1.
a)
[𝑁𝐻3 ]2
𝐾= [𝑁 3
2 ][𝐻2 ]
b)
−𝐸
i) 𝐾 = 𝐾0 𝑒𝑥𝑝 ( 𝑅𝑇𝑎)
𝑇↓
𝑲↑
ii)
[𝐻2 ] ↑
𝑲 𝑵𝒐 𝒄𝒉𝒂𝒏𝒈𝒆
iii)
𝑃↓
𝑲 𝑵𝒐 𝒄𝒉𝒂𝒏𝒈𝒆
iv)
𝑲 𝑵𝒐 𝒄𝒉𝒂𝒏𝒈𝒆
v)
𝑉↓
𝑲 𝑵𝒐 𝒄𝒉𝒂𝒏𝒈𝒆
c)
i)
BLEND 1
1 𝑚𝑜𝑙𝑒
𝑚𝑜𝑙𝑒 𝑁𝑎𝐻𝐶𝑂3 = 6𝑔𝑁𝑎𝐻𝐶𝑂3 × 84.007𝑔 = 0.071
1 𝑚𝑜𝑙𝑒
𝑚𝑜𝑙𝑒 𝐶6 𝐻8 𝑂7 = 6𝑔𝐶6 𝐻8 𝑂7 × = 0.031
192.124𝑔
Limiting reagent
0.071
= 0.024 𝐿. 𝑅.
3
0.031
= 0.031
31
Then
3 𝑚𝑜𝑙𝑒𝐶𝑂2
𝑚𝑜𝑙𝑒 𝐶𝑂2 = 0.071 𝑚𝑜𝑙𝑒 𝑁𝑎𝐻𝐶𝑂3 × = 0.071𝑚𝑜𝑙𝑒 𝐶𝑂2
3𝑚𝑜𝑙𝑒 𝑁𝑎𝐻𝐶𝑂3
BLEND 2
𝑚𝑜𝑙𝑒 𝑁𝑎𝐻𝐶𝑂3 = 0.071
1 𝑚𝑜𝑙𝑒
𝑚𝑜𝑙𝑒 𝐶6 𝐻8 𝑂7 = 2𝑔𝐶6 𝐻8 𝑂7 × = 0.010
192.124𝑔
Limiting reagent
0.071
= 0.024
3
0.010
= 0.010 𝐿. 𝑅.
1
3 𝑚𝑜𝑙𝑒𝐶𝑂2
𝑚𝑜𝑙𝑒 𝐶𝑂2 = 0.010 𝑚𝑜𝑙𝑒 𝐶6 𝐻8 𝑂7 × = 0.030𝑚𝑜𝑙𝑒 𝐶𝑂2
3𝑚𝑜𝑙𝑒 𝐶6 𝐻8 𝑂7
Answer: Blend 1 generates more CO2.
ii)
44𝑔 𝐶𝑂2
0.071 𝑚𝑜𝑙𝑒𝐶𝑂2 = 3.124 𝑔 𝐶𝑂2
1 𝑚𝑜𝑙𝑒𝐶𝑂2
2.
a)
A tablet 𝐶9 𝐻8 𝑂4 = 300𝑚𝑔
V 𝐻2 𝑂 = 500 𝑚𝑙
4 tablets 𝐶9 𝐻8 𝑂4 = 4 × 300𝑔 = 1200𝑚𝑔
i)
𝑔 1200𝑚𝑔 1𝑔 𝑔
𝑚𝐿
= 500𝑚𝑙 × 1000𝑚𝑔 = 0.0024 ⁄𝑚𝑙
ii)
𝑚𝑜𝑙𝑒 𝑔 1000 𝑚𝑙 1𝑚𝑜𝑙𝑒 𝐶9 𝐻8 𝑂4
𝐿
= 0.0024 𝑚𝑙 × 1𝑙
× 180.16 𝑔
= 0.0133 𝑚𝑜𝑙𝑒⁄𝑙
b)
𝑚𝑜𝑙𝑒
[𝑁𝑎𝑂𝐶𝑙] = 0.0125
𝐿
i)
𝑔 0.0125𝑚𝑜𝑙𝑒 74.44𝑔 𝑔
𝑚𝐿
= 𝐿
× 1𝑚𝑜𝑙𝑒 = 0.9305 ⁄𝑙
ii)
𝑚𝑜𝑙𝑒 𝑁𝑎𝑂𝐶𝑙 74.44𝑔𝑁𝑎𝑂𝐶𝑙 1𝐿
𝑔𝑁𝑎𝑂𝐶𝑙 = 0.0125 𝐿
× 1 𝑚𝑜𝑙𝑒 𝑁𝑎𝑂𝐶𝑙 × 1000 𝑚𝑙 × 275 𝑚𝑙 𝑁𝑎𝑂𝐶𝑙
𝑔𝑁𝑎𝑂𝐶𝑙 = 0.256 𝑔
iii)
0.0125 × 𝑉 = 250 × 0.005
𝑉 = 100 𝑚𝐿
iv)
50 × 0.0125 = 𝐶 × 180
𝐶 = 0.00347𝑀
3.
