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Elodia [21]
3 years ago
15

Calculate the solubility of o2 in water at a partial pressure of o2 of 120 torr at 25 ̊c. the henry's law constant for o2 at 25

̊c is 1.3 x 10-3 mol/l atm. how do you expect the solubility to change if the temperature were decreased?

Chemistry
2 answers:
Dimas [21]3 years ago
7 0

A. The solubility of O₂ in water : S = 2.052 . 10⁻⁴ mol/L

B. If the temperature were decreased then the solubility of O₂ will increase

<h3>Further eplanation </h3>

In the transfer from the gas phase to the liquid phase diffusion process occurs.  

The solubility of the gas itself decreases with increasing temperature  

The partial pressure of the gas itself is influenced by the concentration of the gas and temperature  

Henry's Law states that the solubility of a gas is proportional to its partial pressure  

Can be formulated  

S = kH. P.  

S = gas solubility, mol / L  

kH = Henry constant, mol / L.atm  

P = partial gas pressure  

The partial gas pressure is proportional to the percentage of the gas volume in the mixture  

Henry's law constant = 1.3 x 10⁻³mol / l atm.  

P = 120 torr = 0.15789 atm

So  

S = 1.3 x 10⁻³ . 0.15789

S = 2.052 . 10⁻⁴ mol/L

If the temperature were decreased then the solubility of O₂ will increased

<h3>Learn more  </h3>

grams of CO will dissolve in 1.00 l of water  

brainly.com/question/7007748  

the partial pressure of hydrogen gas  

brainly.com/question/7996674  

the solubility of a gas  

brainly.com/question/1221249  

Keywords: Henry's Law, partial pressure, gas solubility  

Vladimir79 [104]3 years ago
6 0

Answer:

1) 2.054 x 10⁻⁴ mol/L.

2) Decreasing the temperature will increase the solubilty of O₂ gas in water.

Explanation:

1) The solubility of O₂ gas in water:

  • We cam calculate the solubility of O₂ in water using Henry's law: <em>Cgas = K P</em>,
  • where, Cgas is the solubility if gas,
  • K is henry's law constant (K for O₂ at 25 ̊C is 1.3 x 10⁻³ mol/l atm),
  • P is the partial pressure of O₂ (P = 120 torr / 760 = 0.158 atm).
  • Cgas = K P = (1.3 x 10⁻³ mol/l atm) (0.158 atm) = 2.054 x 10⁻⁴ mol/L.

2) The effect of decreasing temperature on the solubility O₂ gas in water:

  • Decreasing the temperature will increase the solubilty of O₂ gas in water.
  • When the temperature increases, the solubility of O₂ gas in water will decrease because the increase in T will increase the kinetic energy of gas particles and increase its motion that will break intermolecular bonds and escape from solution.
  • Decreasing the temperature will increase the solubility of O₂ gas in water will because the kinetic energy of gas particles will decrease and limit its motion that can not break the intermolecular bonds and increase the solubility of O₂ gas.


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6 0
2 years ago
CH3OH can be synthesized by the following reaction.
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B: There is 21.56 liter of CO needed

Explanation:

Step 1: Data given

CO(g)+2H2(g)?CH3OH(g)

For 1 mol of CO consumed, we need 2 moles of H2 to produce 1 mol of CH3OH

Molar mass of CO = 28.01 g/mol

Molar mass of H2 = 2.02 g/mol

Molar mass of CH3OH = 32.04 g/mol

Step 2: What volume of H2 gas (in L), measured at 754mmHg and 90?C, is required to synthesize 23.0g CH3OH?

Pressure = 754 mmHg = 0.992 atm

Temperature = 90°C = 363 Kelvin

mass of CH3OH produced = 23.0 grams

Step 3: Calculate moles of CH3OH

Moles CH3OH = mass CH3OH / Molar mass CH3OH

Moles CH3OH = 23.0 grams / 32.04 g/mol

Moles CH3OH = 0.718 moles

Step 4: Calculate moles of H2

For 1 mol of CO consumed, we need 2 moles of H2 to produce 1 mol of CH3OH

For 0.718 moles CH3OH produced, we have 2*0.718 moles =1.436 moles of H2 and 0.718 moles of CO

Step 5: Calculate volume of H2

p*V = n*R*T

with p = the pressure = 0.992 atm

with V the volume = TO BE DETERMINED

with n = the number of moles = 1.436 moles H2

with R = the gasconstant = 0.08206 L*atm/ K*mol

with T = the temperature = 363 Kelvin

V = (n*R*T)/p

V = (1.436*0.08206*363)/0.992

V = 43.12 L

Step 6: Calculate volume of CO

p*V = n*R*T

with p = the pressure = 0.992 atm

with V the volume = TO BE DETERMINED

with n = the number of moles = 0.718  moles CO

with R = the gasconstant = 0.08206 L*atm/ K*mol

with T = the temperature = 363 Kelvin

V = (n*R*T)/p

V = (0.718*0.08206*363)/0.992

V = 21.56 L

A: There is 43.12 Liter of H2 needed

B: There is 21.56 liter of CO needed

3 0
3 years ago
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