Learn more about the properties, composition, and structure of substances (elements and compounds) with these Chemistry questions and answers. This Test can be used by students preparing for Chemistry in JAMB, WAEC, NECO or Post UTME.
4:1
2:1
1:2
1:4
Correct answer is C
No explanation has been provided for this answer.
40.0 cm3
35.7 cm3
28.4 cm3
25.2 cm3
Correct answer is C
Y1 = 30cm3 T1 = 27 + 273 = 300K and P1 = (780 - 10)mm Hg i.e the pressure minus the vapour pressure of water
P2 = 760mm Hg and T2 = 7 + 273 = 280K
From (P2V2) / (T2) = (P1V1) / (T1)
(770 x 30) / (300) = (760 x O2) / (280)
V2 = (770 x 30x 280) / (300 x 760)
V2 = 28.4 cm3
250 cm3
150 cm3
100 cm 3
50 cm3
Correct answer is A
C3H8 + 502 → 4H2O + 3CO2.
From the equation above 1 dm3 of C3H8 require 5dm3 of O2 at s.t.p.
∴ 50cm3 of C3H8 require x = (50) / (1000) x 5 x 1
x = (250) / (1000)
x = 250cm
X(s) + CuSO4(aq) → Cu(s) + XSO4(aq)
X(s) + 2CuSO4(aq) → 2Cu(s) + X(SO4)2(aq)
2X(s) + CuSO4(aq) → Cu(s) + X2SO4(aq)
2X(s) + 3CuSO4(aq) → 3Cu(s) + X2(SO4)3(aq)
Correct answer is A
No explanation has been provided for this answer.
0.89 mol
1.90 mol
3.80 mol
5.70 mol
Correct answer is A
\(V_1\) = \(10dm^3\), \(V_2\) = ?, \(P_1\) = 4atm = 4 x 760 = 3040mmHg, \(P_2\) = 760mmHg \(T_1\) = 273ºc = 273 + 273 = 546K, \(T_2\) = 273K
Using the general gas law = \(\frac{P_1 V_1}{ T_1} = \frac{P_2 V_2}{T_2}\)
\(V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}\)
\(V_2 = \frac{ 3040 \times10 \times273}{760 \times546}\)
\(V_2 = 20dm^{-3}\)
1mole of a gas = \(22. 4dm^{3}\)
x = \(20.0dm^{3}\)
= 0.89mol.