The following is produced when propane (C₃H₈) is combusted completely : H₂O
<h3>Further explanation </h3>
Complete combustion of Hydrocarbons with Oxygen will be obtained by CO₂ and H₂O compounds.
If O₂ is insufficient there will be incomplete combustion produced by CO and H and O
Hydrocarbon combustion reactions (especially alkanes) 
For combustion of propane C₃H₈ (n = 3) ⇒ completely(excess O₂) :
C₃H₈+5O₂⇒3CO₂+4H₂O
The products of combustion : CO₂ and H₂O
Answer: The mass of the gas is 18.3 g/mol.
Explanation:
To calculate the rate of diffusion of gas, we use Graham's Law.
This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:




Squaring both sides and solving for 

Hence, the molar mas of unknown gas is 18.3 g/mol.
Answer:
10425 J are required
Explanation:
assuming that the water is entirely at liquid state at the beginning , the amount required is
Q= m*c*(T final - T initial)
where
m= mass of water = 25 g
T final = final temperature of water = 100°C
T initial= initial temperature of water = 0°C
c= specific heat capacities of water = 1 cal /g°C= 4.186 J/g°C ( we assume that is constant during the entire temperature range)
Q= heat required
therefore
Q= m*c*(T final - T initial)= 25 g * 4.186 J/g°C * (100°C- 0°C) = 10425 J
thus 10425 J are required
Answer:
The answer to your question is: 4.5 %
Explanation:
0.126 moles of AgNO3
mass percent = ?
mass of water = 475 g
Formula
weight percent = weight of solute / weight of solution x 100
Weight of solute
MW AgNO3 = 108 + 14 + (16 x 3)
= 108 + 14 + 48
= 170 g
170 g of AgNO3 ------------------- 1 mol
x --------------------- 0.126 moles
x = (0.126 x 170) / 1 = 21.42 g of AgNO3
Weight percent = 21.42/475 x 100
= 4.5 %