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
Still chill still chill still chill
It would emit energy in most of the cases in form of light
There are two naturally occurring isotopes of gallium: mass of Ga-69 isotope is 68.9256 amu and its percentage abundance is 60.11%, let the mass of other isotope that is Ga-71 be X, the percentage abundance can be calculated as:
%Ga-71=100-60.11=39.89%
Atomic mass of an element is calculated by taking sum of atomic masses of its isotopes multiplied by their percentage abundance.
Thus, in this case:
Atomic mass= m(Ga-69)×%(Ga-69)+X×%(Ga-71)
From the periodic table, atomic mass of Ga is 69.723 amu.
Putting the values,
Thus,
Rearranging,
Therefore, mass of Ga-71 isotope is 70.9246 amu.
