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leonid [27]
2 years ago
8

O2 oxygen is diatomicO + O O=O Double Bond

Chemistry
1 answer:
krek1111 [17]2 years ago
7 0

Answer:

O 100

Explanation:

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The Haber Process is the main industrial procedure to produce ammonia. The reaction combines nitrogen from air with hydrogen mai
Firdavs [7]

Answer:

A) N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g).

B) Kc=0.0933.

C) 0.9 mol.

D) Increasing both temperature and pressure.

Explanation:

Hello,

In this case, given the information, we proceed as follows:

A)

N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)

B) For the calculation of Kc, we rate the equilibrium expression:

Kc=\frac{[NH_3]^2}{[N_2][H_2]^3}

Next, since at equilibrium the concentration of ammonia is 0.6 M (0.9 mol in 1.5 dm³ or L), in terms of the reaction extent x, we have:

[NH_3]=0.6M=2*x

x=\frac{0.6M}{2}=0.3M

Next, the concentrations of nitrogen and hydrogen at equilibrium are:

[N_2]=\frac{1.5mol}{1.5L}-x=1M-0.3M=0.7M

[H_2]=\frac{4mol}{1.5L}-3*x=2.67M-0.9M=1.77M

Therefore, the equilibrium constant is:

Kc=\frac{(0.6M)^2}{(0.7M)*(1.77M)^3}\\ \\Kc=0.0933

C) In this case, the equilibrium yield of ammonia is clearly 0.9 mol since is the yielded amount once equilibrium is established.

D) Here, since the reaction is endothermic (positive enthalpy change), one way to increase the yield of ammonia is increasing the temperature since heat is reactant for endothermic reactions. Moreover, since this reaction has less moles at the products, another way to increase the yield is increasing the pressure since when pressure is increased the side with fewer moles is favored.

Best regards.

7 0
3 years ago
I will give 15 points
zheka24 [161]

Answer:

B and C is your answer

Explanation:

Hope I helped, Sorry if I'm wrong

3 0
3 years ago
Read 2 more answers
Use the reaction N2(g) + 3H2(g) 2NH3(g) to answer the following questions
Veronika [31]

Answer:

K

C

=

[N

2

][H

2

]

3

[NH

3

]

2

N

2

+3H

2

⇌2NH

3

=

(1.5×10

−2

)(3×10

−2

)

3

(1.2×10

−2

)

2

=

1.5×10

−2

×27×10

−6

1.44×10

−4

=

40.5×10

−4

1.44

=0.035×10

4

5 0
3 years ago
Which of the following can you determine from mass spectrometry data
bonufazy [111]

Answer:

different isotopes of na element

7 0
3 years ago
1.
german

Order of metals from least reactive to most reactive: B <C <A <D

<h3>Further explanation</h3>

Reducing agents are substances that experience oxidation  

Oxidizing agents are substances that experience reduction

The metal activity series is expressed in voltaic series  

<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au  </em>

The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent  

The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing agent

So that the metal located on the left can push the metal on the right in the redox reaction  

Let's analyze the statement in the problem

I. Only A, C and D react with 1 mol/L HCl to give H₂(e)

M + HCl ⇒ MCl + H₂(MCl : alkali, MCl₂ : alkaline earth)

A, C and D can react with 1 mol / L HCl, meaning metals A, C and D are located to the left of element H (more reactive), and B in the right of element H

II. When A is added to solutions of the other metal ions, metallic B and C are  formed but not D.

This means that metal A is more reactive than metals B and C, while D is more reactive than A, so metal D is the most reactive

3 0
2 years ago
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