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e-lub [12.9K]
3 years ago
5

Enter your answer in the provided box. The partial pressure of CO2 gas above the liquid in a bottle of champagne at 20°C is 4.5

atm. What is the solubility of CO2 in champagne? Assume the Henry's law constant is the same for champagne as it is for water: at 20°C, kH = 3.7 × 10−2 mol/(L·atm).
Chemistry
1 answer:
sineoko [7]3 years ago
6 0

<u>Answer:</u> The molar solubility of carbon dioxide gas is 0.17 M

<u>Explanation:</u>

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{CO_2}=K_H\times p_{liquid}

where,

K_H = Henry's constant = 3.7\times 10^{-2}mol/L.atm

p_{CO_2} = partial pressure of carbonated drink = 4.5 atm

Putting values in above equation, we get:

C_{CO_2}=3.7\times 10^{-2}mol/L.atm\times 4.5atm\\\\C_{CO_2}=0.17mol/L=0.17M

Hence, the molar solubility of carbon dioxide gas is 0.17 M

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Answer:

Explanation:

AgCl      ⇄       Ag⁺       +       Cl⁻

m                       m                   m

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[ Ag⁺ ] [ Cl⁻ ] = 1.6 x 10⁻¹⁰

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3 years ago
The concentration of dye in Solution A is 26.609 M. You have 13 mL of water at your disposal to make the dilutions. The solution
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6,613 M

Explanation:

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The concentration of Solution B is:

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Thus, concentration of solution C is:

9,552 M ×  = 6,613 M

I hope it helps!

8 0
3 years ago
A horizontal cylinder equipped with a frictionless piston contains 785 cm3 of steam at 400 K and 125 kPa pressure. A total of 83
guapka [62]

Answer:

a. 478.69 K

b. 939.43 cm^{3}

c. 19.30 J

d. 64.5J

Explanation:

From the question, we can identify the following;

V_{o} = 785cm^{3} = 0.000785 m^{3}

T_{o} = 400K

P_{o} = 125 Kpa =  125 000 Pa

Using the ideal gas equation,

PV = nRT

where R is the molar gas constant = 8.314 m^{3}⋅Pa⋅K^{-1}⋅mol^{-1}

Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol

a. Steam temperature in K

To calculate this, we use the constant pressure process;

q = nΔH

Where q is 83.8J according to the question

Thus;

83.8 = 0.03 × [34980 + 35.5T_{1} - (34980 + 35.5T_{o})]

83.8 = (0.03 × 35.5) (T_{1} - 400K)

83.8 = 1.065 (T_{1}  - 400)

78.69 = (T_{1}  - 400)

T_{1} = 400 + 78.69

T_{1}  = 478.69 K

b. Final cylinder volume

To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)

V_{1}/T_{1} = V_{o}/T_{o}

V_{1}  =  V_{o}T_{1}/T_{o}

V_{1}   = (785 × 478.69)/400

V_{1}   = 939.43 cm^{3}

c. Work done by the system

Mathematically, the work done by the system is calculated as follows;

w = P(V_{1}- V_{o}) = 125 KPa ( 939.43 - 785) = 19.30 J

d. Change in internal energy of the steam in J

ΔU = q - w = 83.8 - 19.3 = 64.5J

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saveliy_v [14]

Answer:

D- beaking bread

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