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makkiz [27]
3 years ago
5

A gas occupies a volume of 60 L at a temperature of 0.5 K. What will the volume be at 4 K?

Chemistry
1 answer:
Flauer [41]3 years ago
3 0

Answer:

480 L

Explanation:

In order to solve this question, you should be familiar with gas laws. (I will attach a picture showing all of them under my answer.) In this question in particular, however, we only need Charles's Law because we're dealing with temperature and volume.

As we can see, Charles's Law is:

\frac{V_{1} }{T_{1} } = \frac{V_{2} }{T_{2} }

or, initial volume over initial temperature equals final volume over final temperature.

In this question, 60 L is our <u>initial volume,</u> and 0.5 K is our <u>initial temperature</u> (K being Kelvin). We are only given 4 K as our <u>final temperature</u>. We are asked to solve for the <u>final volume</u>. Let's set up the equation and solve for V_{2}:

--------------------------------------------------------------------------------------------------------------

(60) / (0.5) = V_{2} / (4)

↓

120 = V_{2} / 4

×4           ×4

↓

V_{2} = 480 L

--------------------------------------------------------------------------------------------------------------

There's our answer! Feel free to comment if you have any questions about my answer :)

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Please review the attachment
astra-53 [7]

Answer: The correct answer is -297 kJ.

Explanation:

To solve this problem, we want to modify each of the equations given to get the equation at the bottom of the photo. To do this, we realize that we need SO2 on the right side of the equation (as a product). This lets us know that we must reverse the first equation. This gives us:

2SO3 —> O2 + 2SO2 (196 kJ)

Remember that we take the opposite of the enthalpy change (reverse the sign) when we reverse the equation.

Now, both equations have double the coefficients that we would like (for example, there is 2S in the second equation when we need only S). This means we should multiply each equation (and their enthalpy changes) by 1/2. This gives us:

SO3 —>1/2O2 + SO2 (98 kJ)

S + 3/2O2 —> SO3 (-395 kJ)

Now, we add the two equations together. Notice that the SO3 in the reactants in the first equation and the SO3 in the products of the second equation cancel. Also note that O2 is present on both sides of the equation, so we must subtract 3/2 - 1/2, giving us a net 1O2 on the left side of the equation.

S + O2 —> SO2

Now, we must add the enthalpies together to get our final answer.

-395 kJ + 98 kJ = -297 kJ

Hope this helps!

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Explanation:

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Calculate the mass of water vapour present in a room of volume 400 m3 that contains air at 27 °C on a day when the relative humi
Alex73 [517]

Answer:

Mass of water = 6251. 706g or 6.25Kg

Explanation:

Relative humidity = (actual vapor pressure/saturation vapor pressure) * 100%

Actual vapor pressure, Pw = relative humidity * saturation vapor pressure

Pw = 60% * (35.6 *0.001)atm = 0.0216atm

Note: 1mbar = 0.001atm

Using the ideal gas equation: PV=nRT; where P = Pw= 0.02136atm, V= (400

* 1000)dm^3, R= 0.082 atmdm^3/kmol, T= (27+273)K, n = number of moles

Note: 1m^3 = 1000dm^3,R is the molar gas constant.

Making n subject of the formula, n = PV/RT

n= (0.02136 * 400000)/(0.082 * 300) = 347.317 moles

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