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Airida [17]
4 years ago
15

The ka of benzoic acid is 6.30 ⋅ 10-5. The ph of a buffer prepared by combining 50.0 ml of 1.00 m potassium benzoate and 50.0 ml

of 1.00 m benzoic acid is ________.
Chemistry
1 answer:
Free_Kalibri [48]4 years ago
6 0
<h3><u>Answer;</u></h3>

pH = 4.20

<h3><u>Explanation;</u></h3>

pKa = -log(6.30 × 10^-5)

pKa = 4.20  

Moles of Benzoic acid = volume × molarity

                                      = 0.050L × 1.00M

                                      = 0.050moles benzoic acid  

Moles of the salt = 0.050L ×  1.00M

                           = 0.050 moles salt  

Therefore;

0.050mols / 0.1 L = 0.50M  

0.050mols / 0.1 L = 0.50M  

Thus;

pH = 4.20 + log(0.50/0.50)

pH = 4.20

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= 529.9422

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Explanation:

5  moles Sodium Carbonate

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For which of the following processes would you expect there to be an increase in entropy? Ag+(aq) + Cl-(aq) AgCl(s) H2O(g) H2O(l
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When 0.250 moles of a gas is placed in a container at 298K, it exerts a pressure of 93kPa. What is the volume of the container?
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On combustion, 1.0 L of a gaseous compound of hydrogen, carbon, and sulfur gives 2.0 L of CO2, 3.0 L of H2O vapor, and 1.0 L of
Murrr4er [49]

Answer:

The empirical formula of the organic compound is  = C_2H_6S_1

Explanation:

At STP, 1 mole of gas occupies 22.4 L of volume.

Moles of CO_2 gas at STP occupying 2.0 L = n

n\times 22.4L=2.0L

n=\frac{2.0 L}{22.4 L}=0.08929 mol

Moles of carbon in 0.08920 mol = 1 × 0.08920 mol = 0.08920 mol

Moles of H_2O gas at STP occupying 3.0 L = n'

n'\times 22.4L=3.0L

n'=\frac{3.0 L}{22.4 L}=0.1339 mol

Moles of hydrogen in 0.1339 moles of water vapor = 2 × 0.1339 mol = 0.2678 mol

Moles of SO_2 gas at STP occupying 1.0 L = n''

n''\times 22.4L=1.0L

n''=\frac{1.0 L}{22.4 L}=0.04464 mol

Moles of sulfur in 0.04464 mol = 1 × 0.04464 mol = 0.04464 mol

Moles of carbon , hydrogen and sulfur constituent of that organic compound .

Moles of carbon in 0.08920 mol = 1 × 0.08920 mol = 0.08920 mol

Moles of hydrogen in 0.1339 moles of water vapor = 2 × 0.1339 mol = 0.2678 mol

Moles of sulfur in 0.04464 mol = 1 × 0.04464 mol = 0.04464 mol

For empirical; formula divide the least number of moles from all the moles of elements.

carbon = \frac{0.08920 mol}{0.04464 mol}=2

Hydrogen =  \frac{0.2678 mol}{0.04464 mol}=6

Sulfur = \frac{0.04464 mol}{0.04464 mol}=1

The empirical formula of the organic compound is  = C_2H_6S_1

3 0
3 years ago
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