2017 - JAMB Chemistry Past Questions and Answers - page 5
41
A particle that contains a protons, 10 neutrons and 10 electrons is
A
positive ion
B
neutral atom of a metal
C
neutral atom of a non-metal
D
negative ion
correct option: d
protons = 9
neutrons = 10
electrons = 10
Electronic configuration = 2, 8
Ground state Electronic configuration=2, 7
It means that the atom has gained an electron thereby making it have a negative ion.
When an atom donates an electron, it becomes positively charged.
When an atom accepts an electron, it becomes negatively charged
Users' Answers & Commentsneutrons = 10
electrons = 10
Electronic configuration = 2, 8
Ground state Electronic configuration=2, 7
It means that the atom has gained an electron thereby making it have a negative ion.
When an atom donates an electron, it becomes positively charged.
When an atom accepts an electron, it becomes negatively charged
42
Ethene is prepared industrially by
A
Reforming
B
Polymerization
C
Distillation
D
Cracking
correct option: d
Ethene is produced from cracking which involves breaking up large hydrocarbon molecules into smaller and more useful bits. This is achieved by using high temperatures and pressures without a catalyst.
C15H32 → 2C2H4 + C3H6 + C8H14
......Δheat...ethene........propene.........octane
Users' Answers & CommentsC15H32 → 2C2H4 + C3H6 + C8H14
......Δheat...ethene........propene.........octane
43
Calculate the amount in moles of a gas which occupies 10.5 dm3 at 6 atm and 30oC [R = 082 atm dm3 K-1 mol-1]
A
2.536
B
1.623
C
4.736
D
0.394
correct option: a
For an ideal gas PV = nRT
Amount in moles = n
Volume v = 10.5dm3
Pressure P = 6atm
Temperature T = 30°C + 273 = 303k
R, Gas constant = 0.082 atmdm3k-1 mol
Recall from ideal gas equation
pv = nRT
n = \(\frac{RV}{RT}\)
n = \(\frac{6 \times 105}{0.082 \times 303}\)
n= 2.536mol
Users' Answers & CommentsAmount in moles = n
Volume v = 10.5dm3
Pressure P = 6atm
Temperature T = 30°C + 273 = 303k
R, Gas constant = 0.082 atmdm3k-1 mol
Recall from ideal gas equation
pv = nRT
n = \(\frac{RV}{RT}\)
n = \(\frac{6 \times 105}{0.082 \times 303}\)
n= 2.536mol
44
If 100cm3 of oxygen pass through a porous plug is 50 seconds, the time taken for the same volume of hydrogen to pass through the same porous plug is? [O = 16, H = 1]
A
10.0s
B
12.5s
C
17.7s
D
32.0s
correct option: b
Rate of diffusion \(\frac{\alpha_1}{Density}\) or \(\frac{1}{\sqrt{Mm}}\)
Rate = \(\frac{1}{time}\)
\(\frac{1}{time}\) \(\frac{\alpha_1}{Density}\) or \(\frac{1}{\sqrt{Molarmass}}\)
Time \(\alpha \sqrt{Density}\) or \(\sqrt{Molarmass}\)
At constant volume of 100cm3
\(\frac{t_{o2}}{\sqrt{Mn_{o2}}}\) = \(\frac{t_{n2}}{\sqrt{1 \times 2}}\)
\(\frac{50}{4 \sqrt{2}}\) = \(\frac{t_{n2}}{\sqrt{2}}\)
tn2= 12.5s
Users' Answers & CommentsRate = \(\frac{1}{time}\)
\(\frac{1}{time}\) \(\frac{\alpha_1}{Density}\) or \(\frac{1}{\sqrt{Molarmass}}\)
Time \(\alpha \sqrt{Density}\) or \(\sqrt{Molarmass}\)
At constant volume of 100cm3
\(\frac{t_{o2}}{\sqrt{Mn_{o2}}}\) = \(\frac{t_{n2}}{\sqrt{1 \times 2}}\)
\(\frac{50}{4 \sqrt{2}}\) = \(\frac{t_{n2}}{\sqrt{2}}\)
tn2= 12.5s
45
Due to the high reactivity of sodium, it is usually stored under
A
water
B
mercury
C
paraffin
D
phenol
correct option: c
Na is kept under kerosene (paraffin) to avoid reactivity with air.
Paraffin is also known as Kerosene.
Na(sodium) is kept in kerosene to prevent it from coming in contact with oxygen and moisture. If this happens, it will react with the moisture present in air and form sodium hydroxide.
Users' Answers & CommentsParaffin is also known as Kerosene.
Na(sodium) is kept in kerosene to prevent it from coming in contact with oxygen and moisture. If this happens, it will react with the moisture present in air and form sodium hydroxide.
46
The furring of kettles is caused by the presence in water of
A
calcium hydrogentrioxocarbonate (IV)
B
calcium trioxocarbonate (IV)
C
calcium tetraoxosulphate (VI)
D
calcium hydroxide
correct option: b
Furring of kettles is caused by the temporary hardness in water. Temporary hardness in water is caused by calcium and magnesium trioxocarbonate (IV)
CaCO3 causes the furring of kettles
Users' Answers & CommentsCaCO3 causes the furring of kettles
47
Water for town supply is chlorinate to make it free from
A
bad colour
B
bacteria
C
temporary hardness
D
permanent hardness
48
Ca(OH)2(s) + 2NH4Cl(g) → CaCl2(s) + 2H2O2(l) + X. In the reaction above X is
A
NO2
B
NH3
C
N2O
D
NO2
correct option: b
Balancing the chemical equation.
Ca[OH]2 + 2NH4Cl → CaCl2 + 2H2O + 2NH3
X = NH3
Users' Answers & CommentsCa[OH]2 + 2NH4Cl → CaCl2 + 2H2O + 2NH3
X = NH3
49
X(g) + 3Y(g) ---- 2z(g) H = +ve. if the reaction above takes place at room temperature, the G will be
A
negative
B
zero
C
positive
D
indeterminate
correct option: d
\(\begin{array}{c|c}
\text{Enthalpy Change [ΔH]} & \text{Entropy Change [ΔS]} & \text{Gibbs free Energy[ΔG]} \\ \hline
\text{Positive} & \text{Positive} & \text{depends on T, may be + or -} \ \hline \text{Negative} & \text{Positive} & \text{always negative} \ \hline \text{Negative} & \text{Negative} & \text{depends on T, may be + or -} \ \hline \text{Positive} & \text{Negative} & \text{always positive} \end{array}\)
ΔG= ΔH − TΔS.
To determine whether ΔG will be positive or negative, the value of ΔH(change in enthalpy) and ΔS (change in entropy) must be given. Likewise the temperature.
Users' Answers & Comments\text{Enthalpy Change [ΔH]} & \text{Entropy Change [ΔS]} & \text{Gibbs free Energy[ΔG]} \\ \hline
\text{Positive} & \text{Positive} & \text{depends on T, may be + or -} \ \hline \text{Negative} & \text{Positive} & \text{always negative} \ \hline \text{Negative} & \text{Negative} & \text{depends on T, may be + or -} \ \hline \text{Positive} & \text{Negative} & \text{always positive} \end{array}\)
ΔG= ΔH − TΔS.
To determine whether ΔG will be positive or negative, the value of ΔH(change in enthalpy) and ΔS (change in entropy) must be given. Likewise the temperature.