A mole of oxygen and a mole of hydrogen (at STP) have all of the following in common EXCEPT
1 answer:
Answer:
Root mean squared velocity is different.
Explanation:
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In this case, since we have a mixture of oxygen and nitrogen at STP, which is defined as a condition whereas T = 298 K and P = 1 atm, we can infer that these gases have the same temperature, pressure, volume and moles but a different root mean squared velocity according to the following formula:
Since they both have a different molar mass (MM), nitrogen (28.02 g/mol) and oxygen (32.02 g/mol), thus we infer that nitrogen would have a higher root mean squared velocity as its molar mass is less than that of oxygen.
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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