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olchik [2.2K]
3 years ago
11

A student weighs an empty flask and stopper and finds the mass to be 53.256 g. She then adds about 5 mL of an unknown liquid and

heats the flask in a boiling water bath at 98.8°C. After all the liquid is vaporized, she removes the flask from the bath, stoppers it, and lets it cool. After it is cool, she momen- tarily removes the stopper, then replaces it and weighs the flask and condensed vapor, obtaining a mass of 53.780 g. The volume of the flask is known to be 231.1 mL. The absolute atmospheric pressure in the laboratory that day is 728 mm Hg. a. What was the pressure of the vapor in the flask in atm?
Chemistry
1 answer:
Delvig [45]3 years ago
3 0

Answer:

a. The pressure in the flask open to the atmosphere during the vaporization of the unknown liquid is equal to the prevailing atmospheric pressure equivalent to 0.957734 atm

Explanation:

The mass of the empty flask and stopper, m = 53.256 g

The volume of the unknown liquid she adds = 5 mL

The temperature of the water in which she heats up the flask = 98.8 °C = 371.95 K

The mass of the flask and the condensed vapor = 53.780 g

The volume of the flask, V = 231.1 mL

The atmospheric pressure, P = 728 mm Hg

a. We note that the student stoppers the flask after all the liquid has evaporated. Therefore, given that the flask was open to the atmospheric pressure as the liquid evaporates, the pressure of the vapor in the flask is equal to the prevailing atmospheric pressure, or 728 mmHg

Using a calculator, 728 mm Hg is equivalent to 0.957734 atm.

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Pressure is the force exerted by a substance on its container. Of the three phases of matter which one would exert the highest p
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Gaseous phase exert more pressure on container.

Explanation:

Pressure:

It is the force exerted by the substance on its container.

Mathematical expression:

P = F/A

P = pressure

F = force

A = area

We know that there are phases of matter solid liquid and gas.

Gases don't have definite volume and shape and take the shape and volume of container in which it present.  Molecule of gases randomly move everywhere and occupy all available space.  The molecules of gas randomly move collide with each other and also the wall of container their kinetic energy also increased because of this collision and pressure is produced

While in case of liquid molecules are packed and they can not move freely thus did not exert pressure like gaseous molecules.

In case of solids molecules are tightly packed and form more compact structure. They can not move thus did not exert pressure to the wall of container.

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Molecule of gases randomly move everywhere and occupy all available space.

Gases don't have definite volume and shape and take the shape and volume of container in which it present.

Their densities are very low as compared to the liquid and solids.

Gas molecules are at long distance from each other therefore by applying pressure gases can be compressed.

The very weak inter molecular forces are present between gas molecules.

Properties of Liquid:

Liquid have definite volume but don,t have definite shape.

Their densities are high as compared to the gases but low as compared to the solids.

In liquid, molecules are close to each other and have greater inter molecular forces as compared to the gas molecules.

Properties of solids:

Solids have definite volume and shape.

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What type of consumer are bears?
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4 0
3 years ago
g 2BrO3- + 5SnO22-+ H2O5SnO32- + Br2+ 2OH- In the above reaction, the oxidation state of tin changes from to . How many electron
Archy [21]

Answer:

In the above reaction, the oxidation state of tin changes from 2+ to 4+.

10 moles of electrons are transferred in the reaction

Explanation:

Redox reaction is:

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

SnO₂²⁻ → SnO₃²⁻

Tin changes the oxidation state from +2 to +4. It has increased it so this is the oxidation from the redox (it released 2 e⁻). We are in basic medium, so we add water in the side of the reaction where we have the highest amount of oxygen. We have 2 O on left side and 3 O on right side so we add 1 water on the right and we complete with OH⁻ in the opposite side to balance the H.  

SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O <u>Oxidation</u>

BrO₃⁻ →  Br₂

First of all, we have unbalance the bromine, so we add 2 on the BrO₃⁻. We have 6 O in left side and there are no O on the right, so we add 6 H₂O on the left. To balance the H, we must complete with 12OH⁻. Bromate reduces to bromine at ground state, so it gained 5e⁻. We have 2 atoms of Br, so finally it gaines 10 e⁻.

6H₂O + 10 e⁻ + 2BrO₃⁻ →  Br₂ + 12OH⁻ <u>Reduction</u>

In order to balance the main reaction and balance the electrons we multiply  (x5) the oxidation and (x1) the reduciton

(SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O) . 5

(6H₂O + 10 e⁻ + 2BrO₃⁻ →  Br₂ + 12OH⁻) . 1

5SnO₂²⁻ + 10OH⁻ + 6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ + 5SnO₃²⁻ + 10e⁻ + 5H₂O

We can cancel the e⁻ and we substract:

12OH⁻ - 10OH⁻ = 2OH⁻ (on the right side)

6H₂O - 5H₂O = H₂O (on the left side)

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

6 0
3 years ago
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