How many phosphorus atoms are present in a $25.0 \cdot g$ $25.0*g$ mass of $phosphorus pentoxide$ $"phosphorus pentoxide"$ ?

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How many phosphorus atoms are present in a $25.0 \cdot g$ $25.0*g$ mass of $phosphorus pentoxide$ $"phosphorus pentoxide"$ ?

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Answer 1 · 2016-10-18T19:45:51

$0.352 moles of phosphorus atoms are present in this quantity.$ $"0.352 moles of phosphorus atoms are present in this quantity."$

Explanation:

Number of moles of $P_{2} O_{5}$ $P_2O_5$ $=$ $=$ $\frac{25.0 \cdot g}{141.95 \cdot g \cdot m o l^{- 1}}$ $(25.0*g)/(141.95*g*mol^-1)$ $=$ $=$ $0.176 \cdot m o l$ $0.176*mol$ , with respect to $P_{2} O_{5}$ $P_2O_5$ .

There are thus $2 \times 0.176 \cdot m o l$ $2xx0.176*mol$ of $phosphorus$ $"phosphorus"$ .

$P_{2} O_{5}$ $P_2O_5$ is an empirical formula. The actual molecule is $P_{4} O_{10}$ $P_4O_10$ , which is an extremely hygroscopic white powder. Do the different formulations make a difference to the number of phosphorus atoms present?

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How many phosphorus atoms are present in a $25.0 \cdot g$ $25.0*g$ mass of $phosphorus pentoxide$ $"phosphorus pentoxide"$ ?

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How many phosphorus atoms are present in a 25.0*g25.0⋅g25.0*g mass of "phosphorus pentoxide"phosphorus pentoxide"phosphorus pentoxide"?

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How many phosphorus atoms are present in a $25.0 \cdot g$ $25.0*g$ mass of $phosphorus pentoxide$ $"phosphorus pentoxide"$ ?