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Chapter One and Two Review WS Answers
Chapter One and Two Review answers:
Part A:
- 3
- 4
- 3
- 3
- 1
- 5
- 1
- 3
- 3
- 5
Part B:
- 959.2 m2
- 5.7 cm2
- 0.06324 g/mL
- 720 g
- 0.020 mL
- 520 km
Part C:
- 4.5 x 10-4
- 5.64 x 109
- 0.00002905
- 9200
Part D:
- 4.530 x 105
- 1.500 x 108
Part E:
- neither
- chemical
- physical
- physical
- chemical
- chemical
Part F:
- Heterogeneous mixture
- Homogeneous mixture
- pure substance
- homogeneous mixture
- pure substance
- pure substance
- heterogeneous mixture
- pure substance
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Chapter 3 and 24 review ws
Answers to chapter 3 and 24 review worksheet
PAGE TWO
Symbol
No. of p+
No. of n0
Mass no.
No. of e-
Atomic no.
12
C
6
6
6
12
6
6
11
B
5
5
6
11
5
5
9
Be
4
4
5
9
4
4
37
Cl
17
17
20
37
17
17
Name
Protons
Neutrons
Electrons
Mass Number
Boron-11
5
6
5
11
Chromium-52
24
28
24
52
Sulfur-32
16
16
16
32
Tin-120
50
70
50
120
SKIP QUESTIONS: 1, a, and b
c. At # 27; 59-Mass #
d. Cr-53 protons-24 neutrons-29 electrons-24
PAGE THREE
- B
- D
- C
- B
- A
PAGE FOUR
1.
a. 39 days
b. protons-53 neutrons-73 electrons-53
2. 259.2 g
3. a. 227 4 223
Ac ¨ ƒ¿ + Fr
89 2 87
b. 14 0 14
C ¨ ƒÀ + N ()
6 -1 7
4. Fr is not stable because it is greater than 83 and N is stable
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Chapter 10 review answers
Answers to Mole Review Worksheet:
1. a. 180.0 amu b. 55.85 amu
2. a. one molecule of glucose contains 6 atoms of carbon, 12 atoms of hydrogen, and 6 atoms of oxygen.
b. 1.5 x 10 24 atoms
3. a. 36.5 g b. 16.0 g c. 48.0 g d. 132.1 g
4. (5.88 moles) 3.54 x 1024 molecules
5. a. 26.35 g x 1mol = 0.313 moles MgCO3
84.3 g
b. 18.0 mol x 102.0 g = 1836 g Al2O3
1 mol
c. 125.52 g x 1 mol = 0.9678 mol NiCl2
129.7 g
d. 23.2 mol x 154.0 g =3573 g CCl4
1 mol
6. Al2O3 contains 52.9% Al. 52.9% of 200. g is 105.8 g of Al
Al2O3 = 102. 0 g Al = 2 x 27.0 = 54.0 g/102.0g x 100 = 52.9 %
200 g compound x 52.9 g Al = 105.8 g Al
100 g compound
7. 73.86% Hg
HgCl2 = 271.6 g ; Hg = 200.6g/271.6g x 100 = 73.86%B
8. C4H9
(reduce 8:18 to 4:9)
9. NiC2N2
Ni = 52.6 g x 1 mol = 0.896 mol = 1
58.7 g 0.896
C = 21.9 g x 1 mol = 1.825 mol = 2
12.0 g 0.896
N = 25.5 g x 1 mol = 1.821 mol = 2
14.0 g 0.896
10. CH2O
C = 7.20 g x 1 mol = 0.600 mol = 1
12.0 g 0.600
H = 1.20 g x 1 mol = 1.20 mol = 2
1.0 g 0.600
O = 9.60 g x 1 mol = 0.600 mol = 1
16.0 g 0.600
11. C11H22O11
n = Molecular formula molar mass (given) = 330 g =11….11 x (CH2O) = C11H22O11
Empirical formula molar mass (calculated) 30g
12. Fe2O3 this compound is called iron(III) oxide
Fe = 18.61 g x 1 mol = 0.334 mol = 1 x 2 = 2
55.8 g 0.334
O = 26.61 g – 18.61 g = 8.00 g x 1 mol = 0.500 mol = 1.5 x 2 = 3
16.0 g 0.334
