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3 years ago

What is meant by a solvation

Chemistry

1 answer:

Vladimir79 [104]3 years ago

6 0

Solvation is the process in which molecules of a solvent attract the particles of a solute.

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Rina8888 [55]

Answer:

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Explanation:

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3 years ago

Energy in an ecosystem flow from consumers to producers.True of False

Alenkinab [10]

False - A producer always provides food for the consumer.

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4 years ago

Read 2 more answers

The volume of 0.6305 kg of a liquid is 0.430 L. What is its density in g/mL

motikmotik

Answer: 1.466 g/mL

Explanation:

$\frac{0.6305 \text{kg}}{0.430 \text{L}} \cdot \frac{1000 \text{g}}{1 \text{kg}} \cdot \frac{1 \text{L}}{1000 \text{mL}} = 1.466 \,\frac{\text{g}}{\text{mL}}$

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2 years ago

What is an ion? What are the two ways in which an ion can be made?

denpristay [2]

Answer: Ions are formed by the addition of electrons to, or the removal of electrons from, neutral atoms or molecules or other ions; by combination of ions with other particles; or by rupture of a covalent bond between two atoms in such a way that both of the electrons of the bond are left in association with one of the ...

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4 years ago

Given the following equilibrium constants: Kb B(aq) + H2O(l) ⇌ HB+(aq) + OH−(aq) 1/Kw H+(aq) + OH−(aq) ⇌ H2O(l) What is the equi

bija089 [108]

Answer: The value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

Explanation:

The given chemical equations follows:

Equation 1: $B(aq.)+H_2O(l)\rightleftharpoons HB^+(aq.)+OH^-(aq.);K_b$

Equation 2: $H^+(aq.)+OH^-(aq.)\rightleftharpoons H_2O(l);\frac{1}{K_w}$

The net equation follows:

$B(aq.)+H^+(aq.)\rightleftharpoons HB^+(aq.);K_c$

As, the net reaction is the result of the addition of first equation and the second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=K_1\times K_2$

We are given:

$K_1=K_b$

$K_2=\frac{1}{K_w}$

Putting values in above equation, we get:

$K_c=K_b\times \frac{1}{K_w}=\frac{K_b}{K_w}$

Hence, the value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

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3 years ago