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LenKa [72]
3 years ago
7

The solubility of a gas in a liquid increases with increasing pressure. To understand the above statement, consider a familiar e

xample: cola. In cola and other soft drinks, carbon dioxide gas remains dissolved in solution as long as the can or bottle remains pressurized. As soon as the lid is opened and pressure is released, the CO2 gas is much less soluble and escapes into the air. The relationship between pressure and the solubility of a gas is expressed by Henry's law: S=kP, where S is concentration in M, k is the Henry's law constant in units of mol/(L⋅atm), and P is the pressure in atm. Note: Since temperature also affects the solubility of a gas in an liquid, the Henry's law constant is specific to a particular gas at a particular temperature. The following table provides some information on carbon dioxide solubility in water. S (mol/L) P (atm) k (mol⋅L−1⋅atm−1) T (∘C) 3.80×10−2 1.00 20.0 6.60×10−2 20.0 1.00 3.50×10−2 25.0 Part A What is the Henry's law constant for CO2 at 20∘C?
Chemistry
2 answers:
weeeeeb [17]3 years ago
5 0

Answer:

gukyfkuyk

Explanation:

Tanzania [10]3 years ago
3 0

Answer:

The Henry's law constant for CO2 is 3.8x10⁻²mol/L atm

Explanation:

please, the solution is in the attached Word file

Download docx
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In a 1.0-liter container there are, at equilibrium, 0.10 mole h2, 0.20 mole n2, and 0.40 mole nh3. what is the value of kc for t
bogdanovich [222]
The reaction involved here would be written as:

2N2 + 3H2 = 2NH3

The equilibrium constant of a reaction is the ratio of the concentrations of the products and the reactants when in equilibrium. The expression for the equilibrium constant of this reaction would be as follows:

Kc = [NH3]^2 / [N2]^2[H2]^3
Kc = 0.40^2 / (0.20)^2 (0.10)^3 
Kc = 4000
6 0
3 years ago
Identify the activity that belongs in the field of chemistry.
SVEN [57.7K]
D. All of the above. Developing medicine, analyzing compounds and producing new product such as plastic all have to deal with chemistry. 
3 0
2 years ago
Read 2 more answers
A 3.3 g sample of sodium hydrogen carbonate is added to a solution of acetic acid weighing 10.3 g. The two substances react, rel
Zanzabum

Answer:

1.73g of CO2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

NaHCO3 + CH3COOH → CH3COONa + H2O + CO2

Next we shall determine the masses of NaHCO3 and CH3COOH that reacted and the mass of CO2 produced from the balanced equation. This is illustrated below:

Molar mass of NaHCO3 = 23 + 1 + 12 + (16x3) = 84g/mol

Mass of NaHCO3 from the balanced equation = 1 x 84 = 84g

Molar mass of CH3COOH = 12 + (3x1) + 12 + 16 + 16 + 1 = 60g/mol

Mass of CH3COOH from the balanced equation = 1 x 60 = 60g

Molar mass of CO3 = 12 + (2x16) = 44g/mol

Mass of CO2 from the balanced equation = 1 x 44 = 44g

From the balanced equation above,

84g of NaHCO3 reacted with 60g of CH3COOH to produce 44g of CO2.

Next, we shall determine the limiting reactant of the reaction. This is illustrated below:

From the balanced equation above,

84g of NaHCO3 reacted with 60g of CH3COOH.

Therefore, 3.3g of NaHCO3 will react with = (3.3 x 60)/84 = 2.36g of CH3COOH.

From the above illustration, we can see that only 2.36g of CH3COOH out of 10.3g given reacted completely with 3.3g of NaHCO3. Therefore, NaHCO3 is the limiting reactant while CH3COOH is the excess reactant.

Finally, can determine the mass of CO2 produced during the reaction.

In this case the limiting reactant will be used because it will produce the mass yield of CO2 as all of it were used up in the reaction. The limiting reactant is NaHCO3 and the mass of CO2 produced is obtained as shown below:

From the balanced equation above,

84g of NaHCO3 reacted to produce 44g of CO2.

Therefore, 3.3g of NaHCO3 will react to produce = (3.3 x 44)/84 = 1.73g of CO2.

Therefore, 1.73g of CO2 is released during the reaction.

7 0
2 years ago
PLEASE HELP WHAT IS THE CORRECT ANSWER (if possible let me know why)
ANTONII [103]

Answer:I’m not 100% sure but I think it’s the first answer.

Explanation:

The pressure would increase as the molecules are pushing out word more and more, leading to more space being taken and more pressure. But then again, I’m not sure. I’m just a freshman with below average grades.

8 0
3 years ago
Benzene is a minor component of gasoline. The standard molar enthalpy of formation of benzene C7H16(l) is 48.95 kJ/mol. For the
Radda [10]

Answer:

-3135.47 kJ/mol

Explanation:

Step 1: Write the balanced equation

C₆H₆(l) + 7.5 O₂(g) ⇒ 6 CO₂(g) + 3 H₂O(g)

Step 2: Calculate the standard enthalpy change of the reaction (ΔH°r)

We will use the following expression.

ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)

where,

n: moles

ΔH°f: standard enthalpies of formation

p: products

r: reactants

ΔH°r = 6 mol × ΔH°f(CO₂(g)) + 3 mol × ΔH°f(H₂O(g)) - 1 mol × ΔH°f(C₆H₆(l)) - 7.5 mol × ΔH°f(O₂(g))

ΔH°r = 6 mol × (-393.51 kJ/mol) + 3 mol × (-241.82 kJ/mol) - 1 mol × (48.95 kJ/mol) - 7.5 mol × 0 kJ/mol

ΔH°r = -3135.47 kJ

Since this enthalpy change is for 1 mole of C₆H₆(l), we can express it as -3135.47 kJ/mol.

7 0
2 years ago
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