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2 years ago

Can you help me with this with a solution

Chemistry

1 answer:

LuckyWell [14K]2 years ago

3 0

The complete table is inserted.

A table is given,

Formulas used:

pH= -log(H⁺)

pOH= -log(OH⁻)

pH+ pOH=14

Calculations:

For A: (H⁺)=2×10⁻⁸M

Using the pH formula:

pH= -log(H⁺)=-log(2×10⁻⁸)=7.69

pOH=14 - 7.69=6.3

Calculating OH concentration,

pOH= -log(OH⁻)

6.3= -log(OH⁻)

(OH⁻)=5.011×10⁻⁷M

Hence, the nature of A is basic.

Similarily,

For B,

(OH⁻)=1×10⁻⁷

Using the pH formula:

pOH= -log(OH⁻)= -log(1×10⁻⁷)=7

pH=14-7=7

Calculating H concentration,

pH= -log(H⁺)

7= -log(H⁺)

(H⁺)=1×10⁻⁷M

Hence, the nature of B is neutral.

Similarily,

For C,

pH=12.3

Using the pH formula:

pOH=14-12.3=1.7

Calculating H concentration,

pH= -log(H⁺)

12.3= -log(H⁺)

(H⁺)=5.011×10⁻¹³M

Calculating OH concentration,

pOH= -log(OH⁻)

1.7= -log(OH⁻)

(OH⁻)=1.99×10⁻²M

Hence, the nature of C is Basic.

Similarily,

For D,

pOH=6.8

Using the pH formula:

pH=14-6.8=7.2

Calculating H concentration,

pH= -log(H⁺)

7.2= -log(H⁺)

(H⁺)=6.309×10⁻⁸M

Calculating OH concentration,

pOH= -log(OH⁻)

6.8= -log(OH⁻)

(OH⁻)=1.58×10⁻⁷M

Hence, the nature of D is basic.

Learn more about the acid and bases here:

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the empirical formula for a compound is Mo2Br5Cl3.

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2 years ago

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The rate constant for the decomposition of ethanol on an alumina surface is 24.00 × 10²⁵ M.

The rate law is: rate = 24.00 × 10²⁵ M/s

The integrated rate law is: [C₂H₅OH]t = [C₂H₅OH]₀ - 24.00 × 10²⁵ M/s × t

If the initial concentration of C₂H₅OH was 1.25 × 10²² M, the half-life is 2.60 × 10⁻⁵ s.

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Let's consider the decomposition of ethanol on an alumina surface.

C₂H₅OH(g) ⇒ C₂H₄(g) + H₂O(g)

The plot of [A] vs time (t) resulted in a straight line, which indicates that the reaction follows zero-order kinetics.

The slope, 24.00 × 10²⁵ M/s, represents the rate constant, k.

<h3>What is zero-order kinetics?</h3>

It is a chemical reaction in which the rate of reaction is constant and independent of the concentration of the reacting substances

The rate law for zero-order kinetics is:

rate = 24.00 × 10²⁵ M/s

The integrated rate law for zero-order kinetics is:

[C₂H₅OH]t = [C₂H₅OH]₀ - 24.00 × 10²⁵ M/s × t

Is the time for the amount of substance to decrease by half.

If the initial concentration of C₂H₅OH was 1.25 × 10²² M, we can calculate the half-life [t(1/2)] using the following formula.

t(1/2) = [C₂H₅OH]₀ / 2 × k

t(1/2) = (1.25 × 10²² M) / 2 × (24.00 × 10²⁵ M/s) = 2.60 × 10⁻⁵ s

We can calculate the time required for all the 1.25 × 10²² M C₂H₅OH to decompose using the integrated rate law.

[C₂H₅OH]t = [C₂H₅OH]₀ - 24.00 × 10²⁵ M/s × t

0 M = 1.25 × 10²² M - 24.00 × 10²⁵ M/s × t

t = 5.21 × 10⁻⁵ s

The rate constant for the decomposition of ethanol on an alumina surface is 24.00 × 10²⁵ M.

The rate law is: rate = 24.00 × 10²⁵ M/s

The integrated rate law is: [C₂H₅OH]t = [C₂H₅OH]₀ - 24.00 × 10²⁵ M/s × t

If the initial concentration of C₂H₅OH was 1.25 × 10²² M, the half-life is 2.60 × 10⁻⁵ s.

The time required for all the 1.25 × 10²² M C₂H₅OH to decompose is 5.21 × 10⁻⁵ s.

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