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
Answer:
118MOLS
Explanation:
M1VI=M2V2
THE REST WORK OUT ( BA UNZA STUDENTS HE HE HE )
Answer:
A
Explanation:
There are 2 sodium atoms and 1 oxygen atom.
The percent of O in Cr₂O₃ : 31.58%
<h3>Further explanation</h3>
Given
Cr = 52.00 amu, O = 16.00 amu
Required
The percent of O
Solution
MW Cr₂O₃ = 2 x Ar Cr + 3 x Ar O
MW Cr₂O₃ = 2.52+3.16
MW Cr₂O₃ =152 amu