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Leokris [45]
1 year ago
13

Use Hess's law to calculate AG° rxn using the following information.

Chemistry
1 answer:
PIT_PIT [208]1 year ago
3 0

Answer:

AG rxn=+394.4 kJ

Explanation:

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Which statement accurately describes why air pressure decreases as altitude increases? As you increase altitude the temperature
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Answer:

The correct answer is " As you increase altitude, Earth's gravity is not acting as greatly as when closer to Earth's surface so the weight of air molecules decrease, which also in turn decreases air pressure."

Explanation:

As altitude increases air becomes rarefied. At higher altitudes there is less Earth's gravity acting on the air molecules than at areas close the earth surface. Air pressure is the weight of air acting per unit surface area. At higher altitudes as the weight of air decreases due to reduced gravitational pull the air pressure also decreases.

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Burning fossil fuels creates the ___________ gas carbon dioxide which increases global temperatures.
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Greenhouse gas is your answer it is causing the greenhouse effect of trapping heat in the earth
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If a pork roast must absorb kJ to fully cook, and if only 10% of the heat produced by the barbecue is actually absorbed by the r
kolezko [41]

Answer:

1,033.56 grams of carbon dioxide was emitted into the atmosphere.

Explanation:

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g),\Delta H_{rxn}=-2044 kJ/mol (assuming)

Energy absorbed by pork,E = 1.6\times 10^3 kJ (assuming)

Total energy produced by barbecue = Q

Percentage of energy absorbed by pork = 10%

10\%=\frac{E}{Q}\times 100

Q=\frac{E}{10}\times 100=\frac{1.6\times 10^3 kJ}{10}\times 100=1.6\times 10^4 kJ

Since, it is a energy produced in order to indicate the direction of heat produced we will use negative sign.

Q = -1.6\times 10^4 kJ

Moles of propane burnt to produce Q energy =n

n\times \Delta H_{rxn}=Q

n=\frac{Q}{\Delta H_{rxn}}=\frac{-1.6\times 10^4 kJ}{-2044 kJ/mol}=7.83 mol

According to reaction , 1 mol of propane gives 3 moles of carbon dioxide. then 7.83 moles of will give:

\frac{1}{3}\times 7.83 mol=23.49 mol carbon dioxide gas.

Mass of 23.49 moles of carbon dioxide gas:

23.49 mol × 44 g/mol =1,033.56 g

1,033.56 grams of carbon dioxide was emitted into the atmosphere.

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