Answer: The change in boiling point for 397.7 g of carbon disulfide (Kb = 2.34°C kg/mol) if 35.0 g of a nonvolatile, nonionizing compound is dissolved in it is 
Explanation:
Elevation in boiling point:
where,
= boiling point of solution = ?
= boiling point of pure carbon disulfide= 
= boiling point constant =
m = molality
i = Van't Hoff factor = 1 (for non-electrolyte)
= mass of solute = 35.0 g
= mass of solvent (carbon disulphide) = 397.7 g
= molar mass of solute = 70.0 g/mol
Now put all the given values in the above formula, we get:
Therefore, the change in boiling point is 
<span>explain why the dissolved component does not settle out of a solution -
</span><span>Before saturation, there are attractive forces between solute and solvent. after saturation, the capacity for the attractive forces is reached and no more solute can be dissolved</span>
Answer:
5.8μg
Explanation:
According to the rate or decay law:
N/N₀ = exp(-λt)------------------------------- (1)
Where N = Current quantity, μg
N₀ = Original quantity, μg
λ= Decay constant day⁻¹
t = time in days
Since the half life is 4.5 days, we can calculate the λ from (1) by substituting N/N₀ = 0.5
0.5 = exp (-4.5λ)
ln 0.5 = -4.5λ
-0.6931 = -4.5λ
λ = -0.6931 /-4.5
=0.1540 day⁻¹
Substituting into (1) we have :
N/N₀ = exp(-0.154t)----------------------------- (2)
To receive 5.0 μg of the nuclide with a delivery time of 24 hours or 1 day:
N = 5.0 μg
N₀ = Unknown
t = 1 day
Substituting into (2) we have
[5/N₀] = exp (-0.154 x 1)
5/N₀ = 0.8572
N₀ = 5/0.8572
= 5.8329μg
≈ 5.8μg
The Chemist must order 5.8μg of 47-CaCO3
Answer:
3rd statment
Explanation:
ray 1 and 2 are same vertical line
