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Suppose 25 L to be 25 kg water (almost, -0.2% at room temperature)

then divide 25,000 g by the moalr mass of H2O being 2+16=18 g/mole

you get = 1385 moles (because 1 mole = 18 grams)

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โˆ™ 14y ago
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โˆ™ 14y ago

By filling in this formula with SI units of the symbols R, T, p, V you'll find the number of moles for ALL possible (ideal) gases according to Ideal Gas Law (Boyle's):

n = p*V / R*T or p*V = n*R*T n= number of moles of any gas => 25 mol (your 'known')

p= pressure of (any) gas (Pa) => 1.01325*105 Pa (=1 atm) (Standard)

V= Volume of that gas (m3) => ? (to be calculated)

R= universal Gas constant (J.mol-1.K-1 )=> 8.3145 (universal constant)

T= absolute temperature (K) => 273.15 K (Standard)

With the questioned figures:

V = nRT/p = [25*8.3145*273.15] / [1.01325*105] = 0.56035 (m3) = 560 litres for 25 moles of Ar (or any) gas at (Standard)TP

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โˆ™ 5mo ago

At STP (standard temperature and pressure), 1 mole of any gas occupies 22.4 liters. Therefore, 5.00 moles of Argon would occupy: 5.00 moles x 22.4 L/mole = 112.0 liters.

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โˆ™ 15y ago

1 mole of gas at STP occupies 22.4Lt. of space. So 25 mole will occupie 22.4*25=560lt.

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โˆ™ 11y ago

1 mole = 22.414 liters

So, 5 moles = 112.07 liters

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Q: How many liters of Argon are in5.00 moles at STP?
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