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Alina [70]
2 years ago
9

Describe are seasons like near the equator.

Chemistry
1 answer:
adell [148]2 years ago
3 0

Answer:

Hi! Think of it like this:

You and your family are going to go on a vaction for the holidays, to south america or somewhere else. You´re near the equator so it will be warm. If you were to move there, you might pull out the swimsiutes because most of the time, its useally warm and sunny. They dont really get cold weather. They will have a nice and warm christmas.

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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
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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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