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3 years ago

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An organic compound is 61.5% C, 2.56% H and 35.9% N by mass. 2.00 grams of this gas is entered into a 300.0 mL flask and heated

to 228 degrees C. At these conditions, the gas exerts a pressure of 2670 torr. Find the empirical formula.

1 answer:

pashok25 [27]3 years ago

6 0

Answer:

C₄H₂N₂

Explanation:

First we<u> calculate the moles of the gas</u>, using PV=nRT:

P = 2670 torr ⇒ 2670/760 = 3.51 atm

V = 300 mL ⇒ 300/1000 = 0.3 L

T = 228 °C ⇒ 228 + 273.16 = 501.16 K

3.51 atm * 0.3 L = n * 0.082atm·L·mol⁻¹·K⁻¹ * 501.16 K

n = 0.0256 mol

Now we<u> calculate the molar mass of the compound</u>:

2.00 g / 0.0256 mol = 78 g/mol

Finally we use the percentages given to<em> </em><u>calculate the empirical formula</u>:

C ⇒ 78 g/mol * 61.5/100 ÷ 12g/mol = 4

H ⇒ 78 g/mol * 2.56/100 ÷ 1g/mol = 2

N ⇒ 78 g/mol * 35.9/100 ÷ 14g/mol = 2

So the empirical formula is C₄H₂N₂

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A solid weak acid is weighed, dissolved in water and diluted to exactly 50.00 ml. 25.00 ml of the solution is taken out and is t

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Volume of NaOH $(V_2) = 18.8 \ mL$

Molarity of acid solution $(M_1) = ???$

Molarity of NaOH $(M_2) = 0.10 \ M$

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Making $M_1$ the subject of the formula; we have:

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Total Volume after titration = ( 25 + 18.8 ) m

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Molarity of salt (NaA ) solution = $\frac{number \ of \ moles}{Volume \ (mL)}$

= $\frac{1.88}{43.8}$

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After mixing the two solution ; the volume of half neutralize solution is = 25 mL + 43.8 mL

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Molarity of NaA before mixing $M_1 = 0.0429 \ M$

Volume $(V_1) = 43.8 \ mL$

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Volume $(V_2 ) = 68.8 \ mL$

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Volume = 25 mL

Molarity of acid after mixing = $\frac{0.0752*25}{68.8}$

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$pH = pKa + log \frac{[salt]}{[acid]}$

$3.51= pKa + log \frac{[0.0273]}{[0.0273]} \\ \\ 3.51= pKa + log \ 1 \\ \\ 3.51= pKa + 0 \\ \\ pKa = 3.51$

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