Question

calculate the amount of heat envolve when 200cm3 of 0.2m HCL solution is mixed with 300cm3 of 0.1m NaOH solution​

Answer 1

Explanation:

Answer

500 cm

500 cm 3

500 cm 3 of 0.1 M HCl corresponds to 0.5×0.1=0.05 moles of hydrogen ions.

500 cm 3 of 0.1 M HCl corresponds to 0.5×0.1=0.05 moles of hydrogen ions.200 cm

500 cm 3 of 0.1 M HCl corresponds to 0.5×0.1=0.05 moles of hydrogen ions.200 cm 3

500 cm 3 of 0.1 M HCl corresponds to 0.5×0.1=0.05 moles of hydrogen ions.200 cm 3 of 0.2 M NaOH corresponds to 0.2×0.2=0.04 moles of hydroxide ions.

500 cm 3 of 0.1 M HCl corresponds to 0.5×0.1=0.05 moles of hydrogen ions.200 cm 3 of 0.2 M NaOH corresponds to 0.2×0.2=0.04 moles of hydroxide ions.sodium hydroxide is completely neutralize by Hcl

500 cm 3 of 0.1 M HCl corresponds to 0.5×0.1=0.05 moles of hydrogen ions.200 cm 3 of 0.2 M NaOH corresponds to 0.2×0.2=0.04 moles of hydroxide ions.sodium hydroxide is completely neutralize by HclThe enthalpy of neutralisation is 57.3 kJ/mol. For 0.04 moles, the enthalpy of neutralisation is 0.04×57.3=2.292 kJ

Answer 2

Answer:

hope it helps you

Explanation:

Answer

200 cm*3 of 0.2 M HCl corresponds to 0.2×0.2=0.04 moles of hydrogen ions.

300 cm*3 of 0.1 M NaOH corresponds to 0.3×0.1=0.01 moles of sodium hydroxide ions.

sodium hydroxide is completely neutralize by Hcl

The enthalpy of neutralisation is 57.3 kJ/mol. For 0.04 moles, the enthalpy of neutralisation is 0.04×57.3=2.292 kJ