What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm3 flask at 27 °C ?
Given,
Mass of carbon dioxide = 4.4 g
Molar mass of carbon dioxide= 44g/mol
Mass of methane = 3.2g
Molar mass of methane = 16g/mol
Now amount of methane, nCH4 = 3.2/ 16 = 0.2 mol
& amount of CO2 nCO2 = 4.4/ 44 = 0.1 mol
Also we know ,
Pv = (nCH4 + nCO2) RT
OR P X 9 = (0.2 +0.1) x 0.0821 x 300
Or p = 0.3 x 0.0821 x 300 / 9 = 0.821 atm
Hence, the total pressure exerted by the mixture is 0.821 atm
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(R = 0.083 bar dm3 K–1 mol–1).
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R = 0.083 bar L K–1 mol–1.
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(ii) CH3CH=CH2,
(iii) (CH3)2CO,
(iv) CH2=CHCN,
(v) C6H6
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Calculate the molecular mass of the following:
(i) H2O
(ii) CO2
(iii) CH4
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(a) NaH2PO4
(b) NaHSO4
(c) H4P2O7
(d) K2MnO4
(e) CaO2
(f) NaBH4
(g) H2S2O7
(h) KAl(SO4)2.12 H2O
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(ii) whose value is independent of path
(iii) used to determine pressure volume work
(iv) whose value depends on temperature only.
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a) What is the initial effect of the change on vapour pressure?
b) How do rates of evaporation and condensation change initially?
c) What happens when equilibrium is restored finally and what will be the final vapour pressure?
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(ii) zero
(iii) < 0
(iv) different for each element
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(ii) Δp = 0
(iii) q = 0
(iv) w = 0
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c) Cl2 (g) + 2NO2 (g) ↔ 2NO2Cl (g) Kc = 1.8
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