Which is an example of a renewable energy source?

The chemistry of nitrogen oxides is very versatile. Given the following reactions and their standard enthalpy changes, (1) NO(g)

AVprozaik [17]

Answer:

The heat of reaction for N₂O₃(g) + N₂O₅(s) → 2 N₂O₄(g) is ΔH = -22.2 kJ

Explanation:

Given the following reactions and their standard enthalpy changes:

(1) NO(g) + NO₂(g) → N₂O₃(g) ΔH o rxn = −39.8 kJ

(2) NO(g) + NO₂(g) + O₂(g) → N₂O₅(g) ΔH o rxn = −112.5 kJ

(3) 2 NO₂(g) → N₂O₄(g) ΔH o rxn = −57.2 kJ

(4) 2 NO(g) + O₂(g) → 2 NO₂(g) ΔH o rxn = −114.2 kJ

(5) N₂O₅(s) → N₂O₅(g) ΔH o subl = 54.1 kJ

You need to get the heat of reaction from: N₂O₃(g) + N₂O₅(s) → 2 N₂O₄(g)

Hess's Law states: "The variation of Enthalpy in a chemical reaction will be the same if it occurs in a single stage or in several stages." That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when verified in a single stage.

This law is the one that will be used in this case. For that, through the intermediate steps, you must reach the final chemical reaction from which you want to obtain the heat of reaction.

Hess's law explains that enthalpy changes are additive. And it should be taken into account:

If the chemical equation is inverted, the symbol of ΔH is also reversed.
If the coefficients are multiplied, multiply ΔH by the same factor.
If the coefficients are divided, divide ΔH by the same divisor.

Taking into account the above, to obtain the chemical equation

N₂O₃(g) + N₂O₅(s) → 2 N₂O₄(g) you must do the following:

Multiply equation (3) by 2

(3) 2*[2 NO₂(g) → N₂O₄(g) ] ΔH o rxn = −57.2 kJ*2

4 NO₂(g) → 2 N₂O₄(g) ΔH o rxn = −114.4 kJ

Reverse equations (1) and (2)

(1) N₂O₃(g) → NO(g) + NO₂(g) ΔH o rxn = 39.8 kJ

(2) N₂O₅(g) → NO(g) + NO₂(g) + O₂(g) ΔH o rxn = 112.5 kJ

Equations (4) and (5) are maintained as stated.

(4) 2 NO(g) + O₂(g) → 2 NO₂(g) ΔH o rxn = −114.2 kJ

(5) N₂O₅(s) → N₂O₅(g) ΔH o subl = 54.1 kJ

The sum of the adjusted equations should give the problem equation, adjusting by canceling the compounds that appear in the reagents and the products according to the quantity of each of them.

Finally the enthalpies add algebraically:

ΔH= -114.4 kJ + 39.8 kJ + 112.5 kJ -114.2 kJ + 54.1 kJ

ΔH= -22.2 kJ

The heat of reaction for N₂O₃(g) + N₂O₅(s) → 2 N₂O₄(g) is ΔH = -22.2 kJ

8 0

3 years ago

What are 2 reasons molecules cannot pass easily through the cell membrane? What part of the cell membrane helps them get through

tatuchka [14]

Answer:

The cell membrane's main trait is its selective permeability, which means that it allows some substances to cross it easily, but not others. Small molecules that are nonpolar (have no charge) can cross the membrane easily through diffusion, but ions (charged molecules) and larger molecules typically cannot. A concentration gradient is a just a region of space over which the concentration of a substance changes, and substances will naturally move down their gradients, from an area of higher to an area of lower concentration.

In cells, some molecules can move down their concentration gradients by crossing the lipid portion of the membrane directly, while others must pass through membrane proteins in a process called facilitated diffusion.

Explanation:

there are your two reasons molecules cannot pass easily through the cell membrane and what part of the cell membrane helps them get through is concentration gradient

3 0

3 years ago

Various plant and animal species are on the verge of extinction.

olga2289 [7]

Answer:

1: A.) adapted

2: A.) habitat destruction

Explanation:

Some (few) animals are capable of adjusting their behaviors in relation to human activity and the changes humans make in their habitat. Most are so well adapted to the details of their habitat that they are incapable of surviving when those details change.

"Because all organisms are adapted to their native habitat, human activities that result in habitat destruction are a major reason for their extinction."

_____

Comment on human activities

My friendly local rodent exterminator observed that rodent activity has picked up substantially, since so much human activity has been curtailed by quarantine orders.

3 0

4 years ago

During the discussion of gaseous diffusion for enriching uranium, it was claimed that 235UF6 diffuses 0.4% faster than 238UF6. S

Kay [80]

Answer: The below calculations proves that the rate of diffusion of $^{235}UF_6$ is 0.4 % faster than the rate of diffusion of $^{238}UF_6$

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

We are given:

Molar mass of $^{235}UF_6=349.034348g/mol$

Molar mass of $^{238}UF_6=352.041206g/mol$

By taking their ratio, we get:

$\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\sqrt{\frac{M_{(^{238}UF_6)}}{M_{(^{235}UF_6)}}}$

$\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\sqrt{\frac{352.041206}{349.034348}}\\\\\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\frac{1.00429816}{1}$

From the above relation, it is clear that rate of effusion of $^{235}UF_6$ is faster than $^{238}UF_6$

Difference in the rate of both the gases, $Rate_{(^{235}UF_6)}-Rate_{(^{238}UF_6)}=1.00429816-1=0.00429816$

To calculate the percentage increase in the rate, we use the equation:

$\%\text{ increase}=\frac{\Delta R}{Rate_{(^{235}UF_6)}}\times 100$

Putting values in above equation, we get:

$\%\text{ increase}=\frac{0.00429816}{1.00429816}\times 100\\\\\%\text{ increase}=0.4\%$

The above calculations proves that the rate of diffusion of $^{235}UF_6$ is 0.4 % faster than the rate of diffusion of $^{238}UF_6$

3 0

3 years ago

The membrane of the egg I like the cell membranes because

frez [133]

It’s a thin layer that protects the inside of the egg

5 0

3 years ago

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