How many $hydrogen ATOMS$ $"hydrogen ATOMS"$ are present in a $23.7 \cdot g$ $23.7*g$ mass of $methane$ $"methane"$ ?

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How many $hydrogen ATOMS$ $"hydrogen ATOMS"$ are present in a $23.7 \cdot g$ $23.7*g$ mass of $methane$ $"methane"$ ?

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Answer 1 · 2016-10-03T07:43:15

There are approx. $N_{A} \times 4 \times 2$ $N_Axx4xx2$ hydrogen atoms in such a mass of methane.

Explanation:

In $16.04 \cdot g$ $16.04*g$ methane, there are $N_{A}$ $N_A$ methane molecules, where $N_{A} = Avogadro's number = 6.022 \times 10^{23}$ $N_A="Avogadro's number"=6.022xx10^23$ .

And so $moles of methane$ $"moles of methane"$ $=$ $=$ $\frac{23.7 \cdot g}{12.04 \cdot g \cdot m o l^{- 1}}$ $(23.7*g)/(12.04*g*mol^-1)$ $=$ $=$ $1.97 \cdot m o l$ $1.97*mol$

And thus $atoms of hydrogen$ $"atoms of hydrogen"$ $=$ $=$ $\frac{23.7 \cdot g}{12.04 \cdot g \cdot m o l^{- 1}} \times 4 H atoms \cdot m o l^{- 1} \times N_{A} H atoms$ $(23.7*g)/(12.04*g*mol^-1)xx"4 H atoms"*mol^-1xxN_A" H atoms"$

$≅$ $~=$ $4.80 \times 10^{24} hydrogen atoms$ $4.80xx10^24" hydrogen atoms"$

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How many $hydrogen ATOMS$ $"hydrogen ATOMS"$ are present in a $23.7 \cdot g$ $23.7*g$ mass of $methane$ $"methane"$ ?

1 Answer

Explanation:

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How many hydrogen ATOMS"hydrogen ATOMS" are present in a 23.7⋅g23.7*g mass of methane"methane"?

1 Answer

Explanation:

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How many $hydrogen ATOMS$ $"hydrogen ATOMS"$ are present in a $23.7 \cdot g$ $23.7*g$ mass of $methane$ $"methane"$ ?