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ΔH for CaCO3(s) → CaO(s) + CO2(g) is 176 kJ mol-1 at 1240 K. The ΔU for the change is equal to :
  • a)
    160 kj
  • b)
    165.6 kj
  • c)
    186.3 kj
  • d)
    180.0 kj
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
ΔH for CaCO3(s) →CaO(s)+ CO2(g)is 176 kJ mol-1at 1240 K. Th...
∆H = ∆E + ∆ngRT
176 = ∆E + 1×8.314×1240/1000
∆E = 176 - 10.30 = 165.69 kJ
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ΔH for CaCO3(s) →CaO(s)+ CO2(g)is 176 kJ mol-1at 1240 K. Th...
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ΔH for CaCO3(s) →CaO(s)+ CO2(g)is 176 kJ mol-1at 1240 K. The ΔU for the change is equal to :a)160 kjb)165.6 kjc)186.3 kjd)180.0 kjCorrect answer is option 'B'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about ΔH for CaCO3(s) →CaO(s)+ CO2(g)is 176 kJ mol-1at 1240 K. The ΔU for the change is equal to :a)160 kjb)165.6 kjc)186.3 kjd)180.0 kjCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ΔH for CaCO3(s) →CaO(s)+ CO2(g)is 176 kJ mol-1at 1240 K. The ΔU for the change is equal to :a)160 kjb)165.6 kjc)186.3 kjd)180.0 kjCorrect answer is option 'B'. Can you explain this answer?.
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