I need some chem help :(

What is the most corrosive element in the periodic table?

Fittoniya [83]

Answer:

Fluorine is the most corrosive element in the periodic table.

Explanation:

4 0

2 years ago

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Calculate how many times more soluble Mg(OH)2 is in pure water Based on the given value of the Ksp, 5.61×10−11, calculate the ra

maks197457 [2]

Answer:

molar solubility in water = 2.412 * 10^-4 mol/L

molar solubility of NaOH in 0.130M = 3.32 * 10^-9 mol/L

Mg(OH)2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH

Explanation:

The Ksp refers to the partial solubilization of a mostly insoluble salt. This is an equilibrium process.

The equation for the solubilization reaction of Mg(OH)2 can be given as:

Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)

Ksp can then be given as followed:

Ksp = [Mg^2+][OH^–]²

Step 2: Calculate the solubility in water

Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)

The mole ratio Mg^2+ with OH- is 1:2

So there will react X of Mg^2+ and 2X of OH-

The concentration at equilibrium will be XM Mg^2+ and 2X OH-

Ksp = [Mg^2+][OH^–]²

5.61*10^-11 = X * (2X)² = X *4X² = 4X³

X = 2.412 * 10^-4 mol/L = solubility in water

Step 3: Calculate solubility in 0.130 M NaOH

The initial concentration of Mg^2+ = 0 M

The initial concentration of OH- = 0.130 M

The mole ratio Mg^2+ with OH- is 1:2

So there will react X of Mg^2+ and 2X +0.130 for OH-

The concentration at equilibrium will be XM Mg^2+ and 0.130 + 2X OH-

The value of "[OH–] + 2X" is, because the very small value of X, equal to the value of [OH–] .

Let's consider:

[Mg+2] = X

[OH] = 0.130

Ksp = [Mg^2+][OH^–]²

5.61*10^-11 = X *(0.130)²

5.61*10^-11 = X * (0.130)^2

X = 3.32*10^-9 = solubility in 0.130 M NaOH

Step 4: Calculate how many times Mg(OH)2 is better soluble in pure water.

(2.412*10^-4)/ (3.32*10^-9) = 0.73 * 10^5

Mg(OH)2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH

4 0

3 years ago

Given the chemical reaction: Hg2+(aq) + Cu(s) + Hg(s) + Cu2+(aq)

Pavel [41]

Answer: When the reaction reaches equilibrium, the cell potential will be 0.00 V

Explanation:

Equilibrium state is the state when reactants and products are present but the concentrations does not change with time.

The equilibrium is dynamic in nature and the reactions are continuous in nature. Rate of forward reaction is equal to the rate of backward reaction.

The standard emf of a cell is related to Gibbs free energy by following relation:

$\Delta G=-nFE^0$

The Gibbs free energy is related to equilibrium constant by following relation:

$\Delta G=-2.303RTlog K$

For equilibrium $\Delta G=0$

Thus $0=-nFE^0$

$E^0=0$

Thus When the reaction reaches equilibrium, the cell potential will be 0.00 V

5 0

3 years ago

The value of Ka for phenol (a weak acid), C6H5OH, is 1.00×10-10. Write the equation for the reaction that goes with this equilib

I am Lyosha [343]

Answer:

$Ka=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OH]}$

Explanation:

Hello,

In this case, weak acids are characterized by the fact they do not dissociate completely, it means they do not divide into the conjugated base and acid at all, a percent only, which is quantified via equilibrium. In such a way, the chemical equation representing such incomplete dissociation is said to be:

$C_6H_5OH\rightleftharpoons C_6H_5O^-+H^+$

Thus, we can write the law of mass action, which consider the equilibrium concentrations of all the involved species, which is also known as the acid dissociation constant which accounts for the capacity the acid has to yield hydronium ions:

$K=Ka=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OH]}$

Best regards.

3 0

2 years ago

Who else had to do a final model for the rust unit

ArbitrLikvidat [17]

Which one? Are u saying.

4 0

2 years ago