When 1.960 g of titanium is combusted in a bomb calorimeter, the temperature of the calorimeter increases from 25.00 °c to 86.98 °c. in a separate experiment, the heat capacity of the calorimeter is measured to be 9.84 kj/k. the heat of reaction for the combustion of a mole of ti in this calorimeter is kj/mol.

Answer : The heat of combustion of one mole of Ti is -14897.4 kJ/mole

Explanation :

First we have to calculate the heat gained by the calorimeter.

where,

q = heat gained by calorimeter = ?

c = specific heat =

= final temperature =

= initial temperature =

Now put all the given values in the above formula, we get:

Heat released by the reaction = - Heat gained by calorimeter = -609.9 kJ

Now we have to calculate the heat of combustion of one mole of Ti.

To calculate the number of moles of Ti:

Molar mass of Ti = 47.87 g/mole

Number of moles Ti =

Heat of combustion of one mole of Ti =

Therefore, the heat of combustion of one mole of Ti is -14897.4 kJ/mole

c. because step up transformer steps up the input voltage i.e increases the input voltage  

the number of turns are greater on the secondary side of step up transformer  

transformer does not handle dc current

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :