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hammer [34]
3 years ago
13

A sample of fluorine gas occupies 810 milliliters at 270 K and 1.00 atm. What volume does the gas occupy when the pressure is do

ubled, and the temperature increases to 400 K
Chemistry
2 answers:
sertanlavr [38]3 years ago
8 0

Answer:

The answer to your question is Volume = 600 ml

Explanation:

Data

Volume 1 = 810 ml

Temperature 1 = 270°K

Pressure 1 = 1 atm

Volume 2 = ?

Pressure = 2 atm

Temperature 2 = 400°K

Process

1.- To solve this problem, use the Combine gas law.

               P1V1 / T1 = P2V2 / T2

- Solve for V2

                V2 = P1V1T2 / T1P2

2.- Substitution

                 V2 = (1)(810)(400) / (270)(2)

3.- Simplification

                 V2 = 324000 / 540

4.- Result

                 V2 = 600 ml

CaHeK987 [17]3 years ago
4 0

Answer:

600mL

Explanation:

The following were obtained from the question:

V1 = 810mL

P1 = 1atm

T1 = 270K

V2 =?

P2 = 2atm(since the pressure is doubled)

T2 = 400K

The new volume can be obtained by doing the following :

P1V1/T1 = P2V2/T2

1 x 810/270 = 2 x V2 /400

Cross multiply to express in linear form

270 x 2 x V2 = 810 x 400

540 x V2 = 324000

Divide both side by 540

V2 = 324000/540

V2 = 600mL

The new volume of the gas is 600mL

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How would a collapsing universe affect light emitted from clusters and superclusters? A. Light would acquire a blueshift. B. Lig
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Choice A: Light would acquire a blueshift.

Explanation:

When a universe collapses, clusters of stars start to move towards each other. There are two ways to explain why light from these stars will acquire a blueshift.

Stars move toward each other; Frequency increases due to Doppler's Effect.

The time period t of a beam of light is the same as the time between two consecutive peaks. If \lambda is the wavelength of the beam, and both the source and observer are static, the time period T will be the same as the time it takes for light travel the distance of one \lambda (at the speed of light in vacuum, c).

\displaystyle t = \frac{\lambda}{c}.

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\displaystyle f = \frac{1}{t} = \frac{c}{\lambda}.

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  • Distance from the star when the first peak is sent: d.
  • Time taken for the first peak to arrive: \displaystyle t_1 =\frac{d}{c}.

The star will emit its second peak after a time of. Meanwhile, the distance between the star and the observer keeps decreasing. Let v be the speed at which the star approaches the observer. The star will travel a distance of v\cdot t before sending the second peak.

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The period of the light is t when emitted from the star. However, the period will appear to be shorter than t for the observer. The time period will appear to be:

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