Answer:

Explanation:
Hello,
In this case, the described chemical reaction is:

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

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Answer:
false,soil damage and nutrient loss can be a major problem so it can not gain nutrients quickly
One example of matter could be <em>Light.</em>
Answer:
1.64 moles O₂
Explanation:
Part A:
Remember 1 mole of particles = 6.02 x 10²³ particles
So, the question becomes, how many '6.02 x 10²³'s are there in 9.88 x 10²³ molecules of O₂?
This implies a division of given number of particles by 6.02 x 10²³ particles/mole.
∴moles O₂ = 9.88 x 10²³ molecules O₂ / 6.02 x 10²³ molecules O₂ · mole⁻¹ = 1.64 mole O₂
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Part B needs an equation (usually a combustion of a hydrocarbon).
The answer is A. The volume lines on the bottom are just showing the volume of the beaker. The more dense liquid will sink to the bottom