If the enthalpy change for the combustion 2.00 g of propane, c3h8, was determined to be −102.5 kj by calorimetry, what is the ca
2 answers:
Answer:- -2260 kj.
Solution:- It could easily be solved using dimensional analysis. From given information, the enthalpy change for the combustion of 2.00 grams of propane is -102.5 kj. It asks to calculate the enthalpy change in same conditions for the combustion of 1 mole of propane.
We convert 1 mole of propane to grams using molar mass(44.1 gram per mol) and then calculate the heat using the given information as:
= -2260 kj
Hence, the enthalpy change for the combustion of 1 mol of propane is -2260 kj.
<h2>The molar enthalpy of combustion of propane is -2260 kJ. </h2>
FURTHER EXPLANATION
The molar enthalpy of combustion is the amount of heat released by burning one mole of the substance. It can be determined from the heat of combustion obtained through calorimetry. The following conversion factors may be used:
2.00 g C₃H₈ = -102.5 kJ
1 mol C₃H₈ = 44.097 g
To solve the molar enthalpy of combustion, the following dimensional analysis must be setup:
The least number of significant figures in the given is 3 (2.00 g) so the final answer must also have the 3 significant figures. Hence,
This answer is negative because combustion reactions are exothermic. The magnitude of the molar enthalpy is greater than the sample's (given) because the propane used in calorimetry is much less than 1 mole. Therefore, the final answer is logical.
LEARN MORE
- Learn more about heat of vaporization brainly.com/question/12984871
- Learn more about specific heat capacity brainly.com/question/12976104
- Learn more about calorimetry brainly.com/question/8168263
Keywords: enthalpy, calorimetry, heat of combustion
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Answer:
Explanation:
Given the amount of heat absorbed and the amount of substance in moles, we may calculate the heat of vaporization. Heat of vaporization is defined as the amount of heat per 1 mole of substance required to evaporate that specific substance.
Based on the value of heat of vaporization, we will identify the substance. Firstly, let's calculate the heat of vaporization:
Secondly, let's use any table for heat of vaporization values for substances. We identify that the heat of vaporization of is 18.7 kJ/mol
To compute the energy, Q, needed to melt a certain amount of ice, n, in moles, we have
where the latent heat of fusion for ice is equal to 6.009 kJ/mol.
Now, since we have a 20.0 lb ice, we must first convert its mass to grams. Thus mass = (20.0)(1000)(0.4536) = 9072 g.
To find the number of moles present, we must recall that the molar mass of water (ice) is equal to <span>1.00794(2) + </span><span>15.9994 </span>≈ 18.01 g/mol. Hence, we have
Now, to compute for the molar heat of fusion, Q,
Therefore, the amount of heat needed to melt the 20-lb bag of ice is equal to 3026.9 kJ.
Answer: 3026.9 kJ
where the latent heat of fusion for ice is equal to 6.009 kJ/mol.
Now, since we have a 20.0 lb ice, we must first convert its mass to grams. Thus mass = (20.0)(1000)(0.4536) = 9072 g.
To find the number of moles present, we must recall that the molar mass of water (ice) is equal to <span>1.00794(2) + </span><span>15.9994 </span>≈ 18.01 g/mol. Hence, we have
Now, to compute for the molar heat of fusion, Q,
Therefore, the amount of heat needed to melt the 20-lb bag of ice is equal to 3026.9 kJ.
Answer: 3026.9 kJ