Calculate the enthalpy change when `50 mL` of `0.01 M Ca(OH)_(2)` reacts with `25mL` of `0.01 M HCI`. Given that `DeltaH^(Theta)` neutralisaiton of strong acid and string base is `140 kcal mol^(-1)`

The enthalpy change when 50 mL of 0.01 M Ca(OH)_(2) reacts with 25 mL of 0.01 M HCl is 5 (x + 2) can Given that Delta H^(0) neutralisation of a strong acid and a strong base in 140 kCal mol^(-1) . What is x?

Calculate the enthalpy change when 50 mL of 0.01 M Ca(OH)_2 reacts with 25 mL of 0.01 M HCI. Given that DeltaH_N^@ , of a strong acid and strong base Is 57.1 kJ.

The volume of 0.1 M Ca(OH)_2 required to neutralize 10 mL of 0.1 N HCI

Calculate the PH of the resultant mixture : 10 mL of 0.2 M Ca(OH)_2 + 25 mL of 0.1 M HCl

Calculate the pH of resulting mixtures. 10" ml of " 0.2 M Ca (OH)_(2) + 25 " ml of " 0.1 M HCl

Calculate the pH of the resultant mixtures : 10mL of 0.2 M Ca(OH)_(2)+"25 mL of 0.1 M HCl"

Calculate the PH of the resultant mixture : 10 mL of 0.01 M H_2SO_4 +10 mL of 0.01 M Ca(OH)_2