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kozerog [31]
2 years ago
6

The question is in the picture below

Chemistry
1 answer:
Rus_ich [418]2 years ago
4 0

Answer:

\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3

Explanation:

Hess's Law of Constant Heat Summation states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation is equal to the sum of the enthalpy changes of the other chemical equations. Thus, the reaction that involves the conversion of reactant A to B, for example, has the same enthalpy change even if you convert A to C, before converting it to B. Regardless of how many steps it takes for the reactant to be converted to the product, the enthalpy change of the overall reaction is constant.

With Hess's Law in mind, let's see how A can be converted to 2C +E.

\bf{\text{A} \rightarrow 2\text{B}}                  (Δ\text{H}_1)  -----(1)

Since we have 2B, multiply the whole of II. by 2:

\bf{2\text{B} \rightarrow 2\text{C} +2\text{D}}       (2Δ\text{H}_2) -----(2)

This step converts all the B intermediates to 2C +2D. This means that the overall reaction at this stage is \text{A} \rightarrow 2\text{C} +2\text{D}.

Reversing III. gives us a negative enthalpy change as such:

\bf{2\text{D} \rightarrow \text{E}}                  (-Δ\text{H}_3) -----(3)

This step converts all the D intermediates formed from step (2) to E. This results in the overall equation of \text{A} \rightarrow 2\text{C} +\text{E}, which is also the equation of interest.

Adding all three together:

\text{A} \rightarrow 2\text{C}+\text{E}            (\bf{\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3 })

Thus, the first option is the correct answer.

Supplementary:

To learn more about Hess's Law, do check out: brainly.com/question/26491956

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¿En que se basa la clasificación de los carbohidratos?
lina2011 [118]

Answer:

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espero que te ayude

3 0
3 years ago
Which of these substances is commonly refined by electrolysis?...
Alekssandra [29.7K]
I think brine is the answer
4 0
3 years ago
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A gas has a volume of 100.0 mL at a pressure of 600 mm HG. If the temperature is held consistent, what is the volume of the gas
Amiraneli [1.4K]

Answer: The volume of the gas at a pressure of 800 mm Hg is 75 mL.

Explanation:

Given: V_{1} = 100.0 mL,       P_{1} = 600 mm Hg

V_{2} = ?,        P_{2} = 800 mm Hg

According to Boyle's law, at constant temperature the pressure of given mass of an ideal gas is inversely proportional to volume.

Hence, formula used to calculate new volume of the gas is as follows.

P_{1}V_{1} = P_{2}V_{2}

Substitute the values into above formula.

P_{1}V_{1} = P_{2}V_{2}\\600 mm Hg \times 100.0 mL = 800 mm Hg \times V_{2}\\V_{2} = \frac{600 mm Hg \times 100.0 mL}{800 mm Hg}\\= 75 mL

Therefore, we can conclude that the volume of the gas at a pressure of 800 mm Hg is 75 mL.

6 0
3 years ago
Read 2 more answers
The activation energies are 45. 3 kj mol^-1 for k1 and 69. 8 kj mo1-1 for k2. If the rate constants are equal at 320 k, at what
Katyanochek1 [597]

At temperature 298K, the value of k1/k2 will equal to 2.00

According to Arrhenius's equation

Sometimes the Arrhenius equation is written as k = Ae-^{E/RT}, where k is the rate of the chemical reaction, A is a constant that varies depending on the chemicals involved, E is the activation energy, and R is the universal gas constant, and T is the temperature.

\frac{k1}{k2} = \frac{A_{1} e^{-E_{a1}/ RT} }{A_{2} e^{-E_{a2}/ RT} }

    = \frac{A1}{A2} e^{(E_{a2}-E_{a1)}/RT  }

 

Given

Eₐ₂ = 69.8 × 10³ J/mol

Eₐ₁ = 45.3 × 10³ J/mol

R = 8.314J/kmol³

Now if the rate constant is equal means K1/K2 = 1 at T = 320 K

So if we put the values of K1/K2, Ea2,Ea1, T, and R into the formula

On solving

We get

A1/A2 = 1.001 × 10⁻⁴

Now when K1/K2 = 2

Then if we put the value of A1/A2, Ea1, Ea2, R, K1/K2 in the same equation,

On solving we get

T = 298 K

Hence, At temperature 298K k1/k2 will equal to 2.00

Learn more about Arrhenius's equation here  brainly.com/question/26724488

#SPJ4

5 0
2 years ago
If the acid dissociation constant, Ka, for an acid HA is 8 x 104 at 25°C, what percent of the acid is dissociated
Nesterboy [21]

Answer:

Percent dissociated = 3.92%

Explanation:

The equilibrium of the weak acid, HA, is:

HA ⇄ H⁺ + A⁻

<em>Where Ka = 8x10⁻⁴ is: [H⁺] [A⁻] / [HA]</em>

<em> </em>

As both H⁺ and A⁻ comes from the same equilibrium we can write the concentrations of the species as follows:

[H⁺] = X

[A⁻] = X

[HA] = 0.50M - X

<em>Where X is reaction coordinate</em>

<em />

Replacing:

8x10⁻⁴ = X² / 0.50-X

4x10⁻⁴ - 8x10⁻⁴X - X² = 0

Solving for X:

X = -0.02M. False solution. There is no negative concentrations

X = 0.0196M. Right solution.

Replacing:

[A⁻] = 0.0196M

Percent of the acid that is dissociated is:

[A⁻] / [HA]₀ * 100

<em>[HA]₀ is its initial concentration = 0.50M</em>

0.0196M / 0.50M * 100

<h3>Percent dissociated = 3.92%</h3>
8 0
3 years ago
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