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Anit [1.1K]
3 years ago
9

For the reaction of gaseous nitrogen with gaseous hydrogen to produce gaseous ammonia, the correct equilibrium constant expressi

on is _____
Chemistry
1 answer:
exis [7]3 years ago
7 0

The solution would be like this for this specific problem:

 

First, we need to write out the balanced reaction equation for the problem:<span>

"gaseous nitrogen with gaseous hydrogen to produce gaseous ammonia" 

gaseous nitrogen + gaseous hydrogen = gaseous ammonia 

gaseous nitrogen = N2(g) 
gaseous hydrogen = H2(g) 
gaseous ammonia = NH3(g) 

In here, I’m going to show you how to balance this from start to finish: 

N2(g) + H2(g) ↔ NH3(g) (basic, unbalanced equation that shows reactants and product) 
N2(g) + H2(g) ↔ 2NH3(g) (balances the nitrogens) 
N2(g) + 3H2(g) ↔ 2NH3(g) (balances the hydrogens) => Balanced Equation 

<span>Next, we write the Keq expression of the balanced reaction:

Keq = (∏[products]^n) /(∏[reactants]^n) 
Keq = [NH3]^2 / [[N2][H2]]^3 


Therefore, the correct equilibrium constant expression is Keq = (NH3)2 / (N2)(H2)3.</span></span>

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hichkok12 [17]
Answer:
             <span>Arachidonic Acid and PGE</span>₁<span> are both carboxylic acids with <u>Twenty Carbon</u> atoms. The differences are that Arachidonic acid contains <u>Four <em>cis</em> Double Bonds</u> and no other functional groups, whereas PGE</span>₁<span> has <u>One <em>Trans</em> Double Bond, Two Hydroxyl and One Ketone Functional Groups.</u>. In addition, a part of the PGE</span>₁<span> chain forms a <u>Five Membered Ring</u>.

Structures of Both Arachidonic Acid and PGE</span>₁ are shown Below,

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3 years ago
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4 years ago
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Answer:

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Explanation:

First, convert the 1.35 M to 1.35 mol/L in order for the units to correctly cancel out.

Then, multiply (0.0725 moles Na2CO3/1) times (L/ 1.35 mol).

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3 years ago
Fragilidad de metales​
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Answer: La fragilidad es la propiedad de algunos metales de no poder experimentar deformaciones plásticas, de forma que al superar su límite elástico se rompen bruscamente. La acritud es la propiedad de un metal para aumentar su dureza y su resistencia por el efecto de las deformaciones.

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3 years ago
What volume is occupied by 0.109 molmol of helium gas at a pressure of 0.98 atmatm and a temperature of 307 K
KATRIN_1 [288]

Answer:

2.8 L

Explanation:

From the question given above, the following data were obtained:

Number of mole (n) = 0.109 mole

Pressure (P) = 0.98 atm

Temperature (T) = 307 K

Gas constant (R) = 0.0821 atm.L/Kmol

Volume (V) =?

The volume of the helium gas can be obtained by using the ideal gas equation as follow:

PV = nRT

0.98 × V = 0.109 × 0.0821 × 307

0.98 × V = 2.7473123

Divide both side by 0.98

V = 2.7473123 / 0.98

V = 2.8 L

Thus, the volume of the helium gas is 2.8 L.

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3 years ago
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