Boyle's law states that pressure is inversely proportional to volume of gas at constant temperature
PV = k
where P - pressure , V - volume and k - constant
P1V1 = P2V2
where parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
substituting these values in the equation
1.25 atm x 0.75 L = P x 1.1 L
P = 0.85 atm
final pressure is B) 0.85 atm
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.
D. Cell membrane: surrounds a cell and allows substances to pass in and out