coolayman3823 coolayman3823

19-11-2018
Chemistry

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What is the volume of 19.87mol of ammonium chloride NH4CL at STP

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malika8888 malika8888

19-11-2018

[tex]n = \frac{V}{STP} \\ n = \frac{V}{22.4} \\ 19.87 = \frac{V}{22.4} \\ V = 19.87 \times 22.4 \\ V = 445,088 \: l \: of \: NH4CL \: [/tex]

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