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Ganezh [65]
3 years ago
12

Calcium hydroxide is partially soluble as shown by the

Chemistry
1 answer:
Blizzard [7]3 years ago
4 0

Answer:

4

10

Explanation:

The reaction equation is given as;

            Ca(OH)₂     →    Ca²⁺   + 2OH⁻

Concentration of Ca(OH)₂ = 5 x 10⁻⁵M

         

Unknown:

pOH of the solution  = ?

pH of the solution = ?

Solution:

Solve for the pOH of this solution using the expression below obtained from the ionic product of water;

                 pOH  = ⁻log₁₀[OH⁻]

           Ca(OH)₂             →           Ca²⁺              +              2OH⁻

          1moldm⁻³                         1moldm⁻³                     2 x 1moldm⁻³

         5 x 10⁻⁵moldm⁻³         5 x 10⁻⁵moldm⁻³            2( 5 x 10⁻⁵moldm⁻³ )

                                                                                            1 x 10⁻⁴moldm⁻³

  Therefore;

             pOH = -log₁₀ 1 x 10⁻⁴ = 4

Since

           pOH + pH  = 14

                        pH  = 14  - 4  = 10

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How many moles of HNO3 are present in 450 g of HNO3?
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Which of the following statements describes the correct method of preparation of 1.00 L of a 2.0 M urea solution?
Zolol [24]

Answer:

To prepare 1.00 L of 2.0 M urea solution, we need to dissolve 120 g of urea in enough water to produce a total of 1.00 L solution

Explanation:

Molarity of a solute in a solution denotes number of moles of solute dissolved in 1 L of solution.

So, moles of urea in 1.00 L of a 2.0 M urea solution = 2 moles

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8 0
3 years ago
c. The reaction Br2 (l) --&gt; Br2 (g) has ΔH = 30.91 kJ/mol and ΔS = 93.3 J/mol·K. Use this information to show (within close a
egoroff_w [7]

Answer:

The answer to your question is given below.

Explanation:

From the question given above, the following data were obtained:

Br₂ (l) —> Br₂(g)

Enthalpy change (ΔH) = 30.91 KJ/mol

Entropy change (ΔS) = 93.3 J/mol·K

Boiling temperature (T) =?

Next, we shall convert 30.91 KJ/mol to J/mol. This can be obtained as follow:

1 KJ/mol = 1000 J/mol

Therefore,

30.91 KJ/mol = 30.91 × 1000

30.91 KJ/mol = 30910 J/mol

Thus, 30.91 KJ/mol is equivalent to 30910 J/mol.

Finally, we shall determine the boiling temperature of bromine. This can be obtained as follow:

Enthalpy change (ΔH) = 30910 J/mol

Entropy change (ΔS) = 93.3 J/mol·K

Boiling temperature (T) =?

ΔS = ΔH / T

93.3 = 30910 / T

Cross multiply

93.3 × T = 30910

Divide both side by 93.3

T = 30910 / 93.3

T = 331.29 K

Thus, the boiling temperature of bromine is 331.29 K

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