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LuckyWell [14K]
3 years ago
5

A sample of gas occurs 9.0mL at a pressure of 500mmHg. A new volume of the same sample is at 750mmHg. Use two significant figure

s in answer on volume.
Chemistry
1 answer:
Anit [1.1K]3 years ago
5 0

Answer:

V₂ = 6.0 mL

Explanation:

Given data:

Initial volume = 9.0 mL

Initial pressure = 500 mmHg

Final volume = ?

Final pressure = 750 mmHg

Solution:

According to Boyle's Law

P₁V₁ = P₂V₂

V₂ = P₁V₁ / P₂

V₂ = 500 mmHg × 9.0 mL / 750 mmHg

V₂ = 4500 mmHg .mL / 750 mmHg

V₂ = 6.0 mL

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For the equilibrium PCl5(g) PCl3(g) + Cl2(g), Kc = 2.0 × 101 at 240°C. If pure PCl5 is placed in a 1.00-L container and allowed
Cloud [144]

Answer:

the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M

Explanation:

for the reaction

PCl₅(g) → PCl₃(g) + Cl₂(g)

where

Kc= [PCl₃]*[Cl₂]/[PCl₅] = 2.0*10¹ M = 20 M

and [A] denote concentrations of A

if initially the mixture is pure PCl₅ , then it will dissociate according to the reaction and since always one mole of PCl₃(g) is generated with one mole of Cl₂(g) , the total number of moles of both at the end is the same → they have the same concentration → [PCl₃(g)] = [Cl₂]=0.27 M

therefore

Kc= [PCl₃]*[Cl₂]/[PCl₅] = 0.27 M* 0.27 M /[PCl₅] = 20 M

[PCl₅]  =  0.27 M* 0.27 M / 20 M = 3.64*10⁻³ M

[PCl₅]  = 3.64*10⁻³ M

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6 0
3 years ago
A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. Calculate the pressure when the volume is 1.41 L and the
Vlad1618 [11]

A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. 1.76 atm is the pressure when the volume is 1.41 L and the temperature is 298 K.

<h3>What is Combined Gas Law ?</h3>

This law combined the three gas laws that is (i) Charle's Law (ii) Gay-Lussac's Law and (iii) Boyle's law.

It is expressed as

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

where,

P₁ = first pressure

P₂ = second pressure

V₁ = first volume

V₂ = second volume

T₁ = first temperature

T₂ = second temperature

Now put the values in above expression we get

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

\frac{1.02\ atm \times 2.30\ L}{281\ K} = \frac{P_2 \times 1.41\ L}{298\ K}

P_{2} = \frac{1.02\ atm \times 2.30\ L \times 298\ K}{281\ K \times 1.41\ L}

P₂ = 1.76 atm

Thus from the above conclusion we can say that A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. 1.76 atm is the pressure when the volume is 1.41 L and the temperature is 298 K.

Learn more about the Combined gas Law here: brainly.com/question/13538773

#SPJ4

4 0
2 years ago
Help please. Which mass of urea, CO(NH2)2, contains the same mass of nitrogen as 101.1 g of potassium nitrate?
Masja [62]
Did u mean mass of potassium nitrate or the mass of nitrogen in potassium nitrate

6 0
3 years ago
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