Water gas is produced by
Passing steam over red hot coke
Passing steam and air over red hot.
Burning coke is excess air.
Burning coke is limited supply of air.
The correct descending order of the heat liberated (in kJ) during the neutralisation of the, acids CH3COOH (W), HF (X), HCOOH (Y) and HCN (Z) under identical conditions (Ka of CH3COOH = 1.8 × 10-5, HCOOH =1.8 × 10-10 HCN = 4.9 × 10-10 and HF = 3.2 × 10-4)
Y > X > Z > W
X > Y > W > Z
W > X > Y > Z
Z > W > Y > X
How many times a 0.1 M strong monobasic acid solution should be diluted so that pH of the resulting solution is tripled?
50
10
25
100
The equilibrium pressure for the given reaction, is /4 atm at 400 K. The Kp for the given reaction (in atm2) is
MSO4.2H2O (s) MSO4 (s) + 2H2O (g)
The low solubility of LiF and that of CsI in water are respectively due to which of the following properties of the alkali metal ions?
Higher hydration enthalpy of Li, higher lattice enthalpy of Cs+
Smaller hydration enthalpy of Li, higher lattice enthalpy of Cs+
Smaller lattice enthalpy of Li, higher hydration enthalpy of Cs+
Higher lattice enthalpy of Li, smaller hydration enthalpy of Cs+
D.
Higher lattice enthalpy of Li, smaller hydration enthalpy of Cs+
If hydration energy is greater as compare to lattice energy then the salt is soluble, and if hydration energy is lesser as compared to lattice energy, then the salt is insoluble.
LiF is almost insoluble due to high lattice energy. Both ions in LiF are very small. In this case, hydration energy is very less.
In case of CsI both ions are larger in size. As a result both the ions in CsI are less hydrated therefore, its hydration energy will decrease. Hence, it is insoluble in water.
The relative strength of trichlorides of boron group to accept a pair of electron is given by
GaCl3 > AlCl3 > BCl3
AlCl3 > BCl3 > GaCl3
AlCl3 > GaCl3 > BCl3
BCl3 > AlCl3 > GaCl3
If the ionic product of M (OH)2 is 5 × 10-10, then the molar solubility of M(OH)2 in 0.1 M NaOH is
5 × 10-12 M
5 × 10-8 M
5 × 10-10 M
5 × 10-9 M
Equilibrium constants are given for the following two equilibria.
(i) A2 (g) + B2 (g) 2AB (g); K = 2 × 10-4 L mol-1
(ii) 2AB (g) + C2 (g) 2ABC (g); K = 2 × 10-2 L mol-1
Calculate the equilibrium constant for the following equilibrium.
ABC (g) A2 (g) + B2 (g) + C2 (g)
500 mol1/2 L-1/2
500 mol1/2 L1/2
4 × 10-6 mol1/2 L1/2
200 mol1/2 L-1/2