2008 - JAMB Chemistry Past Questions and Answers - page 2
11
\( _{11} ^{11}Na + _0^1n → _{11}^{24}Na \)
The reaction above is an example of
The reaction above is an example of
A
nuclear fission
B
nuclear fusion
C
artificial transmutation
D
beta decay
correct option: c
Users' Answers & Comments12
Air boiled out of water as steam is richer in
A
nitrogen and oxygen
B
carbon (IV) oxide and oxygen
C
noble gases and carbon (IV) oxide
D
oxygen and noble gases
correct option: b
Users' Answers & Comments13
In the course of purifying water for town supply the water is passed through large settlings tanks, containing sodium aluminate (III) to remove
A
large particles
B
germs
C
fine particles
D
odour
correct option: a
Users' Answers & Comments14
When a few drops of water is added to a blue anhydrous cobalt (II) chloride, the colour changes to
A
white
B
pink
C
red
D
blue
correct option: b
Users' Answers & Comments15
The vulcanizer's solution is prepared by dissolving rubber in
A
ethanol
B
kerosine
C
benzene
D
petrol
correct option: c
Users' Answers & Comments16
117.0g of sodium chloride was dissolved in the solubility in 1.0dm3 of distilled water at 25oC. Determine the solubility in mol dm-3 of sodium chloride at that temperature
[Na = 23, Cl = 35.5]
[Na = 23, Cl = 35.5]
A
1.0
B
2.0
C
3.0
D
4.0
correct option: b
Relative molecular mass of NaCl = 23 + 35.5 = 55.5 g/mol
No. of mole of NaCl = 117/55.5 = 2.1 mol
Users' Answers & CommentsNo. of mole of NaCl = 117/55.5 = 2.1 mol
17
The uncovered raw food that is sold along major roads is likely to contain some amounts of
A
Pb
B
Cu
C
Ag
D
Na
correct option: a
Users' Answers & Comments18
What is the pH of 0.001 mol dm-3 solution of sodium hydroxide
A
14
B
13
C
12
D
11
correct option: d
[H+] [OH-] = 10-14
[H+] = 10-14/[OH-] = 10-14/(1 * 10-3) = 10-11
pH = -Log10[H+]
= - Log10[10-11] = - 1 * -11 Log10[10-11] = 1 = -1 * -11 * 1 = 11
Users' Answers & Comments[H+] = 10-14/[OH-] = 10-14/(1 * 10-3) = 10-11
pH = -Log10[H+]
= - Log10[10-11] = - 1 * -11 Log10[10-11] = 1 = -1 * -11 * 1 = 11
20
0.05 mol dm-3 HCl is neutralized by 25 cm3 NaOH. If the volume of acid used is 32.00cm3, what is the concentration of the base
A
0.016 mol dm-3
B
0.032 mol dm-3
C
0.039 mol dm-3
D
0.064 mol dm-3
correct option: d
HCl + NaOH → NaCl + H2O
CAVA = NA/NB = (0.05 * 32)/(CB * 25)
∴ CB = (0.05 * 32)/25 = 0.064 mol dm-3
Users' Answers & CommentsCAVA = NA/NB = (0.05 * 32)/(CB * 25)
∴ CB = (0.05 * 32)/25 = 0.064 mol dm-3