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MIDTERM EXAM IN
CHEM 121
(CHEMISTRY FOR ENGINEERS)
Test I: What is the oxidation number of nitrogen in each of the following (2 points each)
a
...
Nitrogen Tetra Oxide (N2O4)
2N + 4 (-2) = 0
2N – 8 = 0
2N = 8
N=4

b
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Anmonium Thiosulfate (NH4)2S2O3
2N + 4 (1) + 2 (-2) + 3 (-2) = 0
2N + 4 – 4 – 6 = 0
2N = 6
N=3

1 + N + 3 (-2) = 0
1+N–6=0
N–5=0
N=5
c
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Ammonia (NH3)
N + 3 (1) = 0
N+3=0
N = -3
i
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Nitrous Acid (HNO2)
1 + N + 2 (-2) = 0
1+N–4=0
N–3=0
N=3

d
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K3N
3 (1) + N = 0
3+N=0
N = -3

Test II: Fill in the blanks with formula of compounds composed of the indicated positive and
negative ions
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(30 points)
Sample: Sodium Hydroxide(l) and Sulfuric Acid(l) yield sodium sulfate(l) and water(l)

2NaoH(l) + H2SO4(l) → Na2SO4(l) + 2HOH(l)

1
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Sodium(s) and sulfur hexafluoride(g) yield Sodium Sulfide and Sodium Fluoride (s)

8Na(s) + SF6(g)  6NaF + Na2 S(s)
3
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Sodium Hydroxide(s) plus iron(III) oxide yield Sodium oxide and iron(III) hydroxide(l)

2NaOH(s) + Fe2O3(III)  3Na2O(III) + 2Fe(OH)2(l)
5
...
) yield hydrogen(g) and aluminum chloride(aq)

2Al (s) + 6HCl (aq)  3H2 (g) + 2AlCl3 (aq)

Test IV: Convert the ff
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1x10-2 m → km
1 x 10-2 = 0
...
01 x


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00001 km
= 1 x 10-5 km
2
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32x105 kg → g
7
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= 732000000 g
= 7
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330 lbs → mg
= 330 lbs x

x

= 149,820,000 mg
4
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1 dg x
= 1 cg
5
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11x105 cal
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11 x 105 = 111000 cal
= 111000 cal x
= 464424000 joules
= 4
...
4
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09728
= 4
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09728
= 4
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1
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kg = tons × 907
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5 x 907
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60 kg
8
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67) ×
= (-500 + 459
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33 ×
= -22
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720 m/s → km/h
1m=

𝑘𝑚 ;

1 m/sec =

1 sec =
𝑘𝑚/ℎ𝑟

= 720 x
= 2,592 km/hr
10
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20 oF → oR
T(°R) = T(°F) + 459
...
20°F + 459
...
87 °R

ℎ𝑟
=

𝑘𝑚/ℎ𝑟



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