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OpenStudy (anonymous):
how many molecules are in 72.45g CCl4 ?
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OpenStudy (anonymous):
\[72.45gCCl4*\frac{ 1 mol cCl4}{ 154gCCl4? }*\frac{ 6.022*10^{23}molecules }{ 1molCCl4 }\] answer= 2.833*10^23 molecules
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