3 years ago

PLEASE HELPPPP

Chemistry

1 answer:

USPshnik [31]3 years ago

6 0

Answer:

explain futher pls

Explanation:

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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)

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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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