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astra-53 [7]
2 years ago
9

How many grams Kl are needed to prepare 100.mL of 0.55M solution?​

Chemistry
1 answer:
Makovka662 [10]2 years ago
7 0

Answer:

Explanation:

100mL = 0.1L

0.55 M = mol/0.1 L

mol = 0.055 mol

molar mass of KI = 165.998 g

0.055 * 165.998 = 9.13 g of KI

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<h3>Answer:</h3>

5.71 × 10² nm

<h3>Explanation:</h3>

The product of wavelength and frequency of a wave gives the speed of the wave.

Therefore;

Velocity of wave = Wavelength × Frequency

c = f ×λ

In our case;

Frequency = 5.25 × 10^14 Hz

Speed of light = 2.998 × 10^8m/s

But;

λ = c ÷ f

  = 2.998 × 10^8m/s ÷  5.25 × 10^14 Hz

  = 5.71 × 10^-7 m

But; 1 M = 10^9 nm

Therefore;

wavelength = 5.71 × 10^-7 × 10^9

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The solubility product of PbI2 is 7.9x10^-9. What is the molar solubility of PbI2 in distilled water?
Agata [3.3K]

<u>Answer:</u> The molar solubility of PbI_2 is 1.25\times 10^{-3}mol/L

<u>Explanation:</u>

Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium.

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The balanced equilibrium reaction for the ionization of calcium fluoride follows:

PbI_2\rightleftharpoons Pb^{2+}+2I^-

                s       2s

The expression for solubility constant for this reaction will be:

K_{sp}=[Pb^{2+}][I^-]^2

We are given:

K_{sp}=7.9\times 10^{-9}

Putting values in above equation, we get:

7.9\times 10^{-9}=(s)\times (2s)^2\\\\7.9\times 10^{-9}=4s^3\\\\s=1.25\times 10^{-3}mol/L

Hence, the molar solubility of PbI_2 is 1.25\times 10^{-3}mol/L

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