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nadya68 [22]
3 years ago
10

Question 2 The metal molybdenum becomes superconducting at temperatures below 0.90K. Calculate the temperature at which molybden

um becomes superconducting in degrees Celsius. Round your answer to 2 decimal places.
Chemistry
1 answer:
LenKa [72]3 years ago
5 0

Answer:

Temperature at which molybdenum becomes superconducting is-272.25°C

Explanation:

Conductor are those hard substances which allows path of electric current through them. And super conductors are those hard substances which have resistance against the flow of electric current through them.

As given, molybdenum becomes superconducting at temperatures below 0.90 K.

Temperature in Kelvins can be converted in °C by relation:

T(°C)=273.15-T(K)

Molybdenum becomes superconducting in degrees Celsius.

T(°C)=273.15-0.90= -272.25 °C

Temperature at which molybdenum becomes superconducting is -272.25 °C

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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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