Answer:
90.3 L
Explanation:
Given data:
Volume of water produced = 77.4 L
Volume of oxygen required = ?
Solution:
Chemical equation:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
It is known that,
1 mole = 22.414 L
There are 7 moles of oxygen = 7×22.414 = 156.9 L
There are 6 moles of water = 6×22.414 = 134.5 L
Now we will compare:
H₂O : O₂
134.5 : 156.9
77.4 : 156.9/134.5×77.4 =90.3 L
So for the production of 77.4 L water 90.3 L oxygen is required.
True.
A catalyst is a substancr that increases the rate of a chemical reaction without itself undergoing any permanent chemical change.
Answer:

group 16 period 2 of the periodic table
note: that is not the electronic configuration, that is the Bohr model.
Answer:
Explanation:
what wheres the answer???