole)δh f (N O 4 (g)) + (4 mole)δh f (H (g))] = [ 0 + (4 mole)( 41.8 kj)] [ kj + 0] = [ kj] [+9.66 kj] = kj

9 9 5. Calculate H for the following reaction: (8 pts) MUST SHOW ALL WORK! 5 C (s) + 6 H (g) C 5 H 1 (l) Using the enthalpies of reactions: reverse: C 5 H 1 (l) + 8 O (g) 5 CO (g) +6 H O (g) H = kj (5){C (s) + O (g) CO (g)} H = {393.5 kj}(5) (3){ H (g) + O (g) H O (g)} H = {483.5 kj}(3) 5 CO (g) + 6 H O (g) C 5 H 1 (l) + 8 O (g) H = kj 5 C (s) + 5 O (g) 5 CO (g) H = {393.5 kj}(5) = kj 6 H (g) + 3 O (g) 6 H O (g) H = {483.5 kj}(3) = kj 5 CO (g) + 6 H O (g) + 5 C (s) + 5 O (g) + 6 H (g) + 3 O (g) C 5 H 1 (l) + 8 O (g) + 5 CO (g) + 6 H O (g) 5 C (s) + 6 H (g) C 5 H 1 (l) (exactly the reaction we wanted) H = ( kj) + ( kj) + ( kj) = 173. kj

10 6. The oxygen gas formed in a chemical reaction is collected over water at 30.0ºC at a total pressure of 73 mmhg. What is the partial pressure of the oxygen gas collected in this way? If the total volume of gas collected is 7 ml, what mass of oxygen gas is collected? (8 pts.) T = 30.0 C = K P total = 73 mmhg P HO 31.8 Torr 1atm PO Ptotal PH O 73 Torr 31.8 Torr Torr atm 760 Torr PO V n RT O P V atm 0.7 L O O mole O RT L atm 1mole O n K mole K g g O 10

11 11 Given for Test 3 Gas constant = L atm/mole K = J/mole K 1 cal = J Properties of Water Density = g/ml at 0C Specific heat = ice,.09 J/gK g/ml at 4C water, J/gK g/ml at 5C steam, J/gK g/ml at 100C Heat of fusion = kj/mol Heat of vaporization = kj/mol Properties of Ethanol (C H 5 OH) Melting point = 117.3C Boiling point = 78.5C Heat of fusion = 6.05 cal/g Heat of vaporization = 9,673.9 cal/mol Specific heat = liquid, 7.0 cal/mol gas, 15.7 cal/mol compound H f (kj/mole) compound H f (kj/mole) CH 4 (g) 74.8 C 3 H 8 (g) CO (g) C 4 H 10 (g) CO (g) C 4 H 10 (l) H O (g) 41.8 H O (l) NO(g) NO (g) N O(g) N O 4 (g) C H 4 (g) C H 6 (g) 84.68

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