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.