Answer:
The essence including its given problem is outlined in the following segment on the context..
Explanation:
The given values are:
Moles of CO₂,
x = 0.01962
Moles of water,
Compound's mass,
= 0.4647 g
Let the compound's formula will be:
Combustion's general equation will be:
⇒ 
On putting the estimated values, we get
⇒ 
⇒ 
⇒ 
⇒ 
Now,
x : y : z = 
= 
= 
= 
So that the empirical formula seems to be "C₃H₆O₂".
Here are the answers. A,A,A,C,A,C.
I hope this helps!
Answer:
1. 80g
2. 1.188mole
Explanation:
1. We'll begin by obtaining the molar mass of CH4. This is illustrated below:
Molar Mass of CH4 = 12 + (4x1) = 12 + 4 = 16g/mol
Number of mole of CH4 from the question = 5 moles
Mass of CH4 =?
Mass = number of mole x molar Mass
Mass of CH4 = 5 x 16
Mass of CH4 = 80g
2. Mass of O2 from the question = 38g
Molar Mass of O2 = 16x2 = 32g/mol
Number of mole O2 =?
Number of mole = Mass /Molar Mass
Number of mole of O2 = 38/32
Number of mole of O2 = 1.188mole
Answer : The amount of oxygen gas collected are, 0.217 mol
Explanation :
Using ideal gas equation :
where,
P = pressure of gas = 
(1 atm = 760 torr)
V = volume of gas = 5 L
T = temperature of gas = 
n = number of moles of gas = ?
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in the ideal gas equation, we get:
Thus, the amount of oxygen gas collected are, 0.217 mol
