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Y_Kistochka [10]
3 years ago
5

Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points. nh4cl cobr3 k2so

4

Chemistry
2 answers:
Ira Lisetskai [31]3 years ago
8 0

The freezing order from the largest to the smallest is:

NH4Cl> K2SO4> COBr3

<h3><em>Further explanation </em></h3>

For electrolyte solutions:

ΔTb = Kb.m. i

ΔTf = Kf.m. i

Kb = molal boiling point increase

Kf = molal freezing point constant

m = molal solution

i = van't Hoff factor

i = 1 + (n-1) α

The formula above shows that the freezing point depends on

• molal value

• degree of ionization/dissociation

• number of ions in solution

In the statement of the matter, it is known that there are solutions which have the same concentration and are completely dissociated (α = 1)

So the value of the freezing point drop depends only on the number of ions in the solution

The more ions produced, the greater the value of the freezing point (the freezing point is smaller)

There are several solutions

• 1. NH4Cl ionization:

NH4Cl ---> NH4+ + Cl-

There are 2 ions in the solution

• 2. COBr3 ionization:

COBr3 ---> CO3++ 3Br-

There are 4 ions in the solution

• 3. ionization K2SO4

K2SO4 ---> 2K ++ SO42-

There are 3 ions in the solution

So the order of freezing from the largest to the smallest is:

NH4Cl> K2SO4> COBr3

<h3><em>Learn more </em></h3>

The freezing point of a solution

brainly.com/question/8564755

brainly.com/question/4593922

brainly.com/question/1196173

Keywords: freezing point,  properties, van't Hoff factor

Illusion [34]3 years ago
4 0
Answer: CoBr3 < K2SO4 < NH4 Cl

Justification:

1) The depression of the freezing point of a solution is a colligative property, which means that it depends on the number of particles of solute dissolved.

2) The formula for the depression of freezing point is:

ΔTf = i * Kf * m

Where i is the van't Hoof factor which accounts for the dissociation of the solute.

Kf is the freezing molal constant and only depends on the solvent

m is the molality (molal concentration).

3) Since, you are assuming equal concentrations and complete dissociation of the given solutes, the solute with more ions in the molecular formula will result  in the solution with higher depression of the freezing point (lower freezing point).

4) These are the dissociations of the given solutes:

a) NH4 Cl (s) --> NH4(+)(aq) + Cl(-) (aq) => 1 mol --> 2 moles

b) Co Br3 (s) --> Co(3+) (aq) + 3Br(-)(aq) => 1 mol --> 4 moles

c) K2SO4 (s) --> 2K(+) (aq) + SO4 (2-) (aq) => 1 mol --> 3 moles

5) So, the rank of solutions by their freezing points is:

CoBr3 < K2SO4 < NH4 Cl
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Starting with 195 g Li2O and 106 g H2O, decide which reactant is present in limiting quantities. Given: Li2O+H2O→2LiOH
kkurt [141]

Taking into account the stoicionetry reaction and the definition of limiting reagent, Li₂O is present in limiting quantities.

The balanced reaction is:

Li₂O + H₂O → 2 LiOH

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Li₂O: 1 mole
  • H₂O: 1 mole
  • LiOH: 2 moles

The molar mass of each compound is:

  • Li₂O: 29.88 g/mole
  • H₂O: 18 g/mole
  • LiOH: 23.95 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Li₂O: 1 mole× 29.88 g/mole= 29.88 g
  • H₂O: 1 mole× 18 g/mole= 18 g
  • LiOH: 2 moles× 23.95 g/mole= 47.9 g

On the other side, the limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 18 grams of H₂O reacts with 29.88 grams of Li₂O, 106 grams of H₂O react with how much mass of Li₂O?

mass of Li_{2} O=\frac{29.88 grams of Li_{2} Ox106 grams of H_{2} O}{18 grams of H_{2} O}

mass of Li₂O= 175.96 grams

But 175.96 grams of Li₂O are not available, 195 grams are available. Since you have less moles than you need to react with 106 grams of H₂O, Li₂O will be the limiting reagent.

Finally, Li₂O is present in limiting quantities.

Learn more:

  • <u>brainly.com/question/11621448?referrer=searchResults</u>
  • <u>brainly.com/question/14747537?referrer=searchResults</u>
  • <u>brainly.com/question/11848702?referrer=searchResults</u>

4 0
3 years ago
How many milligrams of CO exist in 12000 liters of air with 10 ppm of CO?
VARVARA [1.3K]

Answer:

The mass of CO in 12,000 Liters of air is 12\times 10^4 mg.

Explanation:

The ppm is the amount of solute (in milligrams) present in one Liter of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

\text{ppm}=\frac{\text{Mass of solute}(mg)}{\text{Volume of solution}(L)}

The ppm concentration of CO = 10 ppm

Mass of CO = m

Volume of air = 12,000 L

10 =\frac{m}{12,000 L}

m=10\times 12,000 mg=120,000 mg=12\times 10^4 mg

The mass of CO in 12,000 Liters of air is 12\times 10^4 mg.

3 0
3 years ago
Calculate the number of atoms in a 3.96 × 103 g sample of silicon.
Crank
Atomic mass Si = 28.0855 u.m.a

28.0855 g -------------------- 6.02x10²³ atoms
3.96x10³ g ------------------- ?? atoms

( 3.96x10³) x 6.02x10²³ / 28.0855 =

8.48x10²⁵ atoms




6 0
3 years ago
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