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One mole of a substance is heated from 300 K to 400 K at constant pressure. The Cp of the substance is given by Cp (JK–1 mole–1) = 5 + 0.1 J. The change in enthalpy in JK–1mol–1 of the substance is (rounded up to first decimal place)
Correct answer is between '9.5,11.0'. Can you explain this answer?
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One mole of a substance is heated from 300 K to 400 K at constant pres...
^-1mol^-1) = 30 + 0.01T, where T is the temperature in Kelvin. Calculate the heat absorbed by the substance during this process.
To calculate the heat absorbed by the substance, we can use the equation:
ΔH = nCpΔT
Where ΔH is the heat absorbed, n is the number of moles of the substance, Cp is the specific heat capacity at constant pressure, and ΔT is the change in temperature.
In this case, we have:
n = 1 mole
Cp = 30 + 0.01T
ΔT = 400 K - 300 K = 100 K
Substituting these values into the equation, we get:
ΔH = (1 mol) x (30 + 0.01 x 300 K) x (100 K)
ΔH = 31 x 1000 J/mol
ΔH = 31 kJ/mol
Therefore, the heat absorbed by the substance during this process is 31 kJ/mol.
To calculate the heat absorbed by the substance, we can use the equation:
ΔH = nCpΔT
Where ΔH is the heat absorbed, n is the number of moles of the substance, Cp is the specific heat capacity at constant pressure, and ΔT is the change in temperature.
In this case, we have:
n = 1 mole
Cp = 30 + 0.01T
ΔT = 400 K - 300 K = 100 K
Substituting these values into the equation, we get:
ΔH = (1 mol) x (30 + 0.01 x 300 K) x (100 K)
ΔH = 31 x 1000 J/mol
ΔH = 31 kJ/mol
Therefore, the heat absorbed by the substance during this process is 31 kJ/mol.
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One mole of a substance is heated from 300 K to 400 K at constant pressure. The Cp of the substance is given by Cp (JK–1 mole–1) = 5 + 0.1 J. The change in enthalpy in JK–1mol–1 of the substance is(rounded up to first decimal place)Correct answer is between '9.5,11.0'. Can you explain this answer?
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One mole of a substance is heated from 300 K to 400 K at constant pressure. The Cp of the substance is given by Cp (JK–1 mole–1) = 5 + 0.1 J. The change in enthalpy in JK–1mol–1 of the substance is(rounded up to first decimal place)Correct answer is between '9.5,11.0'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about One mole of a substance is heated from 300 K to 400 K at constant pressure. The Cp of the substance is given by Cp (JK–1 mole–1) = 5 + 0.1 J. The change in enthalpy in JK–1mol–1 of the substance is(rounded up to first decimal place)Correct answer is between '9.5,11.0'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of a substance is heated from 300 K to 400 K at constant pressure. The Cp of the substance is given by Cp (JK–1 mole–1) = 5 + 0.1 J. The change in enthalpy in JK–1mol–1 of the substance is(rounded up to first decimal place)Correct answer is between '9.5,11.0'. Can you explain this answer?.
One mole of a substance is heated from 300 K to 400 K at constant pressure. The Cp of the substance is given by Cp (JK–1 mole–1) = 5 + 0.1 J. The change in enthalpy in JK–1mol–1 of the substance is(rounded up to first decimal place)Correct answer is between '9.5,11.0'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about One mole of a substance is heated from 300 K to 400 K at constant pressure. The Cp of the substance is given by Cp (JK–1 mole–1) = 5 + 0.1 J. The change in enthalpy in JK–1mol–1 of the substance is(rounded up to first decimal place)Correct answer is between '9.5,11.0'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of a substance is heated from 300 K to 400 K at constant pressure. The Cp of the substance is given by Cp (JK–1 mole–1) = 5 + 0.1 J. The change in enthalpy in JK–1mol–1 of the substance is(rounded up to first decimal place)Correct answer is between '9.5,11.0'. Can you explain this answer?.
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