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27 April, 02:26

If the mouse was in a wire cage and only the weights of the mouse, food, and water were considered, would you come to the same answer as in Part A?

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  1. 27 April, 04:26
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    Here is the full question:

    (A) If a closed container contains a mouse as well as enough food, water, and oxygen for the mouse to live for 3 weeks,

    How much will the container weigh 1 and 2 weeks later after the mouse has eaten, drunk and exercised (respiration is CO2 emission), and why?

    (B) If the mouse was in a wire cage and only the weights of the mouse, food, and water were considered, would you come to the same answer as in (A) and why?

    Explanation:

    (A) The mouse will weigh the same. This is because solids, liquid, and gases cannot escape the closed container. All of the life processes involving reactions conserve the atoms involved. Some of those atoms will appear in the form of gases, some as solids, and others as liquids but all will be retained in the closed container.

    (B) In a wire cage, gases can escape. This means that the weight will not be the same after 1 and 2 weeks. The weight would be less than the original weight of the mouse, it's food, and it's water.
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