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Scorpion4ik [409]
3 years ago
15

What is the ph of a solution of 0.50 m acetic acid?

Chemistry
1 answer:
frosja888 [35]3 years ago
6 0
You need to use the Ka for the acetic acid and the equilibrium equation.

Ka = 1.85 * 10^ -5

Equilibrium reaction: CH3COOH (aq) ---> CH3COO(-) + H(+)

Ka = [CH3COO-][H+] / [CH3COOH]

Molar concentrations at equilibrium

CH3COOH         CH3COO-     H+

 0.50  - x                  x                 x

Ka = x*x / (0.50 - x) = x^2 / (0.50 - x)

Given that Ka is << 1 => 0.50 >> x and 0.50 - x ≈ 0.50

=> Ka ≈ x^2 / 0.50

=> x^2 ≈ 0.50 * Ka = 0.50 * 1.85 * 10^ -5 = 0.925 * 10^ - 5 = 9.25 * 10 ^ - 6

=> x = √ [9.25 * 10^ -6] = 3.04 * 10^ -3 ≈ 0.0030

pH = - log [H+] = - log (x) = - log (0.0030) = 2.5

Answer: 2.5
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Answer:

47911.1 pa

Explanation:

The SI base unit of pressure is pascal, which is N/m^2.

2200 kg is 2200*9.8=21560 N, and 4500 cm^2=4500/10000=0.45 m^2.

So the total pressure exerted on the ground (!!) is 21560/0.45= 47911.1 Pa.

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Which term represents the fixed proportion of elements in a compound
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Chemical formula.

Explanation:

A chemical formula have fix proportion of atoms of elements.

Chemical formula:

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3 years ago
Which is equal to a temperature of 50°F?<br> O 18°C<br> O 46°C<br> O 10°C<br> O 32°C
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C) 10°C

50-32= 18
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8 0
3 years ago
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The last intermediate in citric acid cycle is:_________
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The last intermediate in citric acid cycle is Oxaloacetic acid.

<h3>What is Citric Acid Cycle?</h3>

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Every year, more than two million tons of citric acid are produced. It is frequently used as a flavoring, an acidifier, and a chelating agent.

Citrates, which include salts, esters, and the polyatomic anion present in solution, are derivatives of citric acid. Trisodium citrate is an example of the former; triethyl citrate is an example of an ester.

Learn more about citric acid with the help of the given link:

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7 0
10 months ago
Calcium nitrate and ammonium fluoride react to form calcium fluoride, dinitrogen monoxide, and water vapor. How many grams of di
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Answer:

16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant. Explanation:

Ca(NO_3)_2(aq)+2NH_4F(aq)\rightarrow CaF_2(aq)+2N_2O(g)+4H_2O(g)

Moles of calcium nitrate = \frac{31.3 g}{164 g/mol}=0.1908 mol

Moles of ammonium fluoride = \frac{38.7 g}{37 g/mol}=1.046 mol

According to reaction , 2 moles of ammonium fluoride reacts with 1 mole of calcium nitrate.

Then 1.046 moles of ammonium fluoride will react with :

\frac{1}{2}\times 1.046 mol=0.523 mol calcium nitarte .

This means that ammonium fluoride is in excess amount and calcium nitrate is in limiting amount.

Hence, calcium nitrate is a limiting reactant.

So, amount of dinitrogen monoxide will depend upon moles of calcium nitrate.

So, according to reaction , 1 mole of calcium nitarte gives 2 moles of dinitrogen monoxide gas .

Then 0.1908 moles of calcium nitrate will give:

\frac{2}{1}\times 0.1908 mol=0.3816 molof dinitrogen monoxide gas.

Mass of 0.03816 moles of dinitrogen monoxide gas:

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16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant.

8 0
3 years ago
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