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larisa86 [58]
3 years ago
9

The reaction ag+(aq) + cl−(aq) ↔ agcl(s) has an equilibrium of 1020 (keq = 1020). if you have a beaker containing 1 liter of wat

er with 0.1 mol of agcl(s) in it along with 10−6 m ag+(aq) and 10−15 m cl−(aq), how does the reaction shift?
Chemistry
1 answer:
lord [1]3 years ago
3 0
When Q is equal the initial concentration of the products / the initial concentration of the reactants. 

so, Q = [Ag]*[Cl-]  and we neglected [AgCl] as it is solid

∴ Q = 10^-6 * 10^-5 

       = 10^-11 

now we will compare the value of Q with the value of Keq:

when Q = Keq so, the system is in equilibrium

and when Q > Keq so, the reaction will go forward (shift to right) to achieve equilibrium.

and when Q< Keq so, the reaction will go left (shift to left) to achieve equilibrium.

when Q = 10^-11 and Keq = 10^20

∴Q< Keq 

and the reaction will shift to left.
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If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. We can calculate the concentration of CO₂ using Henry's law.

C = k \times P = \frac{1.65 \times 10^{-3} M }{atm}  \times 4.60 atm = 7.59 \times 10^{-3} M

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\frac{7.59 \times 10^{-3} mol}{L} \times  1.1 L \times \frac{44.01 g}{mol}  = 0.367 g

Now, we will repeat the same procedure for a partial pressure of 1.28 atm.

C = k \times P = \frac{1.65 \times 10^{-3} M }{atm}  \times 1.28 atm = 2.11 \times 10^{-3} M

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If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

Learn more: brainly.com/question/18987224

<em>The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO₂ gas (in g) will be released from 1.1 L of the carbonated water when the partial pressure of CO2 is lowered to 1.28 atm? At 25 ºC, the Henry’s law constant for CO₂ dissolved in water is 1.65 x 10⁻³ M/atm, and the density of water is 1.0 g/cm³.</em>

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