When a 3.25 g sample of solid sodium hydroxide was dissolved in a calorimeter in 100.0 g of water, the temperature rose from 23.9 °c to 32.0 °c. calculate ∆h (in kj/mol) for the solution process: naoh (s) → na+ (aq) + oh- (aq)
use a calorimeter heat capacity of ccal = 15.8 j/°c

Answer : The enthalpy change for the solution is 42.8 kJ/mol

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

where,

q = heat released by the reaction

= heat absorbed by the calorimeter

= heat absorbed by the water

= specific heat of calorimeter =

= specific heat of water =

= mass of water = 100.0 g

= change in temperature =

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

       (1 kJ = 1000 J)

Now we have to calculate the enthalpy change for the solution.

where,

= enthalpy change = ?

q = heat released = 3.5138 kJ

m = mass of NaOH = 3.25 g

Molar mass of NaOH = 40 g/mole

Now,

Therefore, the enthalpy change for the solution is 42.8 kJ/mol

b) mutualism

hope this : )

a electronegativity decreases beacuse the size of nucleus is bigger.