2018
AP Chemistry
Scoring Guidelines
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AP® CHEMISTRY
2018 SCORING GUIDELINES
Question 1

DH rxn


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AP® CHEMISTRY
2018 SCORING GUIDELINES
Question 1 (continued)




 ‒








DH rxn
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AP® CHEMISTRY
2018 SCORING GUIDELINES
Question 1 (continued)


DH rxn


DH rxn

mol
DH rxn
By doubling the volumes, the number of moles of the reactants are
doubled, which doubles the amount of energy produced.
OR
mcΔT
2 mcΔT
.
D
n
2n
Thus the magnitude is the same as calculated in the first experiment
In the second experiment, D






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AP® CHEMISTRY
2018 SCORING GUIDELINES
Question 2

( )
DH 298

DS298

DG298

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AP® CHEMISTRY
2018 SCORING GUIDELINES
Question 2 (continued)


K = e- DG / RT
-
870 J/mol
(8.314 J mol -1 K -1 )(298 K)
K= e
K = 0.70
PNO
PNO 2
PN 2O 3
PN 2O 3

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AP® CHEMISTRY
2018 SCORING GUIDELINES
Question 2 (continued)
1 point is earned for a valid diagram.
1 point is earned for the correct answer.



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AP® CHEMISTRY
2018 SCORING GUIDELINES
Question 2 (continued)

0.100 mol KOH
1000 mL KOH

0.0020 mol HNO2
0.100 L


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AP® CHEMISTRY
2018 SCORING GUIDELINES
Question 3
1 point is earned for a correct electron configuration.
.
Fµ
q1 q2
r2
.





1 point is earned for the correct half-reaction.
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AP® CHEMISTRY
2018 SCORING GUIDELINES
Question 3 (continued)
17 48
´
0 0350
1000
0.000612 mol KMnO4 ´
= 0 000612
4
5 mol Fe2 +
= 0.003059 mol Fe2 +
1 mol KMnO4
1 point is earned for calculating the
number of moles of KMnO
.
1 point is earned for the correct
concentration of
.
2+




The volumetric flask is designed to contain only 25.00 mL precisely.
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1 point is earned for
a valid explanation.
AP® CHEMISTRY
2018 SCORING GUIDELINES
Question 3 (continued)
7.531 g Fe2 O3 ´
1 mol Fe2 O3
= 0.04716 mol Fe2 O3
159.70 g Fe2 O3
0.04716 mol Fe2 O3 ´
0.09431 mol Fe ´
2 mol Fe
= 0.09431 mol Fe
1 mol Fe2 O3
55.85 g Fe
= 5.267 g Fe
1 mol
5.267 g Fe
´ 100 = 78.33%
6.724 g sample
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AP® CHEMISTRY
2018 SCORING GUIDELINES
Question 4

-1
-1
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AP® CHEMISTRY
2018 SCORING GUIDELINES
Question 5




0 0350


=
-
=
( 0 00455) 2
0 0304
= 6 81 ´ 10-4
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AP® CHEMISTRY
2018 SCORING GUIDELINES
Question 5 (continued)
1
2
1
2
-
1
2
1
2

0 035
2
-





0 0175 0 00345
0 0175

2
0 0175



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AP® CHEMISTRY
2018 SCORING GUIDELINES
Question 6

 
 
Ecell

Ecell
 
Ered (cathode) - Ered (anode)

Ered



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AP® CHEMISTRY
2018 SCORING GUIDELINES
Question 6 (continued)
 
D
æ
-ç
è

-ö
æ
÷ø çè 96 485
ö
-÷
ø
(1 54 )

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 
AP® CHEMISTRY
2018 SCORING GUIDELINES
Question 7
0 693

12


   

- 0 069

0
-1
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