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aivan3 [116]
3 years ago
5

What is the new pressure of 150 mL of a gas that is compressed to 50 mL when the original pressure was 3.0 ATM and the temperatu

re is held constant
Chemistry
1 answer:
Arisa [49]3 years ago
8 0

Answer:

\boxed {\boxed {\sf P_2= 9 \ atm}}

Explanation:

Since the temperature is held constant, we only need to focus on the volume and pressure. We will use Boyle's Law, which states the volume of a gas is inversely proportional to the pressure. The formula is:

P_1V_1=P_2V_2

Originally, the gas had a volume of 150 milliliters and a pressure of 3.0 atmospheres. We can substitute these values into the left side of the equation.

3.0 \ atm * 150 \ mL = P_2V_2

The original gas was compressed to a volume of 50 milliliters, but we don't know the volume.

3.0 \ atm *150 \ mL= P_2 * 50 \ mL

Now, we need to solve for the new pressure (P₂). Multiply on the left side first.

450 \ atm*mL= P_2 * 50 \ mL

Since we are solving for the pressure, we need to isolate the variable. It is being multiplied by 50 mL. The inverse of multiplication is division. Divide both sides by 50 mL.

\frac{450 \ atm*mL}{50 \ mL}= \frac{P_2 * 50 \ mL}{50 \ mL}

\frac{450 \ atm*mL}{50 \ mL}= P_2

The units of milliliters will cancel.

\frac{450 \ atm}{50 }= P_2

9 \ atm =P_2

The new pressure is <u>9 atmospheres.</u>

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5 0
3 years ago
onsider the reversible dissolution of lead(II) chloride. P b C l 2 ( s ) − ⇀ ↽ − P b 2 + ( a q ) + 2 C l − ( a q ) PbClX2(s)↽−−⇀
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Answer:

9.34x10^-4

Explanation:

Step 1:

The balanced equation for the reaction.

PbCl2( s ) <=> Pb^2+(aq) + 2Cl^−(aq)

Step 2:

Data obtained from the question:

Mass of PbCl2 = 0.2393 g

Volume = 50mL

concentration of Pb^2+, [Pb^2+] = 0.0159 M

Concentration of Cl^-, [Cl^-] = 0.0318 M

Equilibrium constant, Kc =?

Step 3:

Determination of the number of mole PbCl2.

The number of mole of PbCl2 can be obtained as follow:

Molar Mass of PbCl2 = 207 + (35.5x2) = 278g/mol

Mass of PbCl2 = 0.2393 g

Number of mole =Mass /Molar Mass

Number of mole of PbCl2 = 0.2393/278 = 8.61x10^-4 mole

Step 4:

Determination of Molarity of PbCl2.

At this stage we shall obtain the molarity of PbCl2. This is shown below:

Mole of PbCl2 = 8.61x10^-4 mole

Volume = 50mL = 50/1000 = 0.05L

Molarity of PbCl2 =?

Molarity = mole /Volume

Molarity of PbCl2 = 8.61x10^-4/0.05

Molarity of PbCl2 = 0.01722 M

Step 5:

Determination of the equilibrium constant Kc.

PbCl2( s ) <=> Pb^2+(aq) + 2Cl^−(aq)

The equilibrium constant Kc for the equation above is given by:

Kc = [Pb^2+] [Cl^-]^2 / [PbCl2]

[Pb^2+] = 0.0159 M

[Cl^-] = 0.0318 M

[PbCl2] = 0.01722 M

Kc =?

Kc = [Pb^2+] [Cl^-]^2 / [PbCl2]

Kc = 0.0159 x (0.0318)^2/ 0.01722

Kc = 9.34x10^-4

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Answer:

<span>C</span>

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