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What volume will 20.0g of Argon occupy at STP?
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What volume will 20.0g of Argon occupy at STP?

User G Quintana
by
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1 Answer

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Molar mass Argon = 39.948 g/mol

1 mol ------ 39.948 g
mol ----- 20.0 g

mol = 20.0 * 1 / 39.948

= 0.5006 moles

1 mol --------------------- 22.4 L ( at STP )
0.5006 moles ------------- L

L = 0.5006 * 22.4

= 11.21 L

hope this helps!
User Markmarijnissen
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