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If 0.213 moles of argon occupies a volume of 652 mL at a particular temp and pressure, what volume would it occupy if 0.162 moles of argon were added at the same conditions?

Respuesta :

Neetoo

Answer:

V₂  = 495.89 mL

Explanation:

Given data:

Initial number of moles = 0.213 mol

Initial volume = 652 mL

Final number of moles = 0.162 mol

Final volume = ?

Solution:

V₁/n₁    =    V₂/n₂

By putting values,

652 mL/0.213 mol   =  V₂ /0.162 mol

V₂  = 652 mL 0.162 mol /0.213 mol

V₂  = 105.62 mL.mol /0.213 mol

V₂  = 495.89 mL

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