5 Steps to a 5 AP Chemistry 2019

336 ❯ STEP 5. Build Your Test-Taking Confidence

(c) Rearrange the rate law to: k = (^) +−
Rate
[H 23 SeO][H O][H ]^2221

=

× −−

−

7.4 10 s

[0.100 ][0.100 ][0.100 ]

81

21

M

MMM

=

7.4 × 10 –6 M –1 s–1

This example uses the values from experiment 1. You could use any of the four experiments.

You get 1 point for calculating the value of k. You get 1 additional point for the correct units.

(d) (i) H 2 O 2 is acting as an oxidizing agent.

You get 1 point for this answer.

(ii) The oxidation state of oxygen in –1.

You get 1 point for this answer.

(iii) H 2 O 2 (aq) + 2 H+(aq) + 2 e– → 2 H 2 O(l)

You get 1 point for this answer.

Total your points for the different parts. There are 10 possible points. Subtract 1 point if you did not

report the correct number of significant figures (2) in part (c).

Question 3

(a) (i) 50.00 mL 0.4000 M HCl

 ()























0.4000 mol HCl

L

50.00 mL

1 L

1000  mL

= 0.02000 mol HCl (4 significant figures)

You get 1 point for this answer.

(ii) HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l)

35.25 mL 0.1600 M NaOH

(^) ()





































0.1600 mol NaOH

L

35.25 mL

1 L

1000 mL

1 mol HCl

1 mol NaOH

= 0.005640 mol HCl (4 signifi-

cant figures)

You get 1 point for this answer.

(iii) NH 3 (aq) + HCl(aq) → NH 4 Cl(aq)

()−











0.02000 0.005640 mol HCl 

1 mol NH

1 mol HCl

(^3) = 0.01436 mol NH
3 (4 significant figures)
You get 1 point for this answer. If you miscalculated either of the preceding two answers but use
your results correctly in this step, you still earn this point.
(b) (i)











(0.01436 mol NH) 
1 mol (NH)SO
2 mol NH
132.139 g (NH)SO
(^3) 1 mol (NH )SO
42 4
3
42 4
42 4
= 0.9488 g (NH 4 ) 2 SO 4
(4 significant figures)
You get 1 point for this answer.
(ii) ×
0.9488 g (NH )SO 
1.002 g sample
(^424) 100% = 94.69 % (NH 4 ) 2 SO 4 (4 significant figures)
You get 1 point for this answer. If you got the wrong result for part (b)(i), but used it correctly
here, you still get the point.
20-Moore_PE01_p307-340.indd 336 31/05/18 1:58 pm