please solve asap suppose a 250ml flask is filled with 1.5 mol of co, 1.4 mol of

Question

please solve asap

suppose a 250ml flask is filled with 1.5 mol of co, 1.4 mol of h2O. and 0.80 mol of co2. the following reaction is possible
co(g)+ H2O(g)->co2(g)+H2(g)

equilibrium constant k is 4.74 at temp of flask. cal equilibrium molarity of co.

2) a chemist dissolve 714mg of pure hyrobromic acid in enough water to make up 400ml solution. cal ph of solution

3) the base protonation constant K of ammonia (NH3) is 1.8×10-5. calculaye Ph. of a 0.46M solution of ammonia. at 25 Celsius. round 1 decimal point

calculate the solubilty. of Ag2CO3 in watwr at 25 degress celcuis. ksp is equal to 8.1 ×10-12

Explanation / Answer

CO(g)+ H2O(g) <---->CO2(g)+H2(g)

K = 4.74

a) Molarity of CO

V = 0.25 L

mol CO = 1.5 mol

mol H2O = 1.4 mol

mol CO2 = 0.80

mol H2 = 0

Calculate Molarities

M = mol/V

V = 0.25 L

M CO = 1.5/0.25 = 6

M H2O = 1.4/0.25 = 5.6

M CO2 = 0.80/0.25 = 3.2

M H2 = 0/0.25 = 0

In the equilibrium

M CO = 1.5/0.25 = 6 - x

M H2O = 1.4/0.25 = 5.6 - x

M CO2 = 0.80/0.25 = 3.2 + x

M H2 = 0/0.25 = 0 +x

Substitute in K

K = 4.74

K = [CO2][H2]/[H2O][CO2]

4.74 = (3.2 + x)(x) / (6 - x) / (5.6 - x)

Solve for x

4.74 (33.6-11.6x + x^2) = 3.2x + x^2

159.3-54.98x + 4.74x^2 -3.2x -x^2 = 0

159.3 - (58.18)x + (3.74)x^2 = 0

x = 3.55

Substitute in CO concnetration

M CO = 1.5/0.25 = 6 - x = 6-3.55 = 2.45 M

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