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3 years ago

Complete combustion of 5.90 g of a hydrocarbon produced 19.2 g of CO2 and 5.89 g of H2O. What is the empirical formula for the h

ydrocarbon?

Chemistry

1 answer:

scZoUnD [109]3 years ago

4 0

$44g \ CO_{2} \ \ \ \rightarrow \ \ 12g \ C\\
19,2g \ CO_{2} \rightarrow \ \ \ x\\\\
x=\frac{19,2g*12g}{44g}\approx5,24g \ \ \ \Rightarrow \ \ n=\frac{5,24g}{12\frac{g}{mol}}\approx0,44mol\\\\\\
18g \ H_{2}O \ \ \ \rightarrow \ \ \ 2g \ H\\
5,89g \ H_{2}O \ \rightarrow \ \ y\\\\
y=\frac{5,89g*2g}{18g}\approx0,65g \ \ \ \ \Rightarrow \ \ n=\frac{0,65g}{1\frac{g}{mol}}=0,65mol\\\\\\
n_{C}:n_{H}=0,44:0,65\approx1:1\\\\
CH$

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The pH of 0.10 M solution of an acid is 6. What is the acid dissociation constant of the acid?

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Explanation: Assuming the acid to be monoprotic, the reaction follows:

$HA\rightleftharpoons H^++A^-$

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$pH=-log([H^+])$

$[H^+]=antilog(-pH)$

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As HA ionizes into its ions in 1 : 1 ratio, hence

$[H^+]=[A^-]=10^{-6}M$

As the reaction proceeds, the concentration of acid decreases as it ionizes into its ions, hence the decreases concentration of acid at equilibrium will be:

$[HA]=[HA]-[H^+]$