Heat capacity of water is `18 cal-"degree"^(-1)-mol^(-1)`. The quantity of heat needed to rise temperature of 18 g water by `0.2^(@)C` is x cal. Then amount of `CH_(4(g))` to be burnt to produce X cal heat is `(CH_(4)+2O_(2) rarr CO_(2)+2H_(2)O, Delta H = -200 K.Cal)`A. `1.8 xx 10^(-3) mol`B. `3.6 xx 10^(-5) mol`C. `0.0288 g`D. `0.288 mg`

Heat capacity of water is `18 cal-"degree"^(-1)-mol^(-1)`. The quantit

1.	Heat capacity of water is `18 cal-"degree"^(-1)-mol^(-1)`. The quantity of heat needed to rise temperature of 18 g water by `0.2^(@)C` is x cal. Then amount of `CH_(4(g))` to be burnt to produce X cal heat is `(CH_(4)+2O_(2) rarr CO_(2)+2H_(2)O, Delta H = -200 K.Cal)`A. `1.8 xx 10^(-3) mol`B. `3.6 xx 10^(-5) mol`C. `0.0288 g`D. `0.288 mg`
Answer» Correct Answer - D `Delta U = Zx theta(M)/(W), delta H = Delta U + Delta nRT, Q = mst`

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