Exam 2 A
Chem 111, Section 2 (Martin, 1:25pm) Spring 1998h = 6.626 x 10
-34 J sc = 2.998 x 10
8 m / sh c = 1.986 x 10
-25 J mN
o = 6.022 x 1023 "particles" / molIn the questions below, "MW" refers to molecular weight (molecular mass)
1a. You are manufacturing expensive, hand-built automobiles. You currently have 8 cars ready for delivery to customers, except that they lack tires (each car requires 4 tires and one "skinny" spare). You have in stock 27 regular tires and 9 skinny spares. How many complete cars can you deliver without ordering more tires?
(a) 5 (b
Limiting "reagent" - tires. Can only build 27/4=6.75=>6 cars
2a. Hydrogen and oxygen can be used as a fuel, according to the following reaction:
2 H
2 (g) + O2 (g) Æ 2 H2O (l) 食 = -483.6 kJ/molThis reaction is:
(a
3a. How many moles of O
2 are required to completely react with 6 moles of H2?(a) 6 moles (b
) 3 moles (c) 1 mole (d) 12 moles (e) 24 molesRequire 1 mol O2 per 2 mol H2
4a. In the laboratory, you mix 10.08 g H
2 (MW=2.02 g/mol) with 16.0 g O2 (MW=32.0 g/mol) and provide a spark to ignite the reaction. Assuming that the reaction goes to completion, what mass of water (H2O, MW=18.0 g/mol) is produced?(a) 10.1 g (b) 16.0 g (c
) 18.0 g (d) 36.0 g (e) 26.1 gWe're mixing 5 mol H2 with 0.5 mol O2. O2 is limiting. So we can make (0.5 x 2) mol H2O.
Mass = 1.0 mol x 18.02 g /mol = 18.0 g water
5a. In the laboratory, you mix 2.016 g H
2 (MW=2.016 g/mol) with an excess of O2 (MW=32.0 g/mol) and provide a spark to ignite the reaction. Assuming that the reaction goes to completion, what amount of heat is produced?(a
) 241.8 kJ (b) 483.6 kJ (c) 967.2 kJ (d) 1,934 kJ (e) 2.016 kJThe limiting reagent is 1.0 mol H2. From the above reaction, combustion of 2 mols of H2 releases 483.6 kJ, so combustion of 1 mol results in half that value, 241.8 kJ.
6a. A small pool of water is sitting on your arm on a hot, dry day. If the water is considered the system and your arm the surroundings, which best describes the spontaneous process which occurs?
(a) no heat is transferred
(b) heat is transferred from the system to the surroundings
my arm heats up
(c
) heat is transferred from the surroundings to the systemmy arm cools down
(d) heat is transferred from the system to the surroundings
my arm cools down
(e) heat is transferred from the surroundings to the system
my arm heats up
7a. A 0.13 g sample of a hydrocarbon is burned in oxygen to give 0.44 g CO
2 (MW=44.0 g/mol) and 0.090 g H2O (MW=18.0 g/mol). Which molecular formula is most consistent with this result?(a
) C2H2 (b) C2H4 (c) C2H6 (d) C2H2O (e) C2H4OMoles C = 0.44 g x (mol/44.0 g) = 0.01 mol
Moles H = 2 x (0.090 g x (mol/18.0 g)) = 0.01 mol
Ratio H/C = 0.01/0.01 = 1.0. Empirical formula is CH. C2H2 fits that.
If C2H2 then MW=26.04 g/mol. To get 0.01 mol C, 0.005 mol C2H2 would have reacted.
The mass of C2H2 would then be (26.04 g/mol x 0.005 mol) = 0.13 g <- Observed
If C2H2O then MW=42.04 g/mol. To get 0.01 mol C, 0.005 mol C2H2O would have reacted.
The mass of C2H2O would then be (42.04 g/mol x 0.005 mol) = 0.21 g
8a. A 1.0 g piece of aluminum at 10.0°C is dropped into a beaker of water. The temperature of the water drops from 40° to 20° C. What quantity of heat did the piece of aluminum absorb? Note that the specific heat of aluminum is 0.902 J/gK, while that of water is 4.184 J/gK.
