Answer:
B
Explanation:
I hope it helps you good luck
The answer is C because there are 22 oxygen atoms on the product side so to balance the equation the coefficient needed is 11
Answer:
2.85moles of oxygen gas
Explanation:
Given parameters:
Volume of oxygen gas = 63.8L
Unknown:
Number of moles = ?
Solution:
We assume that the gas is under standard temperature and pressure. To find the number of moles, use the expression below:
1 mole of a gas at STP occupies a volume of 22.4L
So;
63.8L of oxygen gas will take up a volume of
= 2.85moles of oxygen gas
Answer:
8.77 kilo Joules will be the total amount of heat required for both the heating and the vaporizing.
Explanation:
Moles of ethanol of ethanol = 0.200 mol
Heat required to heat 0.200 moles of ethanol = Q = 1.05 kJ
Enthalpy of vaporization of ethanol = 
Heat required to vaporize 0.200 moles of ethanol = Q'

Total heat required to fore heating and the vaporizing :
= Q + Q' = 1.05 kJ + 7.72 kJ = 8.77 kJ
8.77 kilo Joules will be the total amount of heat required for both the heating and the vaporizing.