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Sergeu [11.5K]
2 years ago
6

The volume of a gas is 27.5 mL at 22.0°C and 0.974 atm. What will the volume be at 15.0°C and 0.993 atm? Use Ideal Gas Law (PV =

nRT) to find the appropriate gas law for this problem.
Chemistry
1 answer:
Scorpion4ik [409]2 years ago
4 0

Answer:

26.3 mL

Explanation:

Step 1:

Obtaining an appropriate gas law from the ideal gas equation.

This is illustrated below:

From the ideal gas equation:

PV = nRT

Divide both side by T

PV/T = nR

At this stage, we'll assume the number of mole (n) to be constant.

Note: R is the gas constant.

PV/T = constant.

We can thus, write the above equation as:

P1V1/T1 = P2V2/T2

The above equation is called the general gas equation.

Step 2:

Data obtained from the question. This includes the following:

Initial volume (V1) = 27.5 mL

Initial temperature (T1) = 22.0°C = 22.0°C + 273 = 295K

Initial pressure (P1) = 0.974 atm.

Final temperature (T2) = 15.0°C = 15.0°C + 273 = 288K

Final pressure (P2) = 0.993 atm

Final volume (V2) =..?

Step 3:

Determination of the final volume of the gas using the general gas equation obtained. This is illustrated below:

P1V1 /T1 = P2V2/T2

0.974 x 27.5/295 = 0.993 x V2/288

Cross multiply to express in linear.

295x0.993xV2 = 0.974x27.5x288

Divide both side by 295 x 0.993

V2 = (0.974x27.5x288)/(295x0.993)

V2 = 26.3 mL

Therefore, the new volume of the gas is 26.3 mL

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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
Sodium oxide (Na2O) crystallizes in a structure in which the O2– ions are in a face - centered cubic lattice and the Na+ ions ar
melisa1 [442]

We have to know the number of Na⁺ ions in the unit cell.

The number of Na⁺ ions in the unit cell is (D) 8.

Sodium oxide (Na2O) crystallizes in a structure in which the O2– ions are in a face - centered cubic lattice and the Na+ ions are in tetrahedral holes.

O²⁻ ions are in a face centred cubic lattice, so the number of O²⁻ ions per unit cell is equal to 4. The number of  tetrahedral hole= 2 X 4=8. Na+ ions are present in tetrahedral holes, which indicates there are 8 number of Na+ ions in the unit cell.


3 0
2 years ago
A sample of fluorine gas is confined in a 5.0-L container at 0.432 atm and 37 °C. How many moles of gas are in this sample?
Ahat [919]

Answer:

About 0.08486 moles

Explanation:

PV=nRT, when P is the pressure in atmospheres, V is the volume in liters, n is the number of moles, R is the universal gas constant, and T is the temperature in Kelvin.

0.432 \cdot 5= n \cdot 0.0821 \cdot 310

n\approx 0.08486 moles

Hope this helps!

7 0
2 years ago
Rank the formation of the solutions A, B, and C from the most exothermic to the most endothermic. Rank the enthalpy of solution
Elodia [21]

This is an incomplete question, the table is attached below.

Answer : The correct ranking of the solution from most exothermic to most endothermic will be: A, B and C.

Explanation :

As we know that the intermolecular force of attraction play an important role in the interaction of solute-solute, solute-solvent and solvent solvent solution.

In the solution A, the solute-solute and solvent-solvent interactions are weak. So, their solute-solvent interaction will be strong. That means, the solution will be more exothermic.

In the solution C, the solute-solute and solvent-solvent interactions are strong. So, their solute-solvent interaction will be weak. That means, the solution will be more endothermic.

Thus, the correct ranking of the solution from most exothermic to most endothermic will be: A, B and C.

4 0
2 years ago
A solution is prepared by mixing 50.0 mL of 0.50 M Cu(NO3)2 with 50.0 mL of 0.50 M Co(NO3)2. Sodium hydroxide is added to the mi
vekshin1

Answer:

Cu(OH)₂ will precipitate first, with [OH⁻] = 2.97x10⁻¹⁰ M

Explanation:

The equilibriums that take place are:

Cu⁺² + 2OH⁻ ↔ Cu(OH)₂(s)   ksp = 2.2x10⁻²⁰ = [Cu⁺²]*[OH⁻]²

Co⁺² + 2OH⁻ ↔ Co(OH)₂(s)   ksp = 1.3x10⁻¹⁵ = [Co⁺²]*[OH⁻]²

Keep in mind that <em>the concentration of each ion is halved </em>because of the dilution when mixing the solutions.

For Cu⁺²:

2.2x10⁻²⁰ = [Cu⁺²]*[OH⁻]²

2.2x10⁻²⁰ = 0.25 M*[OH⁻]²

[OH⁻] = 2.97x10⁻¹⁰ M

For Co⁺²:

1.3x10⁻¹⁵ = [Co⁺²]*[OH⁻]²

1.3x10⁻¹⁵ = 0.25 M*[OH⁻]²

[OH⁻] = 7.21x10⁻⁸ M

<u>Because Copper requires less concentration of OH⁻ than Cobalt</u>, Cu(OH)₂ will precipitate first, with [OH⁻] = 2.97x10⁻¹⁰ M

3 0
2 years ago
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