1. Explain, in terms of energy, why there are fewer apex predators than producers or primary

g Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produc

Gala2k [10]

$\text{Ammonia has been studied as an alternative "clean" fuel for internal combustion}$

$\text{engines, since its reaction with oxygen produces only nitrogen and water vapor,}$

$\text{and in the liquid form it is easily transported. An industrial chemist studying this}$

$\text{reaction fills a} \ \mathbf{100 \ L }\ \text{tank with} \ \mathbf{8.6 \ mol} \ \text{of ammonia gas and} \ \mathbf{28 \ mol} \ \ \text{of oxygen gas, }$

$\text{to be} \ \mathbf{2.6\ mol} \ .\ \text{Calculate the concentration equilibrium constant for the combustion of}$

$\text{ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.}$

Answer:

Explanation:

From the correct question above:

The reaction can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

From the above reaction; the ICE table can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

I (mol/L) 0.086 0.28 0 0

C -4x -3x +2x +6x

E 0.086 - 4x 0.28 - 3x +2x +6x

At equilibrium;

The water vapor = $\dfrac{2.6 \ mol}{100 \ L} = 6x$

$x = \dfrac{2.6}{100} \times \dfrac{1}{6}$

$x = 0.00433$

$\text{equilibrium constant} ({k_c}) = \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }$

$\implies \dfrac{(2x)^2 (6x)^6}{(0.086-4x)^4\times (0.28-3x)^3} \\ \\$

Replacing the value of x, we have:

$K_c = \dfrac{4 \times 46,656 \times x^8}{(0.086-4x)^4\times (0.28 -3x)^3} \\ \\ K_c = \dfrac{4 \times 46656 \times (0.00433)^8}{(0.06868)^4(0.26701)^3} \\ \\ K_c = \mathbf{5.4446 \times 10^{-8}}$

$K_c = \mathbf{5.5 \times 10^{-8} \ to \ 2 \ significant \ figures}$

5 0

3 years ago

what happens when a gas obtained by heating Ammonium Chloride and slaked lime are passed through copper sulphate ?

gogolik [260]

Answer:

Consequently, what happens when gas obtained by heating slaked lime and ammonium chloride is passed through copper sulphate solution? The HCl in the gas mixture will form hydrochloric and the H+ will react with some of the NH3(aq), forming NH4^+, and with some of the SO4^2-, forming HSO4^-.

8 0

3 years ago

What is the theoretical yield of NaBr

dolphi86 [110]

The theoretical yield of NaBr given that 2.36 moles of FeBr₃ reacts is 7.08 moles

<h3>Balanced equation </h3>

2FeBr₃ + 3Na₂S → Fе₂S₃ + 6NaBr

From the balanced equation above,

2 moles FeBr₃ reacted to produce 6 moles of NaBr

<h3>How to determine the theoretical yield of NaBr</h3>

From the balanced equation above,

2 moles FeBr₃ reacted to produce 6 moles of NaBr

Therefore,

2.36 moles FeBr₃ will react to produce = (2.36 × 6) / 2 = 7.08 moles of NaBr

Therefore,

Thus, the theoretical yield of NaBr is 7.08 moles

Learn more about stoichiometry:

brainly.com/question/14735801

#SPJ1

7 0

2 years ago

Name at least two benefits of using models in science

STatiana [176]

Answer:

Viewing systems from multiple perspectives.

Discovering causes and effects using model tractability.

Improving system understanding through visual analysis.

Explanation:

Got this from google, lol. But, I put three here just in case you could get extra credit for more than two.

3 0

3 years ago

Read 2 more answers

The combustion of hydrogen and oxygen is commonly used to

posledela

Answer:

6.66 mol

Explanation:

(atm x L) ÷ (0.0821 x K)

(0.875 x 250) ÷ (0.0821 x 400)

=6.66108

5 0

2 years ago