At $STP$ $"STP"$ what is the molar quantity of dinitrogen gas which has a volume of $5.6 \cdot L$ $5.6*L$ ?

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At $STP$ $"STP"$ what is the molar quantity of dinitrogen gas which has a volume of $5.6 \cdot L$ $5.6*L$ ?

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Answer 1 · 2017-10-27T15:29:48

Well, at $STP$ $"STP"$ , the molar volume of an ideal gas is $22.4 \cdot L$ $22.4*L$ ...

Explanation:

And the definitions of $STP$ $"STP"$ seems to vary across curricula; there are a few different standards, of which you have to be mindul....

And so to get the molar quantity of dinitrogen, we take the quotient....

$\frac{volume of gas}{molar volume} = \frac{5.6 \cdot L}{22.4 \cdot L \cdot m o l^{- 1}} = 0.250 \cdot m o l$ $"volume of gas"/"molar volume"=(5.6*L)/(22.4*L*mol^-1)=0.250*mol$ ...how many moles of $nitrogen atoms$ $"nitrogen atoms"$ in this molar quantity?

Answer 2 · 2017-10-27T15:35:43

$0.249 m o l s$ $0.249mols$

Explanation:

At STP, $22.414 l$ $22.414l$ of $N_{2}$ $N_2$ contain moles $= 1$ $= 1$ mole

$5.6 L$ $5.6L$ of $N_{2}$ $N_2$ will contain moles $=$ $=$ $\frac{1 m o l}{22.414liters}$ $(1mol)/"22.414liters"$ $\times$ $xx$ $5.6 l i t e r s$ $5.6liters$

$= 0.249 m o l s$ $= 0.249mols$

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At $STP$ $"STP"$ what is the molar quantity of dinitrogen gas which has a volume of $5.6 \cdot L$ $5.6*L$ ?