(a) 18 J (b) 4.2 J (c) 8.4 J (d
) 9.0 J (e) 16.8 JThe aluminum went from 10° to 20° C. Heat = (0.902 J/gK) x (1.0 g) x (10 K) = 9.0 J
9a. 食
vap for ethanol (C2H5OH, MW=46.07 g/mol) at its boiling point is 38.56 kJ/mol.(a) Boiling 4.6 g ethanol requires more heat input than boiling 3.6 g water
(b
) Boiling 4.6 g ethanol requires less heat than boiling 3.6 g water(c) Condensing 4.6 g ethanol requires more heat than condensing 3.6 g water
(d) Condensing 4.6 g ethanol requires more heat than condensing 3.6 g water
(e) Condensing 4.6 g ethanol releases more heat than condensing 3.6 g water
The table below lists some standard enthalpies of formation at 298 K
Substance 食
f°(kJ/mol)
CO (g) -110.5
CO
2 (g) -393.5CH
3OH (l) -238.7H
2O (l) -285.810a. What is 食° for the following reaction at 298 K:
2 CH
3OH (l) + 2 O2 (g) Æ 2 CO (g) + 4 H2O (l)(a) -157.6 kJ/mol (b) 886.8 kJ/mol (c) 157.6 kJ/mol
(d
食° = [ 2(-110.5) + 4(-285.8) ] - [ 2(-238.7) + 2(0) ] = -886.8 kJ/mol
11a. Both hydrogen (H
2) and ethanol (C2H5OH) have been proposed as alternate fuels for automobiles. 食° for the complete combustion of H2 is -285.8 kJ/mol, while that for ethanol is -1,367 kJ/mol. Remembering that weight contributes to overall fuel efficiency, combustion of which fuel produces more heat per gram?(a
) hydrogen produces more heat per gram(b) ethanol produces more heat per gram
H2 : heat per gram = -285.8 kJ/mol x (mol/2.016g) = 141.8 kJ/g
C2H5OH : heat per gram = -1,367 kJ/mol x (mol/46.07g) = 29.7 kJ/g
12a. Iron (III) oxide ("rust") can be produced from iron and oxygen in a sequence of reactions that can be written as
2 Fe (s) + 6 H
2O (l) Æ 2 Fe(OH)3 (s) + 3 H2 (g) +68.8 kJ/mol2 Fe(OH)
3 (s) Æ Fe2O3 (s) + 3 H2O (l) -35.6 kJ/mol3 H
2 (g) + 3/2 O2 (g) Æ 3 H2O (l) -857.4 kJ/molWhat is 食° for the reaction
4 Fe (s) + 3 O
2 (g) Æ 2 Fe2O3 (s)(a) 1,648 kJ/mol (b) 824.2 kJ/mol (c) -1,682 kJ/mol
(d
Correct balancing requires 2 of first equation, 2 of second equation, and 2 of third equation
食° = 2(+68.6) + 2(-35.6) + 2(-857.4) = -1,648 kJ/mol
13a. A given orbital is labeled by the magnetic quantum number m
l = -1. This could not be a(a) g orbital (b
) s orbital (c) p orbital (d) f orbital (e) d orbital14a. A given orbital has quantum numbers n =3, l = 1, m
l = 1. This orbital must be:(a
) 3p (b) 4d (c) 3d (d) 3s (e) 4p15a. For the hydrogen atom, which of the following transitions produces an emitted photon with the highest frequency?
(a) n = 4
Æ n = 2 1/4 - 1/16 = +0.1875 (emitted)(b
) n = 2 Æ n = 1 1/1 - 1/4 = +0.75 (emitted)(c) n = 2
Æ n = 4 1/16 - 1/4 = -0.1875 (absorbed)(d) n = 1
Æ n = 2 1/4 - 1/1 = -0.75 (absorbed)(e) none emits a photon
16a. Which of the following photons has the highest energy?
(a) radio waves (
l = 0.5 m) (b) microwaves (l = 10-3 m)(c
) near UV (l = 250 nm) (d) orange light (l = 625 nm)(e) infrared (
l = 1000 nm)Lowest wavelength -> highest energy
17a. Radiation in the UV region of the electromagnetic spectrum is quite energetic (remember your sun screen!). If you are bombarded by 1 mol of photons with wavelength
l = 300 nm, how much energy are you being subjected to?(a
) 3.99 x 105 J (b) 6.62 x 10-37 J (c) 3.99 x 10-13 J(d) 6.62 x 10
-28 J (e) 6.62 x 10-19 JE = hc/
l = (1.986 x 10-25 J m / 300 nm) x (109 nm/m) x (6.02 x 1023 /mol) = 3.99 x 105 J
18a. How many nodes does a 3p
x orbital have?(a) 0 (b) 1 (c
) 2 (d) 3 (e) 419a. Which best describes the node in a 2p
x orbital?(a) spherical (b) in the xy plane (c
) in the yz plane(d) in the xz plane (e) there is no node
20a. How many orbitals can have the principle quantum number 3?.
(a) 1 (b) 2 (c) 3 (d) 5 (e
) 9For n=3: l=0, ml=0; l=1, ml=-1,0,1; l=2, ml=-2,-1,0,1,2; So 1 + 3 + 5 = 9
Remember:
You have exam A. Place a "1" in column "P" under "Special Codes"
As a homework assignment, you may earn up to 10% of the points you missed on this exam (eg., if you scored a 60 on the exam, you can earn an extra 4 points), by doing the following:
1) Work through all of the problems at home (consultation with others and with books is OK, but you should answer the questions yourself). Answer all of the questions.
2) Turn in the Op-Scan sheet in class on Wednesday, April 1 (no later!).
Your revised exam will be scored and credit applied proportional to the total number of questions answered correctly (if you do less well than you did on the exam, points will not be taken off). Complete the exam exactly as you did previously (except with all correct answers, of course!).
Answers and scores for the original exam will be available by April 1. Check our course home page.