a) The difference between nuclear radiation and electromagnetic radiation lies in their very
basic nature. Nuclear Radiations are high energy charged particles like alpha rays, beta
rays. Electromagnetic (EM) Radiations are form of energy fields, these are waves of
Electric field and Magnetic Field mutually perpendicular to each other. Thus, they
interact differently with the matter as compared to nuclear radiation.
Now one radiation which is nuclear and electromagnetic is Gamma Rays. A very high
energy electromagnetic radiation that can ionize the atoms and molecules thus a
property of nuclear radiation.
b)
Although Carbon-13 is a less abundant carbon isotope than Carbon-12 (1.07%), it is also
physically stable.
Subatomic particle Carbon-13 Carbon-12
protons 6 6
Neutrons 7 6
electrons 6 6
c) this is because the half-life is a logarithmic equation and only depends on the speed
constant.
For t1/2 = 68.9 years
K=0.01
Then,
For t 1/10 ,
𝑙𝑛10
𝑡1⁄ =
10 𝑘
𝑡1⁄ = 230 𝑦𝑒𝑎𝑟𝑠
10
4.
a)
This is because hydrochloric acid is a strong acid and completely dissociates, while acetic
acid is a weak acid and does not completely dissociate.
For strong acids:
𝑝𝐻 = −𝑙𝑜𝑔(0.1)
𝑝𝐻 = 1
For weak acids
− log(1.8 × 10−5 ) − 𝑙𝑜𝑔(0.1)
𝑝𝐻 =
2
𝑝𝐻 = 2.88
b)
𝑝𝐻 = 14 + 𝑙𝑜𝑔⌈0.0250⌉
𝑝𝐻 = 12.398
c)
i) Hydrofluoric acid is a weak acid due to the strong attraction between the
relatively small F- ion and solvated protons (H3O+), which does not allow it to
dissociate completely in water. Therefore, if we obtain HF in an aqueous
solution, we establish the following equilibrium with only slight dissociation
(Ka(HF) = 6.6x10-4, strongly favors reactants).
ii)
0.1
𝑝𝐻 = 3.17 + 𝑙𝑜𝑔 ( )
0.15
𝑝𝐻 = 2.99
iii) A buffer is made up of a weak acid and it’s conjugate base. The pH of a buffer,
any buffer, depends on the ratio of the concentration of the weak acid and it’s
conjugate base. Diluting the buffer by a factor of ten will dilute both the weak
acid concentration and the conjugate weak base concentration by 10. But that
means the ratio of the two concentrations will not change and neither will the
pH. Of course, there is a limit to how far you can take this. If the concentration
of the buffer is reduced so much that the natural pH of water dominates then
the pH will not remain constant.
iv) Diluting a buffer solution would decrease its buffer capacity. As we know, over
the working range of the buffer. pH changes very little with the addition of acid
or base. Once the buffering capacity is exceeded the rate of pH change quickly
jumps. This occurs because the conjugate acid or base has been depleted
through neutralization. This principle implies that a larger concentration of
conjugate acid or base will have a greater buffering capacity.
5.
a)
i)
1 2 3
hexan-3-ol
ii)
The reason why alcohols have a higher boiling point than alkanes is because the intermolecular
forces of alcohols are hydrogen bonds, unlike alkanes with van der Waals forces as their
intermolecular forces.
iii)
when a secondary alcohol is oxidized, it is converted to a ketone.
Hexan-3-ona
b)
i) It is an amine
ii) It is a tertiary amine; nitrogen has three organic substituents.
iii) N-ethyl-N,1-dimethyl-1-aminobutane
iv) Yes, Carbon number 1
c)
i) Originally, the chemical term “aromatic” referred to the aroma of benzene and
other derivatives of benzene. In the modern context, it refers to the unusually great
resonance stabilization of benzene and closely related derivatives of benzene and
the common reactivity patterns they exhibit as a result of this extraordinary
stabilization.
All protons and carbons are the same, so therefore each spectrum will only have
one signal each. There is a substantial delocalization of negative charge on the ring.
ii)
2-Ethylphenol
