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Pachacha [2.7K]
2 years ago
10

A chemist dissolves of pure potassium hydroxide in enough water to make up of solution. Calculate the pH of the solution. (The t

emperature of the solution is .) Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
Leya [2.2K]2 years ago
8 0

Answer:

12.99

Explanation:

<em>A chemist dissolves 716. mg of pure potassium hydroxide in enough water to make up 130. mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 °C.) Be sure your answer has the correct number of significant digits.</em>

Step 1: Given data

  • Mass of KOH: 716. mg (0.716 g)
  • Volume of the solution: 130. mL (0.130 L)

Step 2: Calculate the moles corresponding to 0.716 g of KOH

The molar mass of KOH is 56.11 g/mol.

0.716 g × 1 mol/56.11 g = 0.0128 mol

Step 3: Calculate the molar concentration of KOH

[KOH] = 0.0128 mol/0.130 L = 0.0985 M

Step 4: Write the ionization reaction of KOH

KOH(aq) ⇒ K⁺(aq) + OH⁻(aq)

The molar ratio of KOH to OH⁻is 1:1. Then, [OH⁻] = 0.0985 M

Step 5: Calculate the pOH

We will use the following expression.

pOH = -log [OH⁻] = -log 0.0985 = 1.01

Step 6: Calculate the pH

We will use the following expression.

pH + pOH = 14

pH = 14 - pOH = 14 -1.01 = 12.99

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Explanation:

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A compound with molecular formula C6H15N exhibits a singlet at d 0.9 ppm (1H), a triplet at 1.10 ppm (3H), a singlet at 1.15 ppm
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Answer:

CH3 - CH2 - NH - C(CH3)3

Explanation:

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From the table, a singlet at 1.15 ppm (9H) is - C(CH3)3.

A singlet at d 0.9 ppm (1H) shows the presence of a secondary amine group, that is -R2NH group.

A triplet at 1.10 ppm (3H) shows that we have;

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Consider the second-order reaction:
kirza4 [7]

Answer:

Initial concentration of HI is 5 mol/L.

The concentration of HI after 4.53\times 10^{10} s is 0.00345 mol/L.

Explanation:

2HI(g)\rightarrow H_2(g)+I_2(g)&#10;

Rate Law: k[HI]^2&#10;

Rate constant of the reaction = k = 6.4\times 10^{-9} L/mol s

Order of the reaction = 2

Initial rate of reaction = R=1.6\times 10^{-7} Mol/L s

Initial concentration of HI =[A_o]

1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2

[A_o]=5 mol/L

Final concentration of HI after t = [A]

t = 4.53\times 10^{10} s

Integrated rate law for second order kinetics is given by:

\frac{1}{[A]}=kt+\frac{1}{[A_o]}

\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}

[A]=0.00345 mol/L

The concentration of HI after 4.53\times 10^{10} s is 0.00345 mol/L.

5 0
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