How many carbon atoms in a $5.85 \cdot g$ $5.85*g$ mass of $C B r_{4}$ $CBr_4$ ?

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How many carbon atoms in a $5.85 \cdot g$ $5.85*g$ mass of $C B r_{4}$ $CBr_4$ ?

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Answer 1 · 2017-08-26T06:34:26

Approx. $1 \times 10^{22} \cdot carbon atoms$ $1xx10^22*"carbon atoms"$ .

Explanation:

We got $5.85 \cdot g$ $5.85*g$ of $C B r_{4}$ $CBr_4$ .

This represents a molar quantity of .................

$\frac{5.85 \cdot g}{331.63 \cdot g \cdot m o l^{- 1}} = 0.0176 \cdot m o l$ $(5.85*g)/(331.63*g*mol^-1)=0.0176*mol$

And of course in such a molar quantity, there are $4 \times 0.0176 \cdot m o l$ $4xx0.0176*mol$ of bromine, and $1 \times 0.0176 \cdot m o l$ $1xx0.0176*mol$ of carbon. Are you with me?

And so we take the product of the molar quantity and Avogadro's number, to get.......

$0.0176 \cdot m o l \times 6.022 \times 10^{23} \cdot m o l^{- 1} = 1.06 \times 10^{22}$ $0.0176*molxx6.022xx10^23*mol^-1=1.06xx10^22$ individual carbon atoms.....

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How many carbon atoms in a $5.85 \cdot g$ $5.85*g$ mass of $C B r_{4}$ $CBr_4$ ?

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How many carbon atoms in a 5.85⋅g5.85*g mass of CBr4CBr_4?

1 Answer

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How many carbon atoms in a $5.85 \cdot g$ $5.85*g$ mass of $C B r_{4}$ $CBr_4$ ?