Question #c5870

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Question #c5870

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Answer 1 · 2017-09-12T04:57:26

$2.020 \cdot 10^{23}$ $2.020* 10^23$ atoms

Explanation:

First, convert grams of chromium to moles:

$17.441 g Cr (\frac{1 mol}{51.996 g Cr}) = 0.335$ $17.441 "g Cr" ("1 mol"/ "51.996 g Cr") = 0.335$ mol Cr

Then convert from moles to atoms using Avogadro's Number:

$0.335 mol Cr (\frac{6.022*10^23 atoms}{1 mol}) = 2.020 \cdot 10^{23}$ $0.335 "mol Cr" ("6.022*10^23 atoms"/ "1 mol") = 2.020*10^23$ atoms Cr

Answer 2 · 2017-09-12T04:58:27

$Number of chromium atoms ≅ 2 \times 10^{23}$ $"Number of chromium atoms"~=2xx10^23$

Explanation:

We work out (i) the molar quantity of chromium....

$Moles of Cr$ $"Moles of Cr"$ $=$ $=$ $\frac{17.441 \cdot g}{51.9961 \cdot g \cdot m o l^{- 1}}$ $(17.441*g)/(51.9961*g*mol^-1)$

$= 0.3354 \cdot m o l$ $=0.3354*mol$ .

And (ii) we know that a mole specifies Avogadro's number of molecules, i.e. $6.022 \times 10^{23} \cdot m o l^{- 1}$ $6.022xx10^23*mol^-1$ , and so we take the product....

$= 0.3354 \cdot m o l \times 6.022 \times 10^{23} \cdot m o l^{- 1} = ? ?$ $=0.3354*molxx6.022xx10^23*mol^-1=??$

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