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mamaluj [8]
2 years ago
11

A solution contains one or more of the following ions: Ag + , Ca 2 + , and Co 2 + . Ag+, Ca2+, and Co2+. Lithium bromide is adde

d to the solution and no precipitate forms. An excess of lithium sulfate is then added to the solution and a precipitate forms. The precipitate is filtered off and lithium phosphate is added to the remaining solution, producing a precipitate. A.Which ions are present in the original solution? Co 2 + Co2+ Ca 2 + Ca2+ Ag + Ag+ B.Write a net ionic equation for the formation of the precipitate observed after the addition of lithium sulfate. Include physical states. net ionic equation: C.Write a net ionic equation for the formation of the precipitate observed after the addition of lithium phosphate. Include physical states. net ionic equation:
Chemistry
1 answer:
Eva8 [605]2 years ago
8 0

Answer:

Since with LiBr no precipitation takes place. So, Ag+ is absent

When we add Li2SO4 to it, precipitation takes place.

Ca2+(aq) + SO42-(aq) ----> CaSO4(s) ...Precipitate

Thus, Ca2+ is present.

When Li3PO4 is added, again precipitation takes place.Reaction is:

Co2+(aq) + PO43-(aq)---->Co3(PO4)2(s) ... Precipitate

A. Ca2+ and Co2+ are present in solution

B. Ca2+(aq) + SO42-(aq) ----> CaSO4(s)

C. 3Co2+(aq) + 2PO43-(aq)---->Co3(PO4)2(s)

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A 0.050 M solution of AlCl3 had an observed osmotic pressure of 3.85 atmatm at 20°C.Calculate the van't Hoff factor iii for AlCl
Alja [10]

Answer:

The actual Van't Hoff factor for AlCl3 is 3.20

Explanation:

Step 1: Data given

Molarity of AlCl3 = 0.050 M

osmotic pressure = 3.85 atm

Temperature = 20 °C

Step 2: Calculate the Van't Hoff factor

AlCl3(aq) → Al^3+(aq) + 3Cl^-(aq)

The theoretical value is 4 ( because 1 Al^3+ ion + 3 Cl- ions) BUT due to the interionic atractions the actual value will be less

Osmotic pressure depends on the molar concentration of the solute but not on its identity., and is calculated by:

π = i.M.R.T

 ⇒ with π = the osmotic pressure = 3.85 atm

⇒ with i = the van't Hoff factor

⇒ with M = the molar concentration of the solution = 0.050 M

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 20 °C = 293.15 Kelvin

i = π /(M*R*T )

i = (3.85) / (0.050*0.08206*293.15)

i = 3.20

The actual Van't Hoff factor is 3.20

6 0
2 years ago
Movement of the ___<br> creates the London dispersion forces.
Tanzania [10]

Answer: electrons

Explanation: moving electrons cause momentarily charge

Distribution on molecule. This distribution induces similar distribution to

Adjacent molecule.

7 0
2 years ago
Consider the solutions, 0.04 m urea [(NH2)2C=O)], 0.04 m AgNO3 and 0.04 m CaCl2. Which has (i) the highest osmotic pressure, (ii
Lana71 [14]

Answer:

i) Highest osmotic pressure: CaCl2

ii) lower vapor pressure : CaCl2

iii) highest boiling point : CaCl2

Explanation:

The colligative properties depend upon the number of solute particles in a solution.

The following four are the colligative properties:

a) osmotic pressure : more the concentration of the solute, more the osmotic pressure

b) vapor pressure: more the concentration of the solute, lesser the vapor pressure.

c) elevation in boiling point: more the concentration of the solute, more the boiling point.

d) depression in freezing point: more the concentration of the solute, lesser the freezing point.

the number of particle produced by urea = 1

the number of particle produced by AgNO3 = 2

the number of particle produced by CaCl2 = 3

As concentrations are same, CaCl2 will have more number of solute particles and urea will have least

i) Highest osmotic pressure: CaCl2

ii) lower vapor pressure : CaCl2

iii) highest boiling point : CaCl2

5 0
2 years ago
A gas sample enclosed in a rigid metal container at room temperature (20.0∘C∘C) has an absolute pressure p1p1p_1. The container
choli [55]

Answer: p2 = 1.06p1

Explanation: pressure increases with temperature increase.

According to Gass law

P1/T1 = P2/T2

T1 = 20°c = 20 +273 = 293k

T2 = 40°c = 40 +373 = 313k

Therefore

P2 = P1T2/T1 = 313P2/293

P2 = 1.06P1

3 0
2 years ago
Perform the following
Ghella [55]

Answer:

1.85 × 10⁻⁶

Explanation:

0.0003 ÷ 162 = 1.851851852 × 10⁻⁶ ⇒ 1.85 × 10⁻⁶

Hope that helps.

4 0
2 years ago
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