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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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slega [8]

Answer:

yes

Explanation:

they di because they have mass and wait

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3 years ago

After he conducted cathode ray tube experiments proving the existence of negatively charged particles we now call electrons, Tho

Lina20 [59]

Answer:

Answer is explained below;

Explanation:

In 1904, after the discovery of the electron, the English physicist Sir J.J. Thomson proposed the plum pudding model of an atom. In this model, the atom had a positively-charged space with negatively charged electrons embedded inside it i.e., like a pudding (positively charged space) with plums (electrons) inside.

In 1911, another physicist Ernest Rutherford proposed another model known as the Rutherford model or planetary model of the atom that describes the structure of atoms. In this model, the small and dense atom has a positively charged core called the nucleus. Also, he proposed that just like the planets revolving around the Sun, the negatively charged electrons are moving around the nucleus.

By conducting a gold foil experiment, Rutherford disproved Thomson's model. In this experiment, positively charged alpha particles emitted from a radioactive source enclosed within a protective lead were used which was then focused into a narrow beam. It was then passed through a slit in front of which a thin section of gold foil was placed. A fluorescent screen (coated with zinc sulfide) was also placed in front of the slit to detect alpha particles which on striking the fluorescent screen would produce scintillation (a burst of light) which was visible through a microscope attached to the back of the screen.

He observed that most of the alpha particles passed straight through the gold foil without any resistance and this implied that atoms contain a large amount of open space. The slight deflection of some of the alpha particles, the large-angle scattering of other alpha particles and even the bouncing back of a very few alpha particles toward the source suggested their interactions with other positively charged particles inside the atom.

So, he concluded that only a dense and positively charged particle such as the nucleus would be responsible for such strong repulsion. Also, the negatively charged electrons electrically balanced the positive nuclear charge and they moved around the nucleus in circular orbits. Between the electrons and nucleus, there was an electrostatic force of attraction just like the gravitational force of attraction between the sun and the revolving planets.

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6 0

3 years ago

Calculate the mass in grams in nine molecules of CH3COOH? Please show how you got your answer.

WINSTONCH [101]

M CH₃COOH: 12u×2 + 1u×4 + 16u×2 = 60u

m 9CH₃COOH: 60u×9 = 540u

(1u ≈ 1,66·10⁻²⁴g)
-----------------------------
1u ------- 1,66·10⁻²⁴g
540u ---- X
X = 540×1,66·10⁻²⁴g
X = 896,4·10⁻²⁴g

7 0

3 years ago

A pump contains 0.5 L of air at 203kPa. You draw back on the piston of the pump until the pressure reads 25.4kPa. What is the vo

densk [106]

<h3> Answer;</h3>

= 4.0 L

<h3>Explanation;</h3>

Boyle's law states that the volume of a fixed mass of a gas is inversely proportional to pressure at a constant temperature.

Therefore; Volume α 1/pressure

Mathematically; V α 1/P

V = kP, where k is a constant;

P1V1 = P2V2

V1 = 0.5 l, P1 =203 kPa, P2 = 25.4 kPa

V2 = (0.5 × 203 )/25.4 

 = 3.996 

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6 0

3 years ago

Given the following data: Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔH = −23 kJ 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ΔH = −39

nalin [4]

Answer:

ΔH° = -11 kj

Explanation:

Step 1: Data given

1) Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔH = −23 kJ

2) 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ΔH = −39 kJ

3) Fe3O4(s) + CO(g) → 3 FeO(s) + CO2(g) ΔH = +18 kJ

Step 2: The balanced equation

FeO + CO → Fe + CO2

Step 3: Calculate ΔH for the reaction FeO(s) + CO(g) → Fe(s) + CO2(g).

To get this equation, we need to combine the 3 equations

We have to multiply the third equation by 2.

2Fe3O4(s) + 2CO(g) → 6 FeO(s) + 2CO2(g) ΔH = +54 kJ

This equation we add to the second equation

3Fe2O3(s) + CO(g) + 2Fe3O4(s) + 2CO(g) → 2Fe3O4(s) + CO2(g) + 6FeO(s) + 2CO2 (g)

3Fe2O3(s) + 3CO(g) → 3CO2(g) + 6FeO(s)

ΔH° = 2*18 + (-39) = 36 - 39 = -3 kJ

This new equation we will divide by 3

3Fe2O3(s) + 3CO(g) → 3CO2(g) + 6FeO(s)

Fe2O3(s) + CO(g) → CO2(g) + 2FeO(s)

ΔH° =-3/3 = -1 kJ

Now we will substract this new equation from the first equation

Fe2O3(s) + 3CO(g) - Fe2O3(s) - CO(g) → 2Fe(s) + 3CO2(g) -2FeO(s) - CO2(g)

2CO(g) + 2FeO(s) → 2Fe(s) + 2CO2(g)

ΔH° = -23kJ +1kJ

ΔH° = -22 kj

The next equation we will divide by 2

2CO(g) + 2FeO(s) → 2Fe(s) + 2CO2(g)

CO(g) + FeO(s) → Fe(s) + CO2(g)

ΔH° = -22kJ /2

ΔH° = -11 kj

7 0

3 years ago