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1 year ago

The molar solubility of cui is 2. 26 × 10-6 m in pure water. Calculate the ksp for cui.

Chemistry

1 answer:

nataly862011 [7]1 year ago

7 0

Ksp(the solubility product constant) = [Cu⁺] [I⁻]

So, the Ksp for Cui would be:

Ksp = (2.26 × 10⁻⁶) (2.26 × 10⁻⁶) = 5.11 x 10⁻¹²

<h3 />

Formula used:

K = $K_{sp} = [A^+]^a [B^-]^b$ , where

Ksp = solubility product constant

A⁺ = cation in an aquious solution

B⁻ = anion in an aqueous solution

a, b = relative concentrations of a and b

<h3>Definition</h3>

The equilibrium constant for a solid material dissolving in an aqueous solution is the SOLUBILITY PRODUCT CONSTANT, Ksp. It stands for the degree of solute dissolution in solution. A substance's Ksp value increases with how soluble it is.

Take into account the general dissolving response (in aqueous solutions) below:

aA(s)⇔cC(aq)+dD(aq)

The molarities or concentrations of the products (cC and dD) must be multiplied in order to find the Ksp. Any product that has a coefficient in front of it must be raised to the power of that coefficient (and also multiply the concentration by that coefficient).

Learn more about the concept of Ksp through the link:

brainly.com/question/1419865

#SPJ4

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Despite the fact that the partial pressure difference is so much smaller for co2, why is there as much co2 exchanged between the

Snezhnost [94]

This is due to the solubility of CO₂ in blood is more than the solubility of Oxygen in blood

7 0

3 years ago

At a certain temperature, 0.740 mol of so3 is placed in a 4.00 l container. so_{3}(g)\rightleftharpoons 2so_{2}(g)+o_{2}(g) at e

Phoenix [80]

Answer : The equilibrium constant kc is 4.76 x 10⁻³

Explanation :

The given equilibrium reaction is

$2SO_{3} (g) \leftrightarrow 2SO_{2} (g) + O_{2} (g)$

Step 1 : Set up ICE table

Let us set up an ICE table for this reaction .

The initial concentration of SO₃ is

$Concentration = \frac{mol}{L} = \frac{0.740mol}{4L} = 0.185 M$

The initial concentrations of products are 0.

Let us assume x is the change .

Please refer the attached picture.

Step 2 : Use the given value to find x

From the ICE table, we can see that at equilibrium, concentration of O₂ is x

But we have been given that , at equilibrium we have 0.190 mol of O₂ .

Let us convert this to concentration unit.

Concentration of O₂ at equilibrium = $\frac{mol}{L} = \frac{0.190mol}{4L} = 0.0475 M$

But concentration of O₂ from the ICE table is x.

Therefore we have x = 0.0475 M

Step 3 : Using x , find equilibrium concentrations

Using this value, let us write the equilibrium concentrations of the given species.

[SO₃]eq = 0.185 M - 2x = 0.185 - 2(0.0475) = 0.09 M

[SO₂]eq = 2x = 0.095 M

[O₂]eq = x = 0.0475 M

Step 4 : Set up equation for kc and solve it

The equilibrium constant kc is calculated as,

$k_{c} = \frac{[SO_{2}]^{2} [O_{2}]}{[SO_{3}]^{2}}$

Let us plug in the above equilibrium values.

$k_{c} = \frac{(0.095)^{2} (0.0475)}{(0.09)^{2}}$

$k_{c} = \frac{0.00042869}{0.09}$

$k_{c} = 4.76 \times 10^{-3}$

The equilibrium constant kc is 4.76 x 10⁻³

3 0

3 years ago

Hydrogen chloride gas is shipped in a container under 5,100 mm Hg of pressure that occupies 20.1 liters at 29°C. How many liters

Agata [3.3K]

The gas will occupy 120 L at STP.

We can use the Combined Gas Laws .

In 1982, chemists redefined STP as 1 bar and 0 °C (273.15 K).

p_1 = 5100 mmHg × (1 atm/760 mmHg) × (1.013 bar/1 atm) = 6.798 bar

T_1 = (29 + 273.15) K = 302.15 K

p_1V_1/T_1 = p_2V_2/T_2

V_2 = V_1 × p_1/p_2 × T_2/T_1

= 20.1 L × (6.798 bar/1 bar) × (273.15 K/302.15 K) = 120 L

7 0

3 years ago

To measure the amount of calcium carbonate in a seashell, an analytical chemist crushes a sample of the shell to a fine powder a

andreyandreev [35.5K]

The given question is complete, the complete question is:

To measure the amount of calcium carbonate (CaCO) in a seashell, an analytical chemist crushes a 4.80 g sample of the shell to a fine powder and titrates it to the endpoint with 515. mL of 0.140 M hydrogen chloride (HCl) solution. The balanced chemical equation for the reaction is: 2HCI(a)Co (a) H2Co,(aq) + 2Cl (aq)

What kind of reaction is this?

If you said this was a precipitation reaction, enter the chemical formula of the precipitate.

If you said this was an acid-base reaction, enter the chemical formula of the reactant that is acting as the base.

If you said this was a redox reaction, enter the chemical symbol of the element that is oxidized

Calculate the mass percent of CaCO in the sample. Be sure your answer has the correct number of significant digits.

Answer:

It is an acid-base reaction and the mass percent of CaCo3 is 75.2%.

Explanation:

A chemical reaction in which an insoluble salt produces from two soluble salts is termed as precipitation reaction.

A reaction in which atleast exchange of one proton takes place between the two species is termed as an acid-base reaction.

A reaction in which any of the element has a change in oxidation state is termed as redox reaction.

In the mentioned reaction, there is a transfer of H⁺, no precipitation is forming, and no change in oxidation state taking place, thus, it is an acid-base reaction.

In the acid-base reaction, the base refers to the species that accepts hydrogen ion or proton. In the given case, CO₃²⁻ is accepting H+ ion to become H₂CO₃. Hence, CO₃²⁻ is the base.

In order to calculate mass percent of CaCO₃, first there is a need to find the moles of HCl reacted for a solution,

Moles = Molarity × Volume (L)

Moles = 0.140 mol/L × 515 × 10⁻³ L

Moles = 0.0721 mol

Now from the balanced equation, one mole of CaCO₃ needs two moles of HCl.

So, moles of CaCO₃ reacted will be,

= 1/2 × 0.0721 = 0.03605 mol

The mass of calcium carbonate taking part in reaction will be,

= Moles × Molar mass

= 0.03605 × 100 gram/mole

= 3.6086 gram

Mass% of CaCO₃ = Mass of CaCO₃/Mass of sample × 100

= 3.6086 grams/4.80 grams × 100

Mass % = 75.2%

3 0

3 years ago

Which of the following is not an example of a molecule A. O₃ B. F C. H₂O₂ D. NCI₃

exis [7]

The correct option is B. F

Hope this helps

-AaronWiseIsBae

4 0

4 years ago