The energy from 0.02 moles of butane is used to heat 328 grams of water. the temperature of the water rose from 298 k to 343 k. (the specific heat capacity of water is 4.18 j/k g.) what is the enthalpy of combustion? a. -61.7 kj b. 1,578.01 j c. 3,084,840.0 j d. 23,513,336 j

Answer : The correct option is, (C) 3,084,840.0 J

Explanation :

Formula used :

where,

= enthalpy of combustion = ?

m = mass of water = 328 g

= specific heat of water=

= final temperature = 343 K

= initial temperature = 298 K

n = moles of butane = 0.02 mole

Now put all the given values in the above formula, we get enthalpy of combustion.

Therefore, the enthalpy of combustion is, 3,084,840.0 J

i believe it is c.) 1575 j.

personal, she chose to do it.