The molar mass (atomic weight ) of sodium is 23.0 grams/mole and the molar mass of sodium azide, NaN3 , is the mass of sodium, 23.0 gram/mole added to the molar mass of three atoms of nitrogen (14.0 x 3 = 42 gram/mole) which equals 65.0 grams/mole. The percentage of sodium is 23.0 /65.0 x 100 % = 35 %
C is the answer I’m pretty sure
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.
The freezing point depression is a colligative property which means that it is proportional to the number of particles dissolved.
The number of particles dissolved depends on the dissociation constant of the solutes, when theyt are ionic substances.
If you have equal concentrations of two solutions on of which is of a ionic compound and the other not, then the ionic soluton will contain more particles (ions) and so its freezing point will decrease more (will be lower at end).
In this way you can compare the freezing points of solutions of KCl, Ch3OH, Ba(OH)2, and CH3COOH, which have the same concentration.
As I explained the solution that produces more ions will exhibit the greates depression of the freezing point, leading to the lowest freezing point.
In this case, Ba(OH)2 will produce 3 iones, while KCl will produce 2, CH3OH will not dissociate into ions, and CH3COOH will have a low dissociation constant.
Answer: Then, you can predict that Ba(OH)2 solution has the lowest freezing point.