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almond37 [142]
3 years ago
11

In a reaction, gaseous reactants form a liquid product. The heat absorbed by the surroundings is 1.1 MJ, and the work done on th

e system is 13.2 kcal. Calculate ΔE (in kJ). Be sure to include the correct sign (+/-). Enter to 0 decimal places.
Chemistry
1 answer:
ratelena [41]3 years ago
4 0

Answer: \Delta E is 1155 kJ

Explanation:

According to first law of thermodynamics:

\Delta E=q+w

\Delta E=Change in internal energy

q = heat absorbed or released

w = work done or by the system

w = work done by the system=-P\Delta V  {Work done on the system is positive as the final volume is lesser than initial volume}

w =13.2kcal=55.2kJ   (1kcal=4.184 kJ)

q = +1.1 MJ = 1100 kJ  (1MJ=1000kJ)   {Heat absorbed by the system is positive}

\Delta E=1100kJ+(55.2)kJ=1155kJ

Thus \Delta E is 1155 kJ

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