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    				| 1. | 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)`A. `14 cal`B. `35 cal`C. `10 cal`D. `7.5 cal` | 
| Answer» Correct Answer - 2 `m.eq.` of base `=50xx0.01xx2=1` `m.eq.` of acid `=25xx0.01" "0.25` `:. Me.eq.` of base reacted `=0.25` `:.` energy released `=(140)/(1000)xx0.25Kcal=(140xx0.25xx1000)/(1000)cal=35cal`. | |