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Lady_Fox [76]
3 years ago
10

PLEASE HELP MEEEE!!!!​

Chemistry
1 answer:
Nonamiya [84]3 years ago
5 0

1. 18 L

2. 20.832 L

<h3>Further explanation</h3>

Given

Combustion of ethane

Required

Volume of CO₂

Solution

Avogadro's Law :

In the same T,P and V, the gas contains the same number of molecules  

So the ratio of gas volume will be equal to the ratio of gas moles  

V₁/n₁=V₂/n₂

1. From equation, mol ratio of C₂H₆ : CO₂ ⇒ 2 : 6

so the volume of CO₂ :

= 6/2 x 6 L

= 18 L

2. mol ethane (MW=32 g/mol) :

= mass : MW

= 10 g : 32 g/mol

= 0.31

mol CO₂ :

= 6/2 x 0.31

= 0.93

The volume CO₂(assume at STP) = 20.832 L

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N₂O is the empirical formula of an oxide of nitrogen containing 63.61% by mass of nitrogen and 36.69% by mass of oxygen.

Empirical formula can be calculated by

Suppose we have 100 g of the substance. That indicates that it has 36.69 grams of oxygen and 63.61 grams of nitrogen.

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Formula used

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