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7 August, 18:55

Calcium carbonate decomposes at 12000 C to form carbon dioxide and calcium oxide. If 25 liters of carbon dioxide are collected at 12750 C, what will the volume of this gas be after it cools to 250 C?

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  1. 7 August, 21:09
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    1L

    Explanation:

    Data obtained from the question include:

    Initial volume (V1) = 25L

    Initial temperature (T1) = 12750°C

    Final temperature (T2) = 250°C

    Final volume (V2) = ... ?

    Next we shall convert from celsius to Kelvin temperature. This is illustrated below:

    T (K) = T (°C) + 273

    Initial temperature (T1) = 12750°C

    Initial temperature (T1) = 12750°C + 273 = 13023K

    Final temperature (T2) = 250°C

    Final temperature (T2) = 250°C + 273 = 523K

    Finally, we can obtain the new volume of the gas by using Charles' law equation as shown below:

    V1/T1 = V2/T2

    25/13023 = V2/523

    Cross multiply to express in linear form

    13023 x V2 = 25 x 523

    Divide both side by 13023

    V2 = 25 x 523 / 13023

    V2 = 1L

    Therefore, the new volume of the gas is 1L.
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