How many neon atoms in a $15.8 \cdot g$ $15.8*g$ mass of neon gas?

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How many neon atoms in a $15.8 \cdot g$ $15.8*g$ mass of neon gas?

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Answer 1 · 2018-01-08T19:14:05

See below..

Explanation:

No. of moles $= \frac{Given mass}{Atomic Mass}$ $=("Given mass")/("Atomic Mass")$ = $\frac{15.8}{20} = 0.79$ $15.8/20=0.79$

No. of atoms = (Avogadro's no.) $\cdot$ $*$ (no. of moles)=
$(6.022 \cdot 10^{23} \cdot 0.79) = 4.75 \cdot 10^{23}$ $(6.022 *10^23 *0.79)=4.75*10^23$

Answer 2 · 2018-01-08T19:20:10

Well, we know that there are $6.022 \times 10^{23}$ $6.022xx10^23$ particles in a mole of such particles....

Explanation:

And we also know (do we?) that a $20.18 \cdot g$ $20.18*g$ mass of neon gas contains $6.022 \times 10^{23}$ $6.022xx10^23$ individual neon atoms....

And so we work out the quotient multiplied by the $Avocado number....$ $"Avocado number...."$

$\frac{15.8 \cdot g}{20.18 \cdot g \cdot m o l^{- 1}} \times 6.022 \times 10^{23} \cdot m o l^{- 1}$ $(15.8*g)/(20.18*g*mol^-1)xx6.022xx10^23*mol^-1$

$≅ 4.5 \times 10^{23} \cdot individual neon atoms$ $~=4.5xx10^23*"individual neon atoms"$ ...

Here I have used $Avogadro's number$ $"Avogadro's number"$ precisely as I would any other collective number: $dozen$ $"dozen"$ ; $Bakers' dozen$ $"Bakers' dozen"$ ; $score$ $"score"$ ; $gross$ $"gross"$ ...

We use $Avogadro's number$ $"Avogadro's number"$ because we know that such a number of neon atoms has a mass of $20.18 \cdot g$ $20.18*g$ ...

Answer 3 · 2018-01-08T19:31:42

$4.71 \cdot 10^{23}$ $4.71 * 10^23$ atoms

Explanation:

Neon has an atomic mass of 20.18. Divide 15.8 grams by the atomic mass to determine the moles of Ne. Then multiply the moles of Ne by Avogadro's number to get the number of atoms.

$(15.8 g \div 20.18$ $(15.8g -: 20.18$ moles $) 6.02 \cdot 10^{23}$ $) 6.02 * 10^23$ atoms = $4.71 \cdot 10^{23}$ $4.71 * 10^23$ atoms

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How many neon atoms in a $15.8 \cdot g$ $15.8*g$ mass of neon gas?

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How many neon atoms in a $15.8 \cdot g$ $15.8*g$ mass of neon gas?