Question #474d7

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Question #474d7

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Answer 1 · 2015-10-25T16:50:55

$1 g$ $1"g"$ helium contains the most particles.

Explanation:

The relative atomic mass in grams of an element contains the number of particles equal to the Avogadro Constant $= 6.02 \times 10^{23} {mol}^{- 1}$ $=6.02xx10^(23)"mol"^(-1)$

So $\frac{m}{A_{r}}$ $m/A_r$ $\times 6.02 \times 10^{23}$ $xx6.02xx10^(23)$ will give you the number of particles, in this case atoms, which are present.

Using approx. $A_{r}$ $A_r$ values:

For $A l$ $Al$ $n = \frac{1}{27} \times 6.02 \times 10^{23} = 2.23 \times 10^{22}$ $n=(1)/(27)xx6.02xx10^(23)=2.23xx10^(22)$

For $C$ $C$ $n = \frac{1}{12} \times 6.02 \times 10^{23} = 5 \times 10^{22}$ $n=1/12xx6.02xx10^(23)=5xx10^(22)$

For $H e$ $He$ $n = \frac{1}{4} \times 6.02 \times 10^{23} = 15 \times 10^{22}$ $n=1/4xx6.02xx10^(23)=15xx10^(22)$

For $F e$ $Fe$ $n = \frac{1}{56} \times 6.02 \times 10^{23} = 1.07 \times 10^{22}$ $n=1/56xx6.02xx10^(23)=1.07xx10^(22)$

So you can see from these data that 1g of He contains the most atoms.

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Question #474d7

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