Answer:
The relative rates of diffusion for methane and oxygen is 1.4142.
Methane gas will be able to travel 1.4142 meter in the same conditions.
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. Mathematically written as:

We are given:
Molar mass of methane gas, m = 16 g/mol
Molar mass of oxygen gas,m' = 32 g/mol
By taking their ratio, we get:


The relative rates of diffusion for methane and oxygen is 1.4142.
If oxygen gas travels 1 meters in time t.
Rate of diffusion of oxygen =
If methane gas travels travels in y meters in time t.
Rate of diffusion of methane=

y = 1.4142 m
Methane gas will be able to travel 1.4142 meter in the same conditions.
Answer:
2
Explanation:
In two reactions energy is released.
1) C₆H₁₂O₆ + 6O₂ → 6H₂O + 6CO₂ + heat
It is cellular respiration reaction.It involves the breakdown of glucose molecule in the presence of oxygen to yield large amount of energy. Water and carbon dioxide are also produced as a byproduct.
Glucose + oxygen → carbon dioxide + water + 38ATP
2) 2H₂ + O₂ → 2H₂O ΔH = -486 kj/mol
The given reaction is formation of water. In this reaction oxygen and hydrogen react to form water and 486 kj/mol is also released.
The reaction in which heat is released is called exothermic reaction.
Exothermic reaction:
The type of reactions in which energy is released are called exothermic reactions.
In this type of reaction energy needed to break the bonds are less than the energy released during the bond formation.
For example:
Chemical equation:
C + O₂ → CO₂
ΔH = -393 Kj/mol
it can be written as,
C + O₂ → CO₂ + 393 Kj/mol
Endothermic reactions:
The type of reactions in which energy is absorbed are called endothermic reactions.
In this type of reaction energy needed to break the bond are higher than the energy released during bond formation.
For example:
C + H₂O → CO + H₂
ΔH = +131 kj/mol
it can be written as,
C + H₂O + 131 kj/mol → CO + H₂
I believe that the answer to the question provided above is that <span>they formed two very different hypotheses since it is what they saw according to what they knew. Also hypothesis is just an intelligent guess.</span>
Hope my answer would be a great help for you. If you have more questions feel free to ask here at Brainly.
The mass percentage is 15.1465%.