We can use this equation for boiling point elevation:
ΔT(b) = i K(b) M
when Δ T(b) is the increase of boiling point of the solution.
and i is ( vant Hoff factor, the number of particles or ions per mole-clue.
and K(b) is boiling point increase constant for the solution ( and for water it is equal 0.52 C° Kg/mol)
We can assume i (vant Hoff factor ) = 1 as the sucrose is nonelectrolyte (not readily ionize).
So for water: Tb° = 100 c° and Kb = 0.52 c° Kg / mol
By substitute at:
ΔTb = i Kb M
∴ = 1 * 0.52 * 3.60 = 1.8432 C°
and when Tb = Tb° + ΔTb
∴ Tb = 100 + 1.8432 = 101.8432 C°
Mercury bc it is melting and getting closer
It has different moleculs sorry I am guessing
Answer:

Explanation:
Hello,
Considering the ideal equation of state:

The moles are defined in terms of mass as follows:

Whereas
the gas' molar mass, thus:

Now, since the density is defined as the quotient between the mass and the volume, we get:

Solving for
:

Thus, the result is given by:
![density=\frac{(1atm)(44g/mol)}{[0.082atm*L/(mol*K)]*298.15K} \\density=1.8g/L=1.8x10^{-3}g/mL](https://tex.z-dn.net/?f=density%3D%5Cfrac%7B%281atm%29%2844g%2Fmol%29%7D%7B%5B0.082atm%2AL%2F%28mol%2AK%29%5D%2A298.15K%7D%20%5C%5Cdensity%3D1.8g%2FL%3D1.8x10%5E%7B-3%7Dg%2FmL)
Best regards.
Answer : 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.
Solution : Given,
Mass of Cu = 300 g
Molar mass of Cu = 63.546 g/mole
Molar mass of
= 183.511 g/mole
- First we have to calculate the moles of Cu.

The moles of Cu = 4.7209 moles
From the given chemical formula,
we conclude that the each mole of compound contain one mole of Cu.
So, The moles of Cu = Moles of
= 4.4209 moles
- Now we have to calculate the mass of
.
Mass of
= Moles of
× Molar mass of
= 4.4209 moles × 183.511 g/mole = 866.337 g
Mass of
= 866.337 g = 0.8663 Kg (1 Kg = 1000 g)
Therefore, 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.