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tankabanditka [31]
4 years ago
14

The substance fluorine has the following properties: normal melting point: 53.5 K normal boiling point: 85.0 K triple point: 1.6

×10-4 atm, 53.4 K critical point: 55 atm, 144.1 K A sample of fluorine at a pressure of 1.00 atm and a temperature of 90.3 K is cooled at constant pressure to a temperature of 49.3 K. Which of the following are true? Choose all that apply The sample is initially a gas. The liquid initially present will vaporize. The final state of the substance is a solid. One or more phase changes will occur. The final state of the substance is a liquid. Submit Answer
Chemistry
1 answer:
mihalych1998 [28]4 years ago
7 0

Answer:

The true statements include;

- The sample is initially a gas.

- The final state of the substance is a solid.

- One or more phase changes will occur.

The untrue/false statements include;

- The liquid initially present will vaporize.

- The final state of the substance is a liquid.

Explanation:

A couple pieces of informatton on Fluorine is imitially provided.

The substance fluorine has the following properties: normal melting point: 53.5 K normal boiling point: 85.0 K triple point: 1.6×10-4 atm, 53.4 K critical point: 55 atm, 144.1 K

So, a question is now attached about a sample of Fluorine. A sample of fluorine at a pressure of 1.00 atm and a temperature of 90.3 K is cooled at constant pressure to a temperature of 49.3 K.

We are then told to examine a group of options to find the ones that are correct/apply.

Taking the options one at a time

- The sample is initially a gas.

The initial state of the Fluorine sample has its temperature at 90.3 K, which is above the gas' boiling point. Hence, the sample can be concluded to initially be a gas.

- The liquid initially present will vaporize.

The sample doesn't initially contain liquid. And even of it did, the temperature is cooled, not heated , Hence, this statement is wrong.

- The final state of the substance is a solid.

The sample of Fluorine moves from a temperature higher than boiling point (85.0 K), with the sample in gaseous form, to one that is at a lower temperature (49.3 K) than the gas' normal melting point (53.5 K).

At temperatures lower than melting point, a substance exists in the solid form. Hence, this statement is true. The final state of the substance is solid.

- One or more phase changes will occur.

In moving from 90.3 K to 49.3 K for the sample and passing through the substance's boiling and melting points (85.0 K and 53.5 K respectively) along the way, it is logical to conclude that there would be one or more phase changes will occur. This statement is true.

- The final state of the substance is a liquid.

This is false as we already established that the final state of the substance is a solid. Hence, this statement is false.

Hope this Helps!!!

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Read 2 more answers
A solution is prepared by dissolving 33.0 milligrams of sodium chloride in 1000. L of water. Assume a final volume of 1000. lite
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Answer:

a. Molarity of NaCl solution = 5.64 * 10⁻⁷ mol/L

b. molarity of Na⁺ = 5.64 * 10⁻⁷ mol/L

c. molarity of Cl⁻ = 5.64 * 10⁻⁷ mol/L

d. Osmolarity = 1.128 osmol

e. mass percent of NaCl = 3.30 * 10⁻⁶ %

f. parts per million NaCl = 0.033 ppm NaCl

g. parts per billion of NaCl = 33 ppb of NaCl

h. From the values obtained from e, f and g, the most convenient to use and understand is parts per billion as it has less of a fractional part to deal with especially since the solute concentration is very small.

Explanation:

Molarity of a solution = number of moles of solute (moles)/volume of solution (L)

where number of moles of solute = mass of solute (g)/molar mass of solute (g/mol)

a. Molarity of NaCl:

molar mass of NaCl = 58.5 g/mol, mass of NaCl = 33.0/1000) g = 0.033g

number of moles of NaCl = 0.033/58.5 = 0.000564 moles

Molarity of NaCl solution = 0.000564/1000 = 5.64 * 10⁻⁷ mol/L

b. Equation for the dissociation of NaCl in solution: NaCl ----> Na⁺ + Cl⁻

From the above equation I mole of NaCl dissociates to give 1 mole of Na⁺ ions,

Therefore molarity of Na⁺ = 1 * 5.64 * 10⁻⁷ mol/L = 5.64 * 10⁻⁷ mol/L

c. From the above equation I mole of NaCl dissociates to give 1 mole of Cl⁻ ions,

therefore molarity of Cl⁻ = 1 * 5.64 * 10⁻⁷ mol/L = 5.64 * 10⁻⁷ mol/L

d. From the above equation, dissociation of NaCl in water produces 1 mol Na⁺ and 1 mole Cl⁻.

Total number of particles produced = 2

Osmolarity of solution = number of particles * molarity of siolution

Osmolarity = 2 * 5.64 * 10⁻⁷ mol/L = 1.128 osmol

e. mass of percent of NaCl = {mass of NaCl (g)/ mass of solution (g)} * 100

density of water = 1 Kg/L

mass of water = 1 Kg/L * 1000 L = 1000 kg

1Kg = 1000 g

Therefore mass of solution in g = 1000 * 1000 = 1 * 10⁶ g

mass percent of NaCl = (0.033/1 * 10⁶) * 100 = 3.30 * 10⁻⁶ %

f. Parts per million of NaCl:

parts per million = 1 mg of solute/L of solution

One thousandth of a gram is one milligram and 1000 ml is one liter, so that 1 ppm = 1 mg per liter = mg/Liter.

Since the density of water is 1kg/L = 1,000,000 mg/L

1mg/L = 1mg/1,000,000mg or one part in one million.

parts per million NaCl = 33.0/1000 L = 0.033 ppm NaCl

g. Parts per billion = 1 µg/L of solution

1 g = 1000 µg

therefore, 33.0 mg = 33.0 * 1000 µg = 3.30 * 10⁴ µg

parts per billion of NaCl = 3.30 * 10⁴ µg/1000 L = 33 ppb of NaCl

h. From the values obtained from e, f and g, the most convenient to use and understand is parts per billion as it has less of a fractional part to deal with especially since the solute concentration is very small.

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