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zavuch27 [327]
3 years ago
6

A balloon containing a sample of helium gas measures 1.50 L at a temperature of 25.0°C is placed in a refrigerator at 36.0°F. Ca

lculate its new volume. Enter your answer in the box provided.
Chemistry
1 answer:
Fiesta28 [93]3 years ago
6 0

Answer:

1,39 L

Explanation:

Charle's Law states that the volume of a fixed amount of gas <em>maintained at constant pressure</em> is directly proportional to the<u> absolute temperature</u> of the gas, for a constant amount of gas we can write:

\frac{V1}{T1}=\frac{V2}{T2}

As the pressure of the balloon doesn't change, we can use Charle's Law to solve the problem. Firs we change the given temperatures to absolute temperature units ( °K), using the following relations:

°K=273+°C

°K=5/9(°F-32)+273

Therefore:

V1=1.5 L, T1=273+25=298°K

V2=?,      T2=5/9(36-32)+273=275,2°K

V2=\frac{V1T2}{T1}=\frac{275,2*1,5}{298}=1,39L

The new volume of the balloon is 1,39 L.

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natural disasters

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these are some examples, hope this helps :)

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Predict whether each substance will dissolve in water.
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4 0
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When a 10 mL graduated cylinder is filled to the 10 mL mark, the mass of the water was measured to be 10.085 g at 20ºC. If the d
Arte-miy333 [17]
The percentage error is the error of the measured value to the true value. To find he percent error, the equation is as follows:

Percent error = |Measured Value - True Value|/True Value  * 100
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</span><em>Percent error = 1.1% </em>
8 0
4 years ago
A 0.4322 g sample of a potassium hydroxide – lithium hydroxide mixture requires 27.10 mL of 0.3565 M HCl for its titration to th
Yuki888 [10]

The mass percent lithium hydroxide in the mixture with potassium hydroxide, calculated from the equivalence point in the titration of HCl with the mixture, is 19.0%.  

The mass percent of lithium hydroxide can be calculated with the following equation:  

\% = \frac{m_{LiOH}}{m_{t}} \times 100    (1)

Where:

m_{t} = m_{KOH} + m_{LiOH} = 0.4322 g   (2)  

We need to find the mass of LiOH.

From the titration, we can find the number of moles of the mixture since the number of moles of the acid is equal to the number of moles of the bases at the equivalence point.    

\eta_{HCl} = \eta_{LiOH} + \eta_{KOH}

0.0271 L*0.3565 \frac{mol}{L} = \eta_{LiOH} + \eta_{KOH}

\eta_{LiOH} + \eta_{KOH} = 9.66 \cdot 10^{-3} \:mol

Since mol = m/M, where M: is the molar mass and m is the mass, we have:

\frac{m_{LiOH}}{M_{LiOH}} + \frac{m_{KOH}}{M_{KOH}} = 9.66 \cdot 10^{-3} \:mol    (3)                                        

Solving equation (2) for m_{KOH} and entering into equation (3), we can find the mass of LiOH:  

\frac{m_{LiOH}}{M_{LiOH}} + \frac{0.4322 - m_{LiOH}}{M_{KOH}} = 9.66 \cdot 10^{-3} \:mol    

\frac{m_{LiOH}}{23.95 g/mol} + \frac{0.4322 g - m_{LiOH}}{56.1056 g/mol} = 9.66 \cdot 10^{-3} \:mol              

Solving for m_{LiOH}, we have:

m_{LiOH} = 0.082 g

Hence, the percent lithium hydroxide is (eq 1):

\% = \frac{0.082 g}{0.4322 g} \times 100 = 19.0 \%  

Therefore, the mass percent lithium hydroxide in the mixture is 19.0%.

Learn more about mass percent here:

  • brainly.com/question/6992535?referrer=searchResults
  • brainly.com/question/5840377?referrer=searchResults

I hope it helps you!                        

5 0
2 years ago
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