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CHE1501 Assignment 3 2020
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- University Of South Africa (Unisa)
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CHE1501 SEMESTER 1UNIQUE NUMBER: 614235 ASSIGNMENT 3
55167764
STUDENT NUMBER:
COURSE CODE: CHE1501
ASSIGNMENT NUMBER: 3
UNIQUE NUMBER: 614235
DUE DATE: 8 May 2020
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, CHE1501 SEMESTER 1
UNIQUE NUMBER: 614235 ASSIGNMENT 3
1. 1a) Chemical.
Chemical changes are characterized by the formation of new chemical species.
Photosynthesis is represented as 12H20 + 6CO2 + Light = C6H12. This shows us
that water, carbon dioxide and light are combined to form glucose and oxygen.
Since new chemical species are formed (glucose and oxygen), photosynthesis is
clearly a chemical change.
1b) Physical.
Physical changes are those where new substances are not formed and the
changes are, generally visible. Boiling, freezing and melting are all examples of
physical changes. H20 in the gas phase, water vapour, becomes H20 in the solid
phase, frost. This shows us that no new substance was formed, thus the changes
made where physical.
1c) Physical.
Physical changes are those where new substances are not formed and the
changes are, generally visible. Boiling, freezing and melting are all examples of
physical changes. Gold in its solid state is melted to become a liquid, the gold is
then pulled into a wire and re-solidifies. This shows us that no new substances
were formed, thus the changes made were physical.
2. Bromine
Symbol – Br
Period – 3
Group – 7 / Halogen Group – 17
Atomic number – 35
Bromine is a non-metal
3. This is an example of ionic bonding. In order to achieve noble gas configurations, the
magnesium atom needs to lose its two valence electrons, while the bromine atom,
which has 7 valence electrons, requires one additional electron to fill its outer shell.
Therefor for the resulting compound to be neutral, two bromine anions must
combine with one magnesium cation to form an ionic bond and thus magnesium
bromide (MgBr2) is formed.
4. The element is Rhodium (Rh)
[Kr] 4d⁶ ion has 42 electrons. As it is a 3+ ion, the parent atom would have 45
electrons. Therefore the element is Rhodium (Rh)
5. CO2 is not an acid itself, since it does not have hydrogen ions (H+), however it does
become an acid when it comes in contact with water. The CO2 dissolves in water to
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, CHE1501 SEMESTER 1
UNIQUE NUMBER: 614235 ASSIGNMENT 3
become a carbonic acid. Carbonic acid (H2CO3) is a weak acid capable of splitting
with its Hydrogen ions.
CO2 + H2O → H2CO3
As some of the carbon dioxide dissolves in the water it forms carbonic acid, a weak
acid that acidifies the solution. Normal rainwater is slightly acidic, with a pH between
7 and 5.6 (because of the presence of dissolved carbon dioxide.)
6. Sulphur has an electronegativity of 2.58 Oxygen has an electronegativity of 3.44.
The electronegativity difference is therefore 0.86
Therefore the electronegativity difference indicates polar covalent bonding. Sulphur
is positively charged and oxygen is negatively charged.
7. The unbalanced chemical equation for the decomposition of nitroglycerin is:
C₃H₅N₃O₉ (l) N2 (g) + CO₂ (g) + H₂O (l) + O₂ (g)
In the past (until the late 1800s), nitroglycerin was used as the main explosive in
construction. However, it is highly sensitive to heat and shock. Dropping or shaking
the bottle of nitroglycerin would cause it to explode. Historically, it resulted in many
deaths over the years, including Alfred Nobel’s brother. Alfred Nobel discovered that
mixing nitroglycerin with silica (diatomaceous earth) would turn the dangerous
liquid into a safe, pliable paste, called dynamite. This paste does not explode due to
shock, but needs detonation. This revolutionized the use of explosives in
construction and reduced the number of consequential deaths.
8. a) HO + O CO + H₂O
₂ ₂
3.56 g ? 2.14g
9.36g
To work
out Carbon we use CO₂ (9,39g) m(CO₂) = 1(2,01)
+ 2(16,00)
= 44,01 g/mol
n(CO₂) = m/m
= 9,39/44,01
= 0,213 mol
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, CHE1501 SEMESTER 1
UNIQUE NUMBER: 614235 ASSIGNMENT 3
0,213 mol CO₂ x 1mol C/ 1 mol CO₂ = 0,213 mol C
؞mass of C (Mc)
n = m/M
m = nM m =
(0,213)(12,01)
(MC)m = 2,562g
To work out Hydrogen we use H₂O (2,14g)
m(H₂O) = 2(1,01) x 1(16,00)
= 18,016 g/mol
n(H₂O) = m/M
= 2,14/18,016
= 0,119 mol
0,119 mol H₂O x 2mol H/ 1 mol H₂O = 0,238 mol H
؞mass of C (Mc)
n = m/M m = nM m =
(0,238)(1,01)
(MH)m = 0,240g
To work out Oxygen (O)
MTotal = MC + MH + MO
3,56 = 2,562 + 0,240 + MO
MO = 3.56 – 2,562 – 0,240
= 0,758g of Oxygen
n= M/M = 0,758/16,00 = 0,047 mol
C n = 0,213 H n = 0,238 O n = 0,047
mol mol mol
= 0,213/0,047 = 0,238/0,047 = 0,047/0,047
= 4,5 =5 =1
n = 4,5 x 2 n=5x2 N=1x2
=9 =10 =2
؞The empirical formula is C₉H₁₀O₂
8. b)
Empirical Molecular
C₉H₁₀O₂ C₉(R)H₁₀(R)O₂(R)
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