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.
Answer:
the answer is Fructose
Explanation:
the reason is because when it brakes down it forms a sort of fructose
Answer:
metal Atom
Explanation:
every transition metal atom are responsible for the flame color. Some metal are also confirmed by flame test.
<span>0.38
You first calculate the total moles by dividing the grams by molecular weight:
45 g N2 / 28.02 g/mol = 1.6 mol N2
40 g Ar / 39.95 g/mol = 1.0 mol
Then you divide the moles of Ar by the total number of moles:
1.0 / (1.6 + 1.0) = 0.38 mol fraction</span>
