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lora16 [44]
3 years ago
12

Calculate the enthalpy of the formation of butane, C4H10, using the balanced chemical equation and the standard value below:

Chemistry
1 answer:
zavuch27 [327]3 years ago
6 0

Answer:

+125.4 KJmol-1

Explanation:

∆H C4H10(g) = -2877.6kJ/mol

∆H C(s)=-393.5kJ/mol

∆H H2(g) = -285.8

∆H reaction= ∆Hproducts - ∆H reactants

∆H reaction= (-2877.6kJ/mol) - [4(-393.5kJ/mol) +5(-285.8)]

∆H reaction= +125.4 KJmol-1

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If you have a solution that contains 46.85 g of codeine, C18H21NO3, in 125.5 g of ethanol, C2H5OH, what is the mole fraction of
irakobra [83]

Answer:

0.22

Explanation:

Given, Mass of C_{18}H_{21}NO_3 = 46.85 g

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The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{46.85\ g}{299.4\ g/mol}

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Molar mass of C_{2}H_{5}OH = 46.07 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{125.5\ g}{46.07\ g/mol}

Moles\ of\ C_{2}H_{5}OH= 0.5535\ mol

So, according to definition of mole fraction:

Mole\ fraction\ of\ codeine=\frac {n_{codeine}}{n_{codeine}+n_{ethanol}}

Mole\ fraction\ of\ codeine=\frac{0.1565}{0.1565+0.5535}=0.22

4 0
3 years ago
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2 years ago
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Answer:

3 atoms of carbon

Explanation:

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