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Submit What is the solubility of Cd3(POA) 2 in water? (Ksp of Cd3(PO4)2 is 2.5 x 10-33) | 1 2 3 +/- . 0 x100

vesna_86 [32]

<u>Answer:</u> The solubility of $Cd_3(PO_4)_2$ in water is $1.18\times 10^{-7}mol/L$

<u>Explanation:</u>

The balanced equilibrium reaction for the ionization of cadmium phosphate follows:

$Cd_3(PO_4)_2\rightleftharpoons 3Cd^{2+}+2PO_4^{3-}$

3s 2s

The expression for solubility constant for this reaction will be:

$K_{sp}=[Cd^{2+}]^3[PO_4^{3-}]^2$

We are given:

$K_{sp}=2.5\times 10^{-33}$

Putting values in above equation, we get:

$2.5\times 10^{-33}=(3s)^3\times (2s)^2\\\\2.5\times 10^{-33}=108s^5\\\\s=1.18\times 10^{-7}mol/L$

Hence, the solubility of $Cd_3(PO_4)_2$ in water is $1.18\times 10^{-7}mol/L$

6 0

3 years ago

Why do your ears “pop” in an airplane?

tino4ka555 [31]

Pressure buildup in ur ear

8 0

3 years ago

Which description best characterizes the motion of particles in a solid?

antoniya [11.8K]

The particles in a solid are tightly packed and locked in place. Although we cannot see it or feel it, the particles are vibrating in place.

As these molecules heat up, they will vibrate more vigorously, and will eventually turn to water, then gas.

5 0

3 years ago

Read 2 more answers

Under identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simulta

scoray [572]

Answer:

70.77 g/mol is the molar mass of the unknown gas.

Explanation:

Effusion is defined as rate of change of volume with respect to time.

Rate of Effusion= $\frac{Volume}{Time}$

Effusion rate of oxygen gas after time t = $E=\frac{4.64 mL}{t}$

Molar mass of oxygen gas = M = 32 g/mol

Effusion rate of unknown gas after time t = $E'=\frac{3.12 mL}{t}$

Molar mass of unknown gas = M'

The rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:

$\text{Rate of effusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

$\frac{E}{E'}=\sqrt{\frac{M'}{M}}$

$\frac{\frac{4.64 mL}{t}}{\frac{3.12 mL}{t}}=\sqrt{\frac{M'}{32 g/mol}}$

M' = 70.77 g/mol

70.77 g/mol is the molar mass of the unknown gas.

8 0

3 years ago

Determine the percent ionization of a 0.225 M solution of benzoic acid. Express your answer using two significant figures.

Assoli18 [71]

Answer:

1.68% is ionized

Explanation:

The Ka of benzoic acid, C₇H₆O₂, is 6.46x10⁻⁵, the equilibrium in water of this acid is:

C₇H₆O₂(aq) + H₂O(l) ⇄ C₇H₅O₂⁻(aq) + H₃O⁺(aq)

Ka = 6.46x10⁻⁵ = [C₇H₅O₂⁻] [H₃O⁺] / [C₇H₆O₂]

<em>Where [] are concentrations in equilibrium</em>

In equilibrium, some 0.225M of the acid will react producing both C₇H₅O₂⁻ and H₃O⁺, the equilibrium concentrations are:

[C₇H₆O₂] = 0.225-X

[C₇H₅O₂⁻] = X

[H₃O⁺] = X

Replacing:

6.46x10⁻⁵ = [X] [X] / [0.225-X]

1.4535x10⁻⁵ - 6.46x10⁻⁵X = X²

1.4535x10⁻⁵ - 6.46x10⁻⁵X - X² = 0

Solving for X:

X = -0.0038. False solution, there is no negative concentrations.

X = 0.00378M. Right solution.

That means percent ionization (100 times Amount of benzoic acid ionized over the initial concentration of the acid) is:

0.00378M / 0.225M * 100 =

<h3>1.68% is ionized</h3>

8 0

3 years ago