How many moles of atoms are in 3.00 g of $^{13} C$ $""^13C$ ?

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How many moles of atoms are in 3.00 g of $^{13} C$ $""^13C$ ?

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Answer 1 · 2016-06-07T04:43:09

There are $\frac{3}{13} \times N_{A} carbon atoms$ $3/13xxN_A" carbon atoms"$ , where $N_{A} = Avogadro's Number, 6.022 \times 10^{23} \cdot m o l^{- 1}$ $N_A="Avogadro's Number, "6.022xx10^23*mol^-1$

Explanation:

In $13 \cdot g$ $13*g$ of $^{13} C$ $""^(13)C$ there are precisely $Avogadro's number of atoms$ $"Avogadro's number of atoms"$ .You have specified $3.00 \cdot g$ $3.00*g$ , thus there are $\frac{3.00 \cdot g}{13.00 \cdot g \cdot m o l} \times 6.022 \times 10^{23} \cdot m o l^{- 1}$ $(3.00*g)/(13.00*g*mol)xx6.022xx10^23*mol^-1$ $=$ $=$ $? ?^{13} C atoms$ $??""^13C " atoms"$

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How many moles of atoms are in 3.00 g of $^{13} C$ $""^13C$ ?

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How many moles of atoms are in 3.00 g of 13C""^13C?

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How many moles of atoms are in 3.00 g of $^{13} C$ $""^13C$ ?