Answer Key 1441 Test 3 chapters 6 & 7 Zumdahl
1. D
2. C ( by definition heat of formation is producing one mole of a compound
from its elements in their standard state. Since you are producing one mole
of the compound, you may have to use fractions for coefficients on the
reactant side.)
3. ∆Hrxn = Hproduces - Hreactants
= [ H2S(g) + 2H2O(l)] – [SO2(g) + 3H2(g)]
= [ -4.8kcal + (2) -68.3 kcal] – [ -71kcal + (3) 0]
= -70.4 kcal
4. 41.8 kcal ( see website help topics Hess’ Law)
5. 36kJ ( see website help topics Hess’ Law)
6. (32gCH4/1)(1 mole CH4/ 16gCH4)( 4.34X 105J/1 mole CH4)
= 8.6 X 105J
Heat is released since the H value was negative which means
exothermic. Generally, answers are expressed without sign if
the question is, what is the energy change or how much heat is
liberated; however, if -8.6 X 105 J had been a choice, select that
for your answer.
7.
∆E = q + w
q is heat, if the volume changes, then the work is
p∆V work and w = -P∆V. So the equation becomes:
∆E = q - P∆V
= -92.2kJ - 40atm(-1.12L)
= -92.2kJ + 44.8 L atm (conversion factor is 101J = 1 Latm)
= -92.2kJ + (44.8Latm/1)(101J / 1 Latm)
= -92.2kJ + 4524 J (1 kJ/1000J)
= -87.7 kJ
8.
∆E = q – P∆V
q = + 1.3 X 108J because it is endothermic
8
= 1.3 X 10 J – (1 atm) ( 4.5 X 106 L – 4.00 X 106 L) ∆ means final – initial
= 1.3 X 108J - .5 X 106Latm(101.3J/1 Latm)
= 1.3 X 108 J – 5.1 X 107 J
= 8 X 107 J
9. Energy = ∆T ( heat capacity)
= (2.25oC)(11.3 kJ/oC) = 25.4 kJ
(25.4 kJ / 0.5269 g)(114.2 g C8H18 / 1 mole C8H18) = 5.5 X 103 kJ/mole
Since heat is released during combustion, the sign would be negative
- 5.5 X 103 kJ/mole.
10. Mass of the solution : D = m/v
m = D V = (1 g/ml)(100ml) = 100g
q = (specific heat)(mass of solution)(∆T)
= ( 4.18 J/gK)(100g)(6.5K) = 2700J or 2.7 kJ
Since it is exothermic it would be expressed as -2.7 kJ
moles of HCl : M = moles/L
moles = M (L) = (1M)(.05 L) = .05 moles of HCl used
Since the same volume of NaOH was used, it is also .05 moles.
∆H = -2.7kJ/0.05 mole HCl = -54 kJ/ mole HCl
Or
(.05 mole HCl / 1 mole HCl) = (-2.7 kJ / X kJ)
X = -54 kJ/ mole HCl
11. -Heat lost = Heat gained
( - is used before heat lost because losing heat is exothermic. For temperatures,
∆ means final – initial temperature)
-(specific heat)(mass)(∆T) = (specific heat)(mass)(∆T)
-(.444J/oCg)(15g)(T2-100) = (4.18 J/oCg)(55g)(T2 – 23oC)
-6.66J/oC(T2-100C) = 229.9J/oC (T2-23oC)
-6.66J/oC T2 + 666J = 229.9J/oC T2 – 5287.7 J
-236.6 J/oC T2 = -5953.9 J
T2 = 25.2oC
12. ν= c / λ= (3 X 1010 cm/s / 9 X 1011s-1 ) = 3.3 X 10-2cm
13. ∆E =-2.178 X 10-18 J [ 1/n2final– 1/n2initial]
= -2.178 X 10-18 J [1/64 – 1/1]
= 2.144 X 10-18 J (The answer is + because as the
electron moves from 1 to 8, energy is absorbed.)
E = hν= hc/λ
= (6.63 X 10-34 J.s)(3 X 108 m/s) / (2.144 X 10-18J)
= 9.3 X 10-8m
= (9.3 X 10-8 m /1)(1 X 109 nm / 1 m) = 93 nm
14.
∆E = -2.178 X 10-18J [1/n2final – 1/n2initial]
= - 2.178 X 10-18 J [1/infinity2* – 1/22]
= 5.45 X 19-19J (answer is + because energy was absorbed in order to
ionize the electron.)
* 1/infinity is approximately 0
E = hc/λ
λ = hc/E = (6.63 X 10-34 Js) (3 X 108 m/s) / (5.45 X 10-19 J)
= 3.65 X 10-7m
= (3.65 X 10-7m /1) ( 1 X 109 nm/ 1 m) = 364 nm
15.
λ = h/mv = 6.63 X 10-34 Js / (9.11 X 10-28g)(8 X 107 cm/s)
Since a J = kg m2/s2 units must be in Kg and m.
= 6.63 X 10-34 Js / [9.11 X 10-28g(1 Kg/1000g)][8 X 107cm/s(1 m/100cm)]
= 9.1 X 10-10 m
16. (a) 1s22s22p63s23p64s23d5 (b) 1s22s22p63s23p63d54s2 (c) [Ar]4s23d5
17. Cu = [Ar]4s13d10 promotes an s electron in order to fill the d sublevel
Cr = [Ar] 4s13d5 promotes an s electron to half-fill the d sublevel
18. Zr2+ = [Kr]4d2
O2- = 1s22s22p6
19. S = 6 1s22s22p63s23p4 ( outer energy level is 3 and has 6 total electrons)
V = 2 [Ar] 4s23d3 (outer energy level is 4 and has two electrons)
20. Sc = 1s22s22p63s23p64s23d1 ( Highest energy is 3d lowest is 1s)
The filling order of electrons is from low to high energy electrons.
21. Ground State- follow Aufbau principle( must fill one level before moving to
the next)
Excited state- moves to a higher level before filling the present level.
Impossible: 1s3 ( s can’t hold 3) and 1s21p1 ( 1p does not exist)
However: 1s22p1 is possible in the excited state.
22. (a) ground state (b) Pauli Exclusion Principle (c) Hund’s Rule
(d) excited state and does not violate any rules
23. 2 ( n= 3 limits it to 28, l = 1 means p and limits it to 6, ml = 1 is an
orbital designation and limits it to 2 electrons)
24. c only ( see allowed quantum numbers under memory list)
25. Outer electron in Al is 3p1. For a p sublevel l = 1, so the magnetic
quantum number, ml, can be l to –l or the answer is 1, 0, -1
26. (a) O (b) N ( violates the trend because N has a half-filled p sublevel.
(c) Be ( violates the trend because Be had a full s sublevel)
27 (a) Rb (b) Br
28. Gamma rays, X rays, blue light, orange light, infrared light, microwaves
29. ν= c/λ
λ = c/ν c = 3 X 108 m/s , ν = 96.3 MHz ( M = mega 1X 106) so
ν = 96.3 X 106 Hz ( unit needs to be Hz which is s-1)
λ =[3 X 108m/s] / [96.3 X 106 Hz or s-1]
= 3.11 m
30. [Xe]6s24f145d106p6