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UNO [17]
3 years ago
6

if a gas sample in a balloon occupies 1.5 L at atmospheric pressure, what would be the pressure (in mmHg) if the volume was redu

ced to 0.8 L?
Chemistry
1 answer:
Nata [24]3 years ago
3 0

Answer:

1425 mmHg.

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 1.5 L

Initial pressure (P1) = 1 atm

Final volume (V2) = 0.8 L

Final pressure (P2) =?

Next, we shall determine the final pressure of the gas by using the Boyle's law equation as follow:

P1V1 = P2V2

1 × 1.5 = P2 × 0.8

1.5 = P2 × 0.8

Divide both side by 0.8

P2 = 1.5/0.8

P2 = 1.875 atm

Finally, we shall convert 1.875 atm to mmHg.

This can be obtained as follow:

1 atm = 760 mmHg

Therefore,

1.875 atm = 1.875 × 760 = 1425 mmHg.

Therefore, the new pressure of the gas is 1425 mmHg.

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3 years ago
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6 0
3 years ago
A 25.0 mL sample of a solution of an unknown compound is titrated with a 0.115 M NaOH solution. The titration curve above was ob
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Answer:

Weak acid

Explanation:

A titration curve is a graphical description of the change in pH of the solution in the conical flask as the reagent is added from the burette. A titration curve can be plotted for the different kinds of acid and base titrations. The volume of the titrant is always plotted as the independent variable and the pH of the solution as the dependent variable. The equivalence point is read off from the titration curve. A titration curve is very important because it shows the pH at various points during the titration.

A weak acid/strong base titration leads to an equivalence point above 7. From the question, we were told that the pH at equivalence point lies around 8. Hence the unknown substance must be a weak acid.

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8 0
3 years ago
Consider the reaction:
goldfiish [28.3K]

Answer:

K = Ka/Kb

Explanation:

P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?

P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka

PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb

K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)

Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)

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Since [PCl₅] = [PCl₅]

From the Ka equation,

[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)

From the Kb equation

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Equating them

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(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)

(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)

Comparing this with the equation for the overall equilibrium constant

K = Ka/Kb

5 0
3 years ago
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