CEM01A1 – TEST 1 (4 March 2024) – Questions + Answers
Question 1:
multiple choice questions. Circle only one letter in each- 2 marks each = 10
1.
Which one of the statements below is true?
a. When two atoms combine to form a chemical compound, they do so in many different
proportions by mass.
b. When two different compounds combine to form an element, they do so in definite
proportions by mass.
c. When two different elements combine to form a chemical compound, they do so in definite
proportions by mass.
d. When two molecules combine in a chemical reaction, a number of different elements can be
generated depending on the masses used.
e. When two different elements combine to form a mixture, they do so in definite proportions by
weight.
Answer: c
Section 0.4
Difficulty Level: medium
2. How many neutrons are there in one atom of 184W?
a) 74
b) 112
c) 258
d) 110
e) 184
Answer: d
3.
Which one of the following is not an intensive property?
a) density
b) temperature
c) melting point
d) mass
e) boiling point
Answer: d
Difficulty Level: medium
The volume of a cylinder is  r2h, where r is the radius and h is the height. What is the derived SI unit
for the volume of a cylinder?
a. kg m/s3
b. liter
c. cm3
d. m3
e. cubic inches
Answer: d
Section 1.4
Difficulty Level: medium
4.
5.
The number of atoms in one formula unit of the substance Cs2SO4·5H2O is
a. 4
b. 17
c. 22
d. 25
e. 33
Answer: c
Section 2.4
Difficulty Level: hard
1.6 The law of constant composition applies to
A. compounds
B. homogeneous mixtures
C. solids
D. heterogeneous mixtures
E. solutions
Answer: A
1.7 Methane and ethane are both made up of carbon and hydrogen. In methane, there are 12.0 g of carbon for
every 4.00 g of hydrogen, a ratio of 3:1 by mass. In ethane, there are 24.0 g of carbon for every 6.00 g of
hydrogen, a ratio of 4:1 by mass. This is an illustration of the law of
A. conservation of mass
B. multiple proportions
C. constant composition
D. octaves
E. conservation of matter
Answer: B
Question 2:
2.1 Give the correct answer, reported to the proper number of significant figures, for the following
calculations:
3 marks
a) (2.05631 × 6.9391136) ÷ 12.59326 = <1.13306393…> = 1.13306
b) (6.217 × 103 ) − (5.23 × 102 ) = 5694
c)
(2.67+19.2)×3.99
1.36+11.366
= <12.726> = 12.7
2.2 Solid antimony has a density of 6.70 g/cm3. Liquid antimony has a density of 6.53 g/cm3. If a 1.34 kg
sample of liquid antimony is allowed to cool into a solid, calculate using dimensional analysis, by how
much will the volume of the sample change?
4 marks
Solution:

The mass of the sample remains the same = 1.34 kg. Calculate the volume of each phase and then calculate the volume difference.
π‘‰π‘™π‘–π‘ž = 1.34 π‘˜π‘” ×
1000 𝑔
1π‘π‘š3
×
= 200. π‘π‘š3
1 π‘˜π‘”
6.70 𝑔
π‘‰π‘ π‘œπ‘™π‘–π‘‘ = 1.34 π‘˜π‘” ×

1000 𝑔
1π‘π‘š3
×
= 205.21 π‘π‘š3
1 π‘˜π‘”
6.53 𝑔

Volume difference = π‘‰π‘™π‘–π‘ž − π‘‰π‘ π‘œπ‘™π‘–π‘‘ = 205.21 − 200 = 5.21 π‘π‘š3

– Sig figs and – DA
Question 3:
What is the volume in cubic millimeters of a 3.54 carat diamond, given that the density of diamond is 3.51
g/mL? (1 carat = 200 mg). Assuming that the diamond is generally spherical (volume of a sphere is V
4 3
pr , calculate using dimensional analysis the diameter of the diamond in cm?
3
=
4 marks
Solution:

æ 200 mg ö æ 1 g ö æ 1 cm3 ö
Volume of diamond = 3.45 carat ç
ç
÷ = 0.197 cm3
÷
ç
÷
è 1 carat ø è 1000 mg ø è 3.51 g ø
V=
4 3
pr
3
4 3
pr
3
r = 0.361 cm 
2r = 0.722 cm
0.197 cm3 =

– Sig figs and – DA
Question 4
A naturally occurring element consists of two isotopes. The data for the isotopes are listed below:
Isotope number:
1
2
Isotope mass
(amu)
68.5257
70.9429
Percentage
abundance
60.226%
x
4.1 Calculate the average atomic mass of this element.
4 marks
Solution:
Percentage abundance for isotope 2 = 100% − 60.226% = 39.774% 
Average atomic mass is then:
68.5257 amu ο‚΄ 0.60226 = 41.270 amu 
70.9429 amu ο‚΄ 0.39774 = 28.219 amu 
41.270 amu + 28.219 amu = 69.489 amu 
– Sig figs
4.2 Use the Periodic Table to identify the element: Gallium / Ga
(1)
4.3 Write the atomic symbol of the heaviest isotope using the 𝐴𝑍𝑋 notation: 71
31πΊπ‘Ž
(2)
Question 5
1 mark each = 8 marks
(a)
Name the following molecular compounds:
(i)
I2O5 diiodine pentoxide
(ii)
S2Cl2 disulphur dichloride
(b)
Write the formulas for the following:
(i)
potassium nitride
K3N
(ii)
selenium tetrachloride
SeCl4
(c)
Name the following ionic compounds:
(i)
CrCl3 οƒ— 4 H2O
chromium(III) chloride tetrahydrate
(ii)
Mn2O3
manganese(III) oxide
(d)
Write formulas for the following:
(i)
copper(II) cyanide
(ii)
potassium permanganate
Cu(CN)2
KMnO4