Question #ca49b

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Question #ca49b

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Answer 1 · 2016-11-09T19:21:41

$m = {33.4 g of F}_{2}$ $m = "33.4 g of F"_2$

Explanation:

Use the following formula, where $N$ $N$ is the number of molecules, $N_{A}$ $N_A$ represents Avogadro's number, and $n$ $n$ represents the number of moles.

$N = N_{A} \cdot n$ $N = N_A * n$

Now, we can substitute the known values.

$5.30 \cdot 10^{23} {molecules F}_{2} = (\frac{6.022 \cdot 10^{23} molecules}{mol}) \cdot n$ $5.30 * 10^(23) " molecules F"_2 = ((6.022 * 10^(23) " molecules")/ ("mol")) *n$

Now solving for the number of moles present.

$n = {0.88010 mol of F}_{2}$ $n = "0.88010 mol of F"_2$

Now we can use the formula, where $n$ $n$ is the number of moles, $m$ $m$ is the mass in grams and $M$ $M$ is the molar mass.

$n = \frac{m}{M}$ $n = m/M$

${0.88010 mol of F}_{2} = \frac{m}{37.996 g/mol} {of F}_{2}$ $"0.88010 mol of F"_2 = m/("37.996 g/mol") " of F"_2$

$:. m = "33.4 g of F"_2$
The answer is rounded to 3 significant digits; as specified by the original question.

Hope this helped :)

Note
I've edited the significant digits in the calculation as the data specified three. Thus the intermediate steps should really carry about one or two intermediate ones. I've carried two to be safe. At the end of the day however, it depends on your instructor or assessment marker.

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Question #ca49b

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