Enthalpy of sublimation of iodine is `24 cal g^(-1)` at `200 ^(@)C`. If specific heat of `I_(2)(s)` and `I_(2)(vap)` are `0.055` and `0.0031 cal g^(-1)K^(-1)` respectively, then enthalpy of sublimation of iodine at `250 ^(@) C` in `cal g^(-1)` is :
Enthalpy of sublimation of iodine is "24 cal g"^(-1)" at " 200^(@)C . If specific heat of l_(2)(s) and l_(2) (vap) are 0.055 and 0.031 respectively, then enthalpy of sublimation of iodine at 250^(@)C in "cal g"^(-1) is:
Enthalpy of sublimation of I_(2)(s) at 200^(@)C is 24J/gm and specific heat of I_(2) (vapor) and I_(2)(s) are 0.013 J/gm-k and 0.045 J/gm-k respectively then enthalpy of sublimation of I_(2(s)) at 250^(@)C is:
The specific heats of iodine vapour and solid are 0.031 and 0.055 cal//g respectively. If heat of sublimation of iodinde is 24cal//g at 200^(@)C , what is its value at 250^(@)C ?
The enthalpy of formation of gaseous iodine is 62.5 kJ "mol"^(-1) . What is the enthalpy of sublimation of iodine at 25^(@)C ?
Calculate the amount of heat required in calorie to change 1 g of ice at -10^(@)C to steam at 120^(@)C . The entire process is carried out at atmospheric pressure. Specific heat of ice and water are 0.5 cal g^(-1) .^(@)C^(-1) and 1.0 cal g^(-1) .^(@)C^(-1) respectively. Latent heat of fusion of ice and vaporization of water are 80 cal g^(-1) and 540 cal g^(-1) respectively. Assume steam to be an ideal gas with its molecules having 6 degrees of freedom. Gas constant R = 2 cal mol^(-1) K^(-1) .
