q.6 How much energy (J) is needed to melt a 7.0 lb bag of ice? q.7 Calculate the energy needed to melt 3.0 g of ice at 0°C, heat it to boiling, boil it, and heat the stear to 125°C? 1
q.6 How much energy (J) is needed to melt a 7.0 lb bag of ice? q.7 Calculate the energy needed to melt 3.0 g of ice at 0°C, heat it to boiling, boil it, and heat the stear to 125°C? 1
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Transcribed Image Text:q.6 How much energy (J) is needed to melt a 7.0 lb bag of ice?
q.7 Calculate the energy needed to melt 3.0 g of ice at 0°C, heat it to boiling, boil it, and heat the stear
to 125°C?
1
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Calculate the energy needed to melt 3.0g
![q.7 Calculate the energy needed to melt 3.0 g of ice at 0°C, heat it to boiling, boil it, and heat the steam to 125°C.
The image contains handwritten calculations related to the problem, breaking down the energy needed to achieve the transitions outlined.
1. There is a formula starting with:
\[((7.00)(1000)(6.01586) = 3173.29)\]
2. Then it continues with:
\[
\left( \frac{3173.29}{1 \text{ mol}} \right)\left( \frac{1 \text{ mol}}{18.01 \text{ g}} \right) = 176.8 \text{ mols}
\]
3. Another part labeled \(c_2\):
\[
(176.30)(6.009) = 1059.3 \text{ kJ}
\]
These calculations represent the different stages of heating and phase changes, calculating energy requirements for each phase transition and temperature change.](https://content.bartleby.com/qna-images/question/8efc1ceb-a8e8-463a-b9b0-209c3b394014/50af9f84-3a83-4a03-ba17-7107d48ccab8/yj7j56k_processed.jpeg)
Transcribed Image Text:q.7 Calculate the energy needed to melt 3.0 g of ice at 0°C, heat it to boiling, boil it, and heat the steam to 125°C.
The image contains handwritten calculations related to the problem, breaking down the energy needed to achieve the transitions outlined.
1. There is a formula starting with:
\[((7.00)(1000)(6.01586) = 3173.29)\]
2. Then it continues with:
\[
\left( \frac{3173.29}{1 \text{ mol}} \right)\left( \frac{1 \text{ mol}}{18.01 \text{ g}} \right) = 176.8 \text{ mols}
\]
3. Another part labeled \(c_2\):
\[
(176.30)(6.009) = 1059.3 \text{ kJ}
\]
These calculations represent the different stages of heating and phase changes, calculating energy requirements for each phase transition and temperature change.
Solution
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