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RUDIKE [14]
3 years ago
5

What is the change in enthalpy associated with the combustion of 530 g of methane (CH4)?Report your answer in scientific notatio

n.Your answer should have two significant figures.Use −890.8kJmol for the molar heat of combustion of methane.
Chemistry
1 answer:
Veseljchak [2.6K]3 years ago
7 0

Answer:

-3.0\times 10^4 kJ is the change in enthalpy associated with the combustion of 530 g of methane.

Explanation:

CH_4+2O_2\rightarrow CO_2+2H_2O,\Delta H_{comb}=-890.8 kJ/mol

Mass of methane burnt = 530 g

Moles of methane burnt = \frac{530 g}{16 g/mol}=33.125 mol

Energy released on combustion of 1 mole of methane = -890.8 kJ/mol

Energy released on combustion of 33.125 moles of methane :

-890.8 kJ/mol\times 33.125 mol=-29,507.75 kJ=-2.950775\times 10^4 kJ

-2.950775\times 10^4 kJ\approx -3.0\times 10^4 kJ

-3.0\times 10^4 kJ is the change in enthalpy associated with the combustion of 530 g of methane.

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Empirical formula: The formula consist of proportions of the elements which is present in the compound or the simplest whole number ratios of atoms.

Now, molecular formula is equal to the product of n (ratio) and empirical formula.

Molecular formula = n\times empirical formula    (1)

molecular formula =C_{12}H_{24}O_{6} (given)

Since, 6 is the smallest subscript in above molecular formula to get the simpler whole number of atoms. Therefore, divide all the subscripts i.e. number of carbon atoms (12), number of hydrogen atoms (24) and number of oxygen atoms (6) by 6.

empirical formula becomes C_{2}H_{4}O

Thus, according to the formula (1)

C_{12}H_{24}O_{6} = 6\times C_{2}H_{4}O

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2 years ago
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If you initially have a volume of 4 L and a temperature of 300K. Then you decreased the volume to 2 L, what is the new temperatu
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Answer:

T₂ = 150 K

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According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

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T₁ = Initial temperature

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T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = T₁V₂/V₁  

T₂ = 300 K × 2L / 4 L

T₂ = 600 L.K / 4 L

T₂ = 150 K

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3 years ago
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