Explanation:
Let us proceed with a reaction
A(aq)+B(aq) C(aq) + D(g)
Taking A, B and C are in aqueous solution. D being a gas. The reaction is carried out in a close vessel. As reaction proceeds, A and B react to give C and D. As the concentration of C and D increases, they react to give back A and B. Now suppose the reaction is carried out in a open vessel so that the gas D can escape in the atmosphere. So the gas D is no longer available to react with C. So reverse reaction will not occur. The forward reaction is forced to reach 100 %
Let us take another example.
A(aq)+B(aq) C(aq) + D(aq)
Here, D is a volatile liquid. As soon as D is formed, it is removed by distillation. So D is no longer available to react with C. So reverse reaction will not occur. The forward reaction is forced to reach 100 %
To answer this question Kc of this reaction OR the conc of reactants nd products at equilibria is required.
Explanation:
Answer:
CuCl2-Ion-dipole forces
CuSO4-Ion-dipole forces
NH3-Dipole-dipole forces
CH3OH-Dipole-dipole forces
Explanation:
Water consists of a dipole. The water molecule contains a positive end and a negative end. The positive ion attracts the negative dipole of water while the positive dipole in water interacts with the negative ion of an ionic substance. This explains the dissolution of ionic substances in water.
Copper II chloride and copper sulphate are ionic substances hence they dissolve by the mechanism described above.
Molecules consisting of dipoles dissolves by interaction of the molecule's dipoles with the dipoles in water. For example, methanol interacts with water through hydrogen bonding which is involves molecular dipoles
Answer:
The molar heat capacity of C₂H₆O = 2.797 J/g·°C
Explanation:
The known parameters are first listed as follows from reading the question
Heat transferred, H = 1328 J
Mass of the heated ethanol, m = 40.4 g
Increase in temperature of the heated gas = ΔT = 13.4 °C
Molar mass of ethanol = 46.07 g/mol
Number of moles of ethanol = 40.4/46.07 = 0.877 moles
The heat required to raise the temperature of 40.4 g of ethanol by 13.4 °C is 1328 J
Therefore we have H = m·c·ΔT
c = specif heat capacity = H ÷ (m·ΔT) = = 2.453 J/g·°C
Therefore the heat required to raise the temperature of one mole of ethanol by one degree C = J/g·°C = 2.797 J/g·°C
=