The moles of oxygen that are needed to produce 13.7 moles of carbon dioxide is 21.17 moles of Oxygen
<u><em>calculation</em></u>
2 C₆H₁₂O + 17 O₂ → 12 CO₂ +12 H₂O
The moles of O₂ is determined using the mole ratio
that is for given equation above O₂ : Co₂ is 17 :12
therefore the moles of O ₂= 13.7 moles x 17/12 =21.17 moles
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:
The rule is especially applicable to carbon, nitrogen, oxygen, and the halogens, but also to metals such as sodium or magnesium. ... All four of these electrons are counted in both the carbon octet and the oxygen octet, so that both atoms are considered to obey the octet rule.
Wind isn’t an external force that acts in the rock cycle.