All categories

3 years ago

Give the mathematical relationship for an unsaturated solution in comparing Q with K sp:__________.

Chemistry

1 answer:

e-lub [12.9K]3 years ago

3 0

Answer:

Q < Ksp

Explanation:

The general equilibrium of a constant product solubility, ksp, is:

AB ⇄ A⁺ + B⁻

<em>Where Ksp is defined as:</em>

Ksp = [A⁺] [B⁻]

When [A⁺] [B⁻] = Ksp, the solution is saturated or oversaturated because there are the maximum amount of ions that solution can dissolve.

When the solution is oversaturated, AB is produced.

Now, in a unsaturated solution, the [A⁺] [B⁻] is less than the maximum amount that can be dissolved. That means:

[A⁺] [B⁻] = Q < Ksp

Q is defined in the same way than Ksp, just in Q the system is not in equilibrium.

Right answer is:

You might be interested in

Pamela wants to learn more about the career preferences of students in her school. She interviews a number of students from the

Viktor [21]

It is too early for students to have career preferences while still in school

(i believe) <span />

4 0

3 years ago

Calculate the number of hydrogen atoms present in 40g of urea, (NH2)2CO

tekilochka [14]

Answer: There are $16.14 \times 10^{23}$ atoms of hydrogen are present in 40g of urea, $(NH_{2})_{2}CO$ .

Explanation:

Given: Mass of urea = 40 g

Number of moles is the mass of substance divided by its molar mass.

First, moles of urea (molar mass = 60 g/mol) are calculated as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{40 g}{60 g/mol}\\= 0.67 mol$

According to the mole concept, 1 mole of every substance contains $6.022 \times 10^{23}$ atoms.

So, the number of atoms present in 0.67 moles are as follows.

$0.67 mol \times 6.022 \times 10^{23} atoms/mol\\= 4.035 \times 10^{23} atoms$

In a molecule of urea there are 4 hydrogen atoms. Hence, number of hydrogen atoms present in 40 g of urea is as follows.

$4 \times 4.035 \times 10^{23} atoms\\= 16.14 \times 10^{23} atoms$

Thus, we can conclude that there are $16.14 \times 10^{23}$ atoms of hydrogen are present in 40g of urea, $(NH_{2})_{2}CO$ .

7 0

2 years ago

Ants live in the hollow thorns of an acacia tree. The tree provides food and shelter for the ants. The ants protect the tree fro

AfilCa [17]

A. Mutualism

Mutualism: Both organism benefit

Commensalism: One organism benefits. One organism is unaffected

Parasitism: One organism benefits. One organism is harmed

7 0

2 years ago

Read 2 more answers

Calculate the standard enthalpy of formation of liquid methanol, CH3OH(l), using the following information: C(graphite) + O2 --&

Darya [45]

Answer : The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation :

Standard formation of reaction : It is a chemical reaction that forms one mole of a substance from its constituent elements in their standard states.

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of $CH_3OH$ will be,

$C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g)$ $\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

(1) $C(graphite)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.5kJ/mole$

(2) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_3=-285.8kJ/mole$

(3) $CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l)$ $\Delta H_1=-726.4kJ/mole$

Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, we get :

(1) $C(graphite)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_1=-393.5kJ/mole$

(2) $2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol$

(3) $CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g)$ $\Delta H_3=726.4kJ/mole$

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)$

$\Delta H=-238.7kJ/mole$

Therefore, the standard enthalpy of formation of methanol is, -238.7 kJ/mole

6 0

3 years ago

Which of the following will have the highest melting point C6H12O6<br>KOI <br>NH3 <br>I2

sveticcg [70]

I believe the compound KOI is meant to be potassium iodate, which is KIO₃. This compound will also be the one with the highest melting point.

This is due to the fact that potassium ion will form ionic bonds with whatever atom or molecule it binds to. Ionic bonds involve the complete transfer of electrons, making them very strong and hard to break, meaning the compound will have the highest melting point.

3 0

3 years ago