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1 year ago

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A 684.6 mL sample of carbon dioxide was heated to 387 K. If the volume of the carbon dioxide sample at 387 K is 933.9 mL, what w

as its temperature at 684.6 mL

1 answer:

erma4kov [3.2K]1 year ago

6 0

According to Charle's law equation, the volume of a sample of carbon dioxide will be 684.6 mL and the temperature will be 283.60 K.

<h3>Charle's law equation</h3>

According to Charles' Law, the volume of a gas container with a certain amount of gas within is precisely proportional to the temperature when the pressure is constant.

<h3>Calculation:</h3>

$V_{1}$ = 684.6 mL

$V_{2}$ = 933.9 mL

$T_{1}$ = ?

$T_{2}$ = 387 K

Using Charle's law equation

$\frac{684.6}{T_{1} } = \frac{933.9}{387}$

$T_{1} = \frac{2,64,862.8}{933.9}$

$T_{1} = 283.60 K$

According to Charle's law equation, the temperature at 684.6 mL volume was 283.60 K.

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3 years ago

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2 years ago

A. Based on the activation energies and frequency factors, rank the following reactions from fastest to slowest reaction rate, a

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Answer:

A) $E_{a} = 350KJ/mol$ , $E_{a} = 50KJ/mol$ , $E_{a} = 50KJ/mol$

A = 1.5× $10^{-7}s^{-1}$ , A = 1.9× $10^{-7} s^{-1}$ , A=1.5× $10^{-7} s^{-1}$

B) 4.469

Explanation:

From Arrhenius equation

$K=Ae^{\frac{E_{a} }{RT} }$

where; K = Rate of constant

A = Pre exponetial factor

$E_{a}$ = Activation Energy

R = Universal constant

T = Temperature in Kelvin

Given parameters:

$E_{a} =165KJ/mol$

$T_{1}=505K$

$T_{2}=525K$

$R=8.314JK^{-1}mol^{-1}$

taking logarithm on both sides of the equation we have;

$InK=InA-\frac{E_{a} }{RT}$

since we have the rate of two different temperature the equation can be derived as:

$In(\frac{K_{2} }{K_{1} } )=\frac{E_{a} }{R}(\frac{1}{T_{1} } -\frac{1}{T_{2} } )$

$In(\frac{K_{2} }{K_{1} } )=\frac{165000J/mol}{8.314JK^{-1}mol^{-1} }.(\frac{1}{505} -\frac{1}{525} )$

$In(\frac{K_{2} }{K_{1} } )$ = 19846.04×7.544× $10^{-5}$ = 1.497

$\frac{K_{2} }{K_{1} }$ = $e^{1.497}$ = 4.469

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3 years ago

One way to determine the degree of saturation of a solid-liquid solution is to drop a crystal of the solute into the solution.

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The answer is D: Saturated.

A saturated solution is one in which the exact maximum amount of solute has been dissolved. So, new solute will not dissolve in the solution. In contrast, an unsaturated solution can hold more solute, so if that option were correct, the crystal would have dissolved.

The other two terms are a bit more complicated. A supersaturated solution is one holding an amount of solute above the sustainable limit. Because of that, when more solute is added, the solution will immediately adjust, and some solute will come out of solution in a precipitate. Because the crystal isn't growing, we can eliminate this option.

A concentrated solution is one holding a relatively large amount of solute. However, you can have concentrated solutions that are saturated and unconcentrated (the word for this is dilute) solutions that aren't saturated. Therefore, we can say that because the crystal doesn't dissolve, this solution is saturated, but we can't say with certainty that it is concentrated.

Because the first three options are invalid, as described above, while the scenario does describe a saturated solution, D is the correct answer.

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3 years ago

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