All categories

2 years ago

What is the molar solubility in water of ag2cro4 ? (the ksp for ag2cro4 is 8 x 10-12. )

Chemistry

2 answers:

nikitadnepr [17]2 years ago

4 0

The molar solubility in water of Ag₂CrO₄ is 2.52 * 10⁻⁴ M .

<h3>How do you define Solubility ?</h3>

The measure of the degree to which a substance gets dissolved in a solvent to become a solution.

The formula for determination of molar solubity is

$\rm K_{sp} = [A^+]^a [B^-]^b$

$\rm K_{sp}$ = solubility product constant

$\rm [A^+]^a$ = cation in an aqueous solution

$\rm [B^-]^b$ = anion in an aqueous solution

a, b = relative concentrations of a and b

We have the data given as

$\rm K_{sp}$ for Ag₂CrO₄ is 8 x 10⁻¹²

For Silver chromate , the solubility would be only one-half of the Ag⁺ concentration.

We denote the solubility of Ag₂CrO₄ as S mol L⁻¹.

Then for a saturated solution, we have

[Ag⁺]=2S

[CrO₂⁴⁻]=S

the relation between the solubility and the solubility product constant depends on the stoichiometry of the dissolution reaction

(2S)²(S)=4S³ = 8 * 10⁻¹²

S=∛ (2*10⁻¹²)

S= 1.26 * 10⁻⁴ M

2S = 2.52 * 10⁻⁴ M

Therefore the molar solubility in water of Ag₂CrO₄ is 2.52 * 10⁻⁴ M .

To know more about Solubility

brainly.com/question/8591226

#SPJ4

qaws [65]2 years ago

3 0

Answer:

¿Cuál será el pH de una disolución saturada de hidróxido de cinc?.

Explanation:

Dato: Kps (Zn(OH)2) = 1,8 x 10-14. Solución: 9,52.

You might be interested in

The first excited electronic energy level of the helium atom is 3.13 x 10-18 J above the ground level. Estimate the temperature

salantis [7]

Answer:

75603.86473 K

Explanation:

Given that:

The 1st excited electronic energy level of He atom = 3.13 × 10⁻¹⁸ J

The objective of this question is to estimate the temperature at which the ratio of the population will be 5.0 between the first excited state to the ground state.

The formula for estimating the ratio of population in 1st excited state to the ground state can be computed as:

$\dfrac{N_2}{N_1} = e ^{^{-\dfrac{(E_2-E_1)}{KT}}} = e ^{^{-\dfrac{(\Delta E)}{KT}}}$

From the above equation:

Δ E = energy difference = 3.13 × 10⁻¹⁸ J

k = Boltzmann constant = 1.38 × 10⁻²³ J/K

$\dfrac{N_2}{N_1} = 0.5$

Thus:

$0.05 =e^{^{ -\dfrac{3.13 \times 10^{-18} \ J}{1.38\times 10^{-23 \ J/K}\times T}}}$

$In (0.05) = { -\dfrac{3.13 \times 10^{-18} \ J}{1.38\times 10^{-23 \ J/K}\times T}}}$

$-3.00 = { -\dfrac{3.13 \times 10^{-18} \ J}{1.38\times 10^{-23 \ J/K}\times T}}}$

$-3.00 = -226811.5942 \times \dfrac{1}{T}$

$T = \dfrac{-226811.5942}{-3.00 }$

T = 75603.86473 K

4 0

3 years ago

I need help asap! !!

tekilochka [14]

The answer is the answer that answers this question which gives u the answer

3 0

3 years ago

Match the action to the effect on the equilibrium position for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g). Match Term Definition Remo

pshichka [43]

Answer:

The answers to the questions are given below.

Explanation:

According to Le Chatelier's principle, if an external constrain such as change in concentration, temperature or pressure is imposed on a chemical system in equilibrium, the equilibrium will shift in order to neutralize the effect.

A. Effective of removing ammonia, NH3.

N2(g) + 3H2(g) ⇌ 2NH3(g)

Removing NH3 from the reaction simply means we are left with more reactants and no product. Therefore, the reactant will react to produce the product. Hence, the equilibrium position will shift to the right.

2. Effect of removing H2

N2(g) + 3H2(g) ⇌ 2NH3(g)

Remoing H2 simply means we have more products and less reactant. Therefore, the product will be convert to reactant. Hence, the equilibrium position will shift to the left.

C. Effect of adding a catalyst.

N2(g) + 3H2(g) ⇌ 2NH3(g)

Catalyst does not affect the equilibrium position. It only creates an alternative path to arrive at the product within a short time. Hence, it has no effect.

7 0

4 years ago

How is Oxygen -18 different from oxygen -16?

STatiana [176]

Answer:The only one difference is that ozone is made up of three oxygen atoms, while the stuff we breathe is made up of only two atoms.

Explanation:

8 0

3 years ago

A 8 kg cat is running 4 m/s. How much kinetic energy does it have

Debora [2.8K]

<span>KE = 1/2mv^2
KE = 1/2(8)4 m/s^2
KE = 4*4
KE = 16 Joules

Kinetic energy would equal 16 J </span>

8 0

3 years ago