Class 11th
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New answer posted
6 months agoContributor-Level 10
Using the equation Kp = Kc (RT)ng
(i) From the given equation, Δng = 3 – 2 = 1,
R = 0.0821 L atm K-1 mol-1
T = 500 K, Kp= 1.8 * 10–2
Thus, Kc = Kp / (RT)ng= (1.8 x 10-2) / (0.0821 L atm K-1 mol-1 x 500 K)
4.4 x 10-4 mol L-1
(ii) From the given equation, Δng =1,
R = 0.0821 L atm K-1 mol-1
T = 1073 K, Kp= 167 atm
Thus, Kc = Kp / (RT)ng= (167 atm) / (0.0821 L atm K-1 mol-1 x 1073 K)
= 1.9 mol L-1
New answer posted
6 months agoContributor-Level 10
(i) The expression for the equilibrium constant is Kc= [NO]2 [Cl2] / [NOCl]2.
(ii) The expression for the equilibrium constant is Kc= [NO2]4 [O2] / [ (2Cu (NO3)2]2 = [NO2]4 [O2].
(iii) The expression for the equilibrium constant is Kc= [CH3COOH] [C2H5OH] / [CH3COOC2H5].
(iv) The expression for the equilibrium constant is Kc= 1 / [Fe3+] [OH]3.
(v) The expression for the equilibrium constant is Kc= [IF5]2 / [F2]5.
New answer posted
6 months agoContributor-Level 10
According to the given condition,
Total pressure of equilibrium mixture = 105 Pa
Partial pressure of iodine atoms (I), PI = 40 % of 105 Pa
= 0.4 x 105 Pa
Partial pressure of iodine molecules (I2), PI2 = 60 % of 105 Pa
= 0.6 x 105 Pa
Kp for the equilibrium = (PI)2/ PI2 = (0.4 x 105 Pa)2 / (0.6 x 105 Pa)
= 2.67 x 104 Pa
New answer posted
6 months agoContributor-Level 10
(a) On increasing the volume of the container, the vapour pressure will initially decrease because the same amount of vapours is now distributed over a larger space.
(b) On increasing the volume of the container, the rate of evaporation will increase initially because now more space is available. Since the amount of the vapours per unit volume decrease on increasing the volume, therefore, the rate of condensation will decrease initially.
(c) Finally, equilibrium will be restored when the rates of the forward and backward processes become equal. However, the vapour pressure will remain unchanged because it depends upon the tempera
New answer posted
6 months agoContributor-Level 10
5.7. Applying PV = nRT
We get P = nRT / V
Given: nCH4 = 3.2 / 16 mol = 0.2 mol
nCO2 = 4.4 /44 mol = 0.1 mol
So,
PCH4 = (0.2 mol) (0.0821 dm3atm K-1 mol-1) (300 K) / (9 dm3)
= 0.55 atm
PCO2= (0.1 mol) (0.0821 dm3atm K-1 mol-1) (300 K) / (9 dm3)
= 0.27 atm
Ptotal =
New answer posted
6 months agoContributor-Level 10
5.6. The chemical equation for the reaction is
2 Al + 2 NaOH + H2O? 2 NaAlO2 + 3H2
i.e. 2 moles of Al give 3 moles of H2 gas.
Moles of aluminium = 270.15g = 5.56*10?3 moles
Moles of H2 produced= 23*5.56*10?3
= 8.33*10?3 moles
Volume of H2 = nRT / P
&nbs
New answer posted
6 months agoContributor-Level 10
5.5. Suppose molecular masses of A and B are MA and MB respectively. Then their number of moles will be
nA= 1/MA nB = 2/MB
Given: PA = 2 bar and PA + PB = 3 bar
=> PB = 1bar
Since, PV = nRT
PA= nART and PBV= nBRT
Therefore, (PA / PB) = (nA / nB)= (MB / 2MA)
=> MB / MA = 2 x PA / PB = 2 x 2 /1 = 4
=> MB = 4 MA
New answer posted
6 months agoContributor-Level 10
Answer: According to ideal gas equation
PV = nRT
Or P= nRT/V
Replacing n by m/M, we get
P M1 x 2 = 28 x 5 (Molecular mass of N2 = 28 g/mol)
or
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6 months agoNew question posted
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