Answer:
molecular weight (Mb) = 0.42 g/mol
Explanation:
mass sample (solute) (wb) = 58.125 g
mass sln = 750.0 g = mass solute + mass solvent
∴ solute (b) unknown nonelectrolyte compound
∴ solvent (a): water
⇒ mb = mol solute/Kg solvent (nb/wa)
boiling point:
- ΔT = K*mb = 100.220°C ≅ 373.22 K
∴ K water = 1.86 K.Kg/mol
⇒ Mb = ? (molecular weight) (wb/nb)
⇒ mb = ΔT / K
⇒ mb = (373.22 K) / (1.86 K.Kg/mol)
⇒ mb = 200.656 mol/Kg
∴ mass solvent = 750.0 g - 58.125 g = 691.875 g = 0.692 Kg
moles solute:
⇒ nb = (200.656 mol/Kg)*(0.692 Kg) = 138.83 mol solute
molecular weight:
⇒ Mb = (58.125 g)/(138.83 mol) = 0.42 g/mol
Answer:
The freezing point will be 
Explanation:
The depression in freezing point is a colligative property.
It is related to molality as:

Where
Kf= 
the molality is calculated as:




Depression in freezing point = 
The new freezing point = 
The change is that the air goes up then forms clouds.
Answer:
First one is 5.0 M ammonia and the Second one ?
Explanation:
The question is incomplete. Here is the complete question.
An atom of lead has a radius of 154 pm and the average orbitalspeed of the electron in it is about 1.8x
m/s. Calculate the least possible uncertainty in a measurement of the speed of an electron in an atom of lead. Write your answer as a percentage of the average speed, and round it to significant 2 digits.
Answer: v% = 0.21 m/s
Explanation: To calculate the uncertainty, use <u>Heisenberg's Uncertainty Principle</u>, which states that: ΔpΔx≥
where h is <u>Planck's constant</u> and it is equal to 6.626.
m²kg/s.
Since p (momentum) is p = m.v:
mΔv.Δx ≥ 
Δv = 
Given that: r = x = 1.54.
m and mass of an electron is m=9.1.
kg
Δv = 
Δv = 0.0376.
As percentage of average speed:
Δv.
.100% =
.10² = 0.021.10 = 0.21%
The least possible uncertainty in a speed of an electron is 0.21%.