1.

A cylinder of gas is assumed to contain 11.6 kg of butane (C_4H_10). If a normal family needs 20000kJ ofenergy per day, then the cylinder will last for : (Given that ΔH for combustion of butane is – 2658kJ)

Answer»

20 DAYS
18 days
26.6 days
32 days

SOLUTION :`DeltaH_"released"=(11.6xx10^3)/58xx2658` KJ
No of DAY =`(11.6xx10^3xx2658)/(58xx20000)`
=26.58
`approx 26.6`


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