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Sagot :
To determine how much energy is consumed by thawing 4.3 grams of ice, we need to understand the underlying process of phase change from solid (ice) to liquid water. The energy required for this phase change is described by the enthalpy of fusion (ΔH_fusion).
Given:
- The enthalpy of fusion (ΔH_fusion) is 6.03 kJ/mol.
- The molar mass of water is 18.02 g/mol.
- The mass of ice is 4.3 g.
### Step-by-Step Solution:
1. Calculate the number of moles of ice:
[tex]\[ \text{Number of moles of ice} = \frac{\text{mass of ice}}{\text{molar mass of water}} = \frac{4.3 \, g}{18.02 \, g/mol} \approx 0.2386 \, \text{moles} \][/tex]
2. Calculate the energy consumed for the thawing process:
[tex]\[ \text{Energy consumed} = \text{number of moles of ice} \times \Delta H_{\text{fusion}} = 0.2386 \, \text{moles} \times 6.03 \, \text{kJ/mol} \approx 1.4389 \, \text{kJ} \][/tex]
### Conclusion:
The correct answer is D) 1.4389 kJ, which means it is a practically correct numerical depiction of the energy consumed by thawing 4.3 grams of ice. The closest option is indeed not listed explicitly among choices A, B, or C. However, if you were to choose from those that involved using the fusion energy and molar mass conversion, by further simplifying the accurate path considered:
A. [tex]\( \approx 4.3 \,g \times 1 \, mol/18.02 \, g \times 6.03 \,kJ/mol = 1.4389 \,kJ\)[/tex].
Thereby, detailed computations as portrayed verified how the standard listed operations evaluate correctly, following the detailed format shown.
Given:
- The enthalpy of fusion (ΔH_fusion) is 6.03 kJ/mol.
- The molar mass of water is 18.02 g/mol.
- The mass of ice is 4.3 g.
### Step-by-Step Solution:
1. Calculate the number of moles of ice:
[tex]\[ \text{Number of moles of ice} = \frac{\text{mass of ice}}{\text{molar mass of water}} = \frac{4.3 \, g}{18.02 \, g/mol} \approx 0.2386 \, \text{moles} \][/tex]
2. Calculate the energy consumed for the thawing process:
[tex]\[ \text{Energy consumed} = \text{number of moles of ice} \times \Delta H_{\text{fusion}} = 0.2386 \, \text{moles} \times 6.03 \, \text{kJ/mol} \approx 1.4389 \, \text{kJ} \][/tex]
### Conclusion:
The correct answer is D) 1.4389 kJ, which means it is a practically correct numerical depiction of the energy consumed by thawing 4.3 grams of ice. The closest option is indeed not listed explicitly among choices A, B, or C. However, if you were to choose from those that involved using the fusion energy and molar mass conversion, by further simplifying the accurate path considered:
A. [tex]\( \approx 4.3 \,g \times 1 \, mol/18.02 \, g \times 6.03 \,kJ/mol = 1.4389 \,kJ\)[/tex].
Thereby, detailed computations as portrayed verified how the standard listed operations evaluate correctly, following the detailed format shown.
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