Answer:
The solutions are listed from the lowest freezing point to highest freezing point.
Mg₃(PO₄)₂ - AlBr₃ - BeBr₂ - KBr
Explanation:
Colligative property of freezing point, is the freezing point depression.
Formula is:
ΔT = Kf . m . i
Where ΔT = T° freezing pure solvent - T° freezing solution
Kf is the cryoscopic constant
m is molality and i is the Van't Hoff factor (number of ions, dissolved in solution)
In this case, T° freezing pure solvent is 0° because it's water, so the Kf is 1.86 °C/m, and m is 0.1 molal.
The i modifies the T° freezing solution. The four salts, are ionic compounds, so for each case:
BeBr₂ → Be²⁺ + 2Br⁻ i = 3
AlBr₃ → Al³⁺ + 3Br⁻ i = 4
Mg₃(PO₄)₂ → 3Mg²⁺ + 2PO₄⁻³ i = 5
KBr → K⁺ + Br⁻ i = 2
Solution of Mg₃(PO₄)₂ will have the lowest temperature of freezing and the solution of KBr, the highest (always lower, than 0°).
Living organisms contain relatively large amounts of oxygen, carbon, hydrogen, nitrogen, and sulfur (thesefive elements are known as the bulk elements), along with sodium, magnesium, potassium, calcium, chlorine, and phosphorus (these six elements are known as macrominerals).
Answer:
Colloidal can not be separated through filtration.
Suspension can be separated through filtration.
Explanation:
Colloidal:
Colloidal consist of the particles having size between 1 - 1000 nm i.e, 0.001- 1μm. While the pore size of filter paper is 2μm. That's why we can not separate the colloidal through the filtration. However it can be separated through the ultra filtration. In ultra filtration the pore size is reduced by soaking the filter paper in gelatin and then in formaldehyde. This is only in case of when solid colloidal is present, if colloid is liquid , there is no solid particles present and ultra filtration can not be used in this case.
Suspension:
The particle size in suspension is greater than 1000 nm. The particles in suspension can be separated through the filtration. These particles are large enough and can be seen through naked eye.
The given question is incorrect. The correct question is as follows.
If 20.0 g of 
and 4.4 g of 
are placed in a 5.00 L container at 
, what is the pressure of this mixture of gases?
Explanation:
As we know that number of moles equal to the mass of substance divided by its molar mass.
Mathematically, No. of moles = 
Hence, we will calculate the moles of oxygen as follows.
No. of moles =
Moles of = 
= 0.625 moles
Now, moles of 
= 0.1 moles
Therefore, total number of moles present are as follows.
Total moles = moles of + moles of
= 0.625 + 0.1
= 0.725 moles
And, total temperature will be:
T = (21 + 273) K = 294 K
According to ideal gas equation,
PV = nRT
Now, putting the given values into the above formula as follows.
P = 
= 
= 
atm
= 3.498 atm
or, = 3.50 atm (approx)
Therefore, we can conclude that the pressure of this mixture of gases is 3.50 atm.
