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xxTIMURxx [149]
3 years ago
15

A chamber with a fixed volume is shown above. The temperature of the gas inside the chamber before heating is 25.2 C and it’s pr

essure is 0.600 atm. The gas is heated with a flame to a temperature of 72.4 C, what is it’s pressure at this temperature
Chemistry
1 answer:
rusak2 [61]3 years ago
3 0

Answer:

Explanation:

Given parameters:

Initial temperature T₁  = 25.2°C  = 25.2 + 273  = 298.2K

Initial pressure  = P₁  = 0.6atm

Final temperature = 72.4°C   = 72.4 + 273  = 345.4K

Unknown:

Final pressure = ?

Solution:

To solve this problem, we use an adaption of the combined gas law where the volume gas is fixed. This simplification results into:

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

where P and T are temperatures, 1 and 2 are initial and final temperatures.

 Input the parameters and solve;

          \frac{0.6}{298.2}   = \frac{P_{2} }{345.4}  

          P₂   = 0.7atm

         

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Answer:

The four quantum number for each electron will be:

1s^{2}

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2s^{2}

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2p^{6}

n=1;l=1;\\m=+1,0.-1 \\s=+\frac{1}{2}/-\frac{1}{2}

3s^{2}

n=3;l=0;m=0;s=+\frac{1}{2}/-\frac{1}{2}

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As the element is neutral, the number of protons will be equal to number of electrons which will be the atomic number of the element.

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Atomic number = 12

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The principal shell is represented by "n"

i) For "s" subshell the value of l =0 (azimuthal quantum number) thus m (magnetic quantum number)= 0

The two electrons in s subshell will have either plus half or minus half spin quantum number

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