Answer:
C the mass of each product formed
Explanation:
To the determine the limiting reactant, it is essential that we have the balanced equation of the reaction from which we can calculate the stochiometry mole ratio of the reactant. After this, we need to calculate the molar mass of the reactants, using the mole from the balanced equation we can calculate each mass of each reactant needed. Finally we need the mass of each reactant using proportion we can calculate the amount needed for the reaction from the masses of the reactant by comparing the mass given against the mass calculated from the balanced equation. After this, the mass that is exhausted or that is finished will be the limiting reactant which is the reactant that finished and caused the reaction to stop.
Answer:
The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Explanation:
Given;
CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol
From the combustion reaction above, it can be observed that;
1 mole of methane (CH₄) released 890 kilojoules of energy.
Now, we convert 59.7 grams of methane to moles
CH₄ = 12 + (1x4) = 16 g/mol
59.7 g of CH₄ 
1 mole of methane (CH₄) released 890 kilojoules of energy
3.73125 moles of methane (CH₄) will release ?
= 3.73125 moles x -890 kJ/mol
= -3320.81 kJ
Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Answer: C
Explanation:
Hot air is a lot less dense for a few reasons. Hot air essentially means the particles have more kinetic energy, and move around a lot more. Cold air is dense because the particles move a lot less, have less energy, and are closer together.
Moles = (6.74*10^23)/(6.02*10^23) =1.119 moles
1.119*44.09=49.36g
Answer:
we only see parts of the lit side as the moon goes around the earth
Explanation:
Unlike the sun, the moon orbits the Earth. This is the reason why we see the <em>different phases of the moon.</em> The reflection of the moon is being illuminated back to us with the help of the sun. So, as the moon circles the Earth, we only see parts of the lit side. Such changes helps us see the moon in different phases such as<em> </em>the <em>Third Quarter, Crescent, New Moon, Full Moon, etc.</em>
For example, during "Full Moon," <em>the moon's entire face is lit up by the sun</em>. Thus, we see the entire moon's lit portion.
Thus, this explains the answer.
