Question #844e5

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Question #844e5

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Answer 1 · 2018-01-01T08:01:04

$1.2 \times 10^{24}$ $1.2 xx 10^24$

Explanation:

1 mol of $C O_{2}$ $CO_2$ = $(12 + 2 \cdot 16)$ $(12 + 2*16)$ g = $44$ $44$ g

And the Simplest Unitary Method Comes!

44 g of $C O_{2} \to$ $CO_2 rarr$ 1 mol

1 g of $C O_{2} \to$ $CO_2 rarr$ $\frac{1}{44}$ $1/44$ mol

88 g of $C O_{2} \to$ $CO_2 rarr$ $\frac{88}{44}$ $88/44$ mol = 2 mol

As, 1 mol = $6.022 \times 10^{23}$ $6.022 xx 10^23$ ,

2 mol = $2 mol \times 6.022 \times 10^{23} l molecules/mol = 12 \times 10^{23} l molecules = 1.2 \times 10^{24} l molecules$ $"2 mol" xx 6.022 xx 10^23 color(white)(l)"molecules/mol"= 12 xx 10^23color(white)(l) "molecules" = 1.2 xx 10^24 color(white)(l)"molecules"$

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Question #844e5

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