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alina1380 [7]
3 years ago
5

If the value of kc at 25oc is 3.7108, and the equilibrium concentrations for n2 and h2 are 0.000105 m and 0.0000542 m, respectiv

ely, use the expression in question 5 to calculate the equilibrium concentration of nh3 .
Chemistry
1 answer:
frez [133]3 years ago
5 0

Equation of decomposition of ammonia:

N2+3H2->2NH3

Euilibrium constant:

Kc=(NH3)^2/((N2)((H2)^3))

As concentration of N2=0.000105, H2=0.0000542

so equation will become:

3.7=(NH3)^2/(0.000105)*(0.0000542)^3

NH3=√(3.7*0.000105*(0.0000542)^3)

NH3=7.8×10⁻⁹

So concentration of ammonia will be 7.8×10⁻⁹.

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Answer:

Carbon Monoxide / Carbon Dioxide / Sulfur and Nitrogen Dioxide

Explanation:

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nasty-shy [4]

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First, let's determine how many moles of oxygen we have.

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We have 3 drops at 0.050 ml each for a total volume of 3*0.050ml = 0.150 ml

Since the density is 1.149 g/mol,

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Now take the formula and solve for V, then substitute the known values and solve.

PV = nRT V = nRT/P V = 0.005386274 mol * 0.082057338 L*atm/(K*mol) * 303.15 K / 1.0 atm V = 0.000441983 L*atm/(K*) * 303.15 K / 1.0 atm V = 0.133987239 L*atm / 1.0 atm V = 0.133987239 L

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8 0
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Answer:

hope this helps

Explanation:

Photosynthesis

Photosynthesis

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7 0
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Calculate the energy required to ionize a hydrogen atom to an excited state where the electron is initially in the n = 5 energy
SVETLANKA909090 [29]

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