An evacuated observation balloon after filling with helium gas had an increase in weight equal to 2 g. The volume of the balloon at STP must be?

Question

THANGUTURU NIRGUNA · Answer

STP value is 22.4 litres.then there is a balloon weight 2 grams.then we have yo multiply the weight *stp value 22.4*2=44.8litres

Yogita Negi · Answer

volume= moles = volume at STP condition/molar volume i.e. 2/4 = vol./22.4 = 2/4*22.4 = 11.2 L ...check out this ans.

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An evacuated observation balloon after filling with helium gas had an increase in weight equal to 2 g. The volume of the balloon at STP must be?

An evacuated observation balloon after filling with helium gas had an increase in weight equal to 2 g. The volume of the balloon at STP must be?

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An evacuated observation balloon after filling with helium gas had an increase in weight equal to 2 g. The volume of the balloon at STP must be?

An evacuated observation balloon after filling with helium gas had an increase in weight equal to 2 g. The volume of the balloon at STP must be?

2 Answers

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Other Related Questions on Physical Chemistry

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