1. A solution was made by dissolving 0.837g of Ba(OH)2 in 100 mL of solution. What is the pOH and pH of the solution? 


2. What is the percentage ionization of 1.0M acetic acid? What is the pH of this solution?

Answers

Answer 1

Answer:

1. pOH = 1.01 and the pH = 12.99.

2. a) %I = 0.417 %, b) pH =  2.34.  

Explanation:

1. The pH of the dissolution of Ba(OH)₂ in water can be calculated using the following equation:

[tex]pH = 14 - pOH = 14 - (-log[OH^{-}]) = 14 + log[OH^{-}][/tex]

To find the pOH, first, we need to calculate the concentration (C) of Ba(OH)₂:

[tex] C_{Ba(OH)_{2}} = \frac{\eta}{V} = \frac{m}{M*V} [/tex]

Where:

η: is the number of moles of Ba(OH)₂

V: is the volume of the solution = 100 ml = 0.100 L

M: is the molar mass of Ba(OH)₂ = 171.34 g/mol

m: is the mass of Ba(OH)₂ = 0.837 g          

[tex] C_{Ba(OH)_{2}} = \frac{m}{M*V} = \frac{0.837 g}{171.34 g/mol*0.100 L} = 0.049 mol/L [/tex]  

Knowing that in 1 mol of Ba(OH)₂ we have two moles of OH⁻, the concentration of OH⁻ is:

[tex] C_{OH^{-}} = \frac{2*\eta_{Ba(OH)_{2}}}{V} = 2*C_{Ba(OH)_{2}} = 2*0.049 mol/L = 0.098 mol/L [/tex]

Now, we can find the pOH and the pH of the solution:

[tex] pOH = -log[OH^{-}] = -log(0.098) = 1.01 [/tex]

[tex] pH = 14 - pOH = 14 - 1.01 = 12.99 [/tex]

2. a) The percentage of ionization (% I) of acetic acid can be calculated using the following equation:

[tex]\% I = \frac{[H^{+}]_{eq}}{[HA]_{0}}*100[/tex]

We need to find the [H⁺] = [H₃O⁺] at the equilibrium:

CH₃COOH(aq) + H₂O(l) → CH₃COO⁻(aq) + H₃O⁺(aq)  

 1.0 - x                                        x                   x

[tex] Ka = \frac{[CH_{3}COO^{-}]*[H_{3}O^{+}]}{[CH_{3}COOH]} [/tex]

Ka: is the dissociation constant of acetic acid = 1.75x10⁻⁵

[tex] 1.75\cdot 10^{-5} = \frac{x^{2}}{1.0 - x} [/tex]

[tex] 1.75\cdot 10^{-5}(1.0 - x) - x^{2} = 0 [/tex]     (1)  

By solving equation (1) for x we have two solutions:

x₁ = -0.00417

x₂ = 0.00417 = [H₃O⁺] = [CH₃COO⁻]  

We will take the positive value to find the percentage ionization:          

[tex]\% I = \frac{[H^{+}]_{eq}}{[HA]_{0}}*100 = \frac{0.00417 M}{1.0 M}*100 = 0.417 \%[/tex]    

b) The pH of the solution is:

[tex] pH = -log([H_{3}O^{+}]) = -log(0.00417) = 2.34 [/tex]

I hope it helps you!          


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Answer:

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Answer:

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Answer:

Complete ionic equation

Explanation:

For every reaction, we could write the molecular equation as well as the ionic equation. The net ionic equation only shows the reaction of the main ions that take part in the reaction. It does not show the reaction of spectator ions.

In order to write the complete ionic equation for any reaction, we must observe the following steps;

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Answers

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Explanation:

According to ideal gas equation:

[tex]PV=nRT[/tex]

P = pressure of gas = 901 mm Hg = 1.18 atm   (760 mm Hg = 1 atm)

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According to stoichiometry:

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Answer:

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Answer:

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Answer:

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Answer:

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Answers

Answer:

approximately 15.1 grams.

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The key to chemistry is to change everything to moles. Then when you have the answer in moles change the answer back to grams, liters, or whatever you want.

change 25 grams of potassium chlorate to moles.

calculate the gram molecular mass of potassium chlorate.

Chlorate is Cl with 3 oxygens. ate = saturated. Chlorine has seven valance electrons when it is saturated six of these electrons are used by oxygen ( 2 electrons per oxygen) leaving only 1 electron.

1 K x 39 grams/mole

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+3 O x 16 grams/ mole

= 122.4 grams / mole Potassium Chlorate

25

122.4

= moles.

2.05 moles of Potassium Chlorate.

There is a 1:1 mole ratio. 1 mole of Potassium Chlorate will produce 1 mole of Potassium Chloride.

2.05 moles of Potassium Chlorate will produce 2.05 moles of Potassium Chloride.

Find the gram molecular mass of Potassium Chloride.

1 K x 39 = 39

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2.05 moles x 74.4 grams/ mole = 15.2 grams

Hope it helps :)

Answer:

9.79g of O2

Explanation:

So this question you are trying to go from g of KCLO3 to grams of oxygen. There is no direct conversion so you have to utilize multiple process.

[tex]25.0g KCLO3 * \frac{1mol KCLO3}{122.55 gKCLO3} * \frac{3mol O2}{2 mol KCLO3} * \frac{32.00g O2}{1mol O2}[/tex]

=9.79 g of O2

So first you go from g of KCLO3 to moles of KCLO3  by using molar mass of KCLO3. Then you use mol:mol ratio from balanced equation to go to moles of O2, and finally use molar mass of O2 to go to grams of O2.

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Answer:

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edg2020

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Answer:

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Answer:u got me stumped

Explanation:

How many ml of solution are needed to make a 5.78 M solution with 2.98 moles of CaO?

Answers

Answer:

515.6 mL

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

The number of mL of the solution needed to make a 5.78 M solution with 2.98 moles of CaO is 516 mL.

The number of moles that is present in a liter of solution is regarded to be the Molarity of the solution that is measured in mol/L.

[tex]\mathbf{Molarity = \dfrac{number\ of \ moles(n)}{Volume \ (in \ liters)}}[/tex]

[tex]\mathbf{Molarity = \dfrac{2.98 moles }{5.78 \ mol/ L}}[/tex]

Molarity = 0.51557 L

If 1 Liter of solution contains 1000 mL, then 0.51557 L will contain:

= 0.516 × 1000 mL

= 516 mL of solution.

Therefore, we can conclude that the number of mL of the solution needed to make a 5.78 M solution with 2.98 moles of CaO is 516 mL.

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Time
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Half-Life
Minutes
Predicted
Simulated
Cycles, n
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0.0
100
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100
1
100
0.5
1
2.
1.0
1.5
3
2.0
4
2.5
5
3.0
6
3.5
7
4.0
8
id

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Answer:

The 2nd one

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Answer: I don't have the last column though.

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Answers

Answer:

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d





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