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densk [106]
3 years ago
7

Consider the reversible reaction in aqueous medium shown below. At equilibrium, the concentrations of A, B, C, and D are 0.0053

M, 0.0085 M, 0.0040 M, and 0.0035 M, respectively. What is the equilibrium constant for the reaction? A+3B⇌3C+D
Chemistry
1 answer:
GuDViN [60]3 years ago
8 0

Answer:

The equilibrium constant for the reaction is 0.0689

Explanation:

<u>Step 1:</u> Data given

The concentrations of A = 0.0053 M

The concentration of B = 0.0085 M

The concentration of C = 0.0040 M

The concentration of D = 0.0035 M

<u>Step 2:</u> The balanced equation:

A+3B⇌3C+D

<u>Step 3:</u> Calculate the equilibrium constant

K = [C]³[D] / [A][B]³

K = (0.0040)³(0.0035) / (0.0053)(0.0085)³

K = 0.0689

The equilibrium constant for the reaction is 0.0689

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True, both of the red and white phosphorus is a pure substance.

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4 0
3 years ago
The formation constant for the reaction ag (aq) 2nh3(aq) ag(nh3)2 (aq) is kf = 1.7 × 107 at 25°c. what is δg° at this temperatur
anzhelika [568]

The value of ΔG° at this temperature is -18034.18 J/mol

Calculation,

Given information

formation constant (Kf)= 1.7 × 10^{7}

Universal gas constant (R) = 8.314 J/K• mol

Temperature = 25° C = 25 °C + 273 = 300 K

Formula used:

ΔG° = -RT㏑Kf

By putting the valur of R,T, Kf we get the value of ΔG°

ΔG° = - 8.314 J/K• mol×300K㏑ 1.7 × 10^{7}

ΔG° = -2494.2㏑ 1.7 × 10^{7} = -18034.18 J/mol

So, change in standard Gibbs's free energy is -18034.18 J/mol

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8 0
2 years ago
What is the heaviest element that is made in nature on the periodic table
ANTONII [103]
It would be uranium
5 0
3 years ago
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How many grams of carbon dioxide are produced from 0.98 mol of Fe3O4?
nexus9112 [7]
Hello

the answer is 43.129310000000004

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4 0
3 years ago
Calculate the solubility of mn(oh)2 in grams per liter when buffered at ph=7.0. assume that buffer capacity is not exhausted
Vanyuwa [196]
When PH + POH = 14 
∴ POH = 14 -7 = 7

when POH = -㏒[OH-]

          7    = -㏒ [OH-]
∴[OH-] = 10^-7

by using ICE table:

           Mn(OH)2(s) ⇄  Mn2+ (aq)  + 2OH-(aq)
initial                              0                     10^-7
change                           +X                      +2X
Equ                                 X                  (10^-7 + 2X)

when Ksp = [Mn2+][OH-]^2

when Ksp of Mn(OH)2 = 4.6 x 10^-14

by substitution:

4.6 x 10^-14 = X*(10^-7+2X)^2  by solving this equation for X

∴ X =2.3 x 10-5 M

∴ The solubility of Mn(OH)2 in grams per liter (when the molar mass of Mn(OH)2 = 88.953 g/mol
= 2.3 x10^-5 moles/L * 88.953 g/mol

= 0.002 g/ L
3 0
4 years ago
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