Answer: P= 1.64 atm
Explanation: solution attached.
Use Ideal gas law
PV= nRT
Derive for P
P= nRT/V R= 0.08205 L.atm/mol.K
Substitute the values.
Answer:
Mass = 55.52 g
Explanation:
Given data:
Number of atoms of Li = 4.81×10²⁴ atom
Number of grams = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen
For Li:
4.81×10²⁴ atom × 1 mol / 6.022 × 10²³ atom
8 moles
Mass in gram:
Mass = number of moles × molar mass
Mass = 8 mol × 6.94 g/mol
Mass = 55.52 g
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.
Answer:
Explanation:
If one mole of carbon monoxide has a mass of 28.01 g and one mole of carbon dioxide has a mass of 44.01 g , it follows that the reaction produces 44.01 g of carbon dioxide for every 28.01 g of carbon monoxide.