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ololo11 [35]
3 years ago
5

If the pOH of a solution is 3.01, what is the concentration of the Mg(OH)2 solution?​

Chemistry
2 answers:
Amanda [17]3 years ago
8 0

Answer:

9.77×10^-4 M

Explanation:

Now you need to recall that pOH=-log [OH^-]

If that be so;

log[OH^-] = - pOH

Then [OH^-] = Antilog (-pOH)

Thus;

[OH^-] = Antilog (-3.01)

[OH^-] = 9.77×10^-4 M

From

Mg(OH)2(s) <--------> Mg^2+(aq) + 2OH^-(aq)

Let the concentration of the solution= x

Since [Mg^2+] = [2OH^-] = 9.77×10^-4 M = x

Therefore, the concentration of the Mg(OH)2 solution is 9.77×10^-4 M

Llana [10]3 years ago
5 0

Answer:

4.885x10^-4 M

Explanation:

Step 1:

Data obtained from the question.

pOH = 3.01

Step 2:

Determination of the concentration of OH ion [OH-]

This is shown below:

pOH = - Log [OH-]

pOH = 3.01

3.01 = - Log [OH-]

-3.01 = Log [OH-]

Take the anti-log of -3.01

[OH-] = 9.77x10^-4 M

Step 3:

Dissociation equation for Mg(OH)2.

we shall write the equation for the dissociation of Mg(OH)2. This is show below

Mg(OH)2 —> Mg^2+ + 2OH^-

Step 4:

Determination of the concentration of the Mg(OH)2 solution.

From the balanced equation above,

1 mole of Mg(OH)2 dissociate to produce 2 moles of OH-.

Therefore xM of Mg(OH)2 will dissociate to produce 9.77x10^-4 M i.e

xM of Mg(OH)2 = (9.77x10^-4)/2

xM of Mg(OH)2 = 4.885x10^-4 M

Therefore, the concentration of the Mg(OH)2 solution is 4.885x10^-4 M

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What are the moles of gas are in a 30 liter scuba canister if the temperature of the canister is 300 K and the pressure is 200 a
Lady_Fox [76]

Answer:

n = 243.605

Explanation:

Given

Pressure, P = 200\ atm

Temperature, T = 300k

Volume, V = 30L

Required

Calculate the number of moles

We'll apply the following formula to solve this question

n = \frac{pV}{RT}

Where

R = 0.0821 L\ atm/(mol\ K)

The above equation is an illustration of the ideal gas law

Substitute values for p, V, R and T in:

n = \frac{pV}{RT}

n = \frac{200 * 30}{0.0821 * 300}

n = \frac{6000}{24.63}

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3 years ago
Consider the reaction: CaCO3(s) à CaO(s) + CO2(g)
Vsevolod [243]

Answer:

131.5 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

First, we will calculate the standard enthalpy of the reaction (ΔH°).

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g) ) - 1 mol × ΔH°f(CaCO₃(s) )

ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)

ΔH° = 179.2 kJ

Then, we calculate the standard entropy of the reaction (ΔS°).

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g) ) - 1 mol × S°(CaCO₃(s) )

ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)

ΔS° = 160.2 J/K = 0.1602 kJ/K

Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.

ΔG° = ΔH° - T × ΔS°

ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K

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Answer:

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Explanation:

We are informed that the reaction is endothermic. An endothermic reaction is one in which energy is absorbed hence ∆H is positive at all temperatures.

Similarly, absorption of energy leads to a decrease in entropy of the reaction system. Hence the change in entropy of the reaction ∆Sreaction is negative at all temperatures.

The change in free energy for the reaction is positive at all temperatures since ∆S reaction is negative then from ∆G= ∆H - T∆S, we see that given the positive value of ∆H, ∆G must always return a positive value at all temperatures.

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