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Amiraneli [1.4K]
3 years ago
6

How much energy would need to be input to completely boil 17.3 moles of ice starting at 0 degrees * C and going to 100°C?

Chemistry
1 answer:
kolezko [41]3 years ago
4 0

Answer:

Q = 934.911\,kJ

Explanation:

The energy needed to boil 17.3 moles of ice is:

Q = (17.3\,mol)\cdot \left[\left(40.5\,\frac{kJ}{mol} \right)+ \left(\frac{4.186}{1000}\,\frac{kJ}{g\cdot ^{\circ}C}  \right)\cdot \left(\frac{18.015\,g }{1\,mol} \right)\cdot (100^{\circ}C-0^{\circ}C)+\left(6\,\frac{kJ}{mol}\right)\right]Q = 934.911\,kJ

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Po_{210} ^{84}  ⟶  Pb_{206} ^{82}+  + He_{4} ^{2}

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The radioactive decay law is  

N=Noe^(-λt)

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No = number of atoms in t=0

N = number of atoms in t=t (now) in this case t=338.8 days  

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The radioactive constant is related to the half-life by the next equation  

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λ= \frac{ln2}{138.4 days}  =0,005008 days^(-1)

No (Atoms of  Po_{210} ^{84}  in t=0)

To get No we need to calculate the number of atoms of  Po_{210} ^{84}   in the initial sample. We have a sample of 0,561 g of PoCl4. If we get the number of moles of PoCl4 in the sample, this will be the number of moles of  Po_{210} ^{84}  in the initial sample.  

This is:

\frac{0,561 g of PoCl4}{350. 79 g of PoCl4 /mol} = 0,001599 mol of  PoCl4

This is the number of mol of  Po_{210} ^{84} in the initial sample.

To get the number of atoms in the initial sample we use the Avogadro’s number = 6.023*10^23  

0,001599 mol of  Po_{210} ^{84} * 6.023*10^23 atoms/ mol of  Po_{210} ^{84} = 9.632 *10^20 atoms of  Po_{210} ^{84}

Atoms after 338.8 days

We use the radioactive decay law to get this value  

N=Noe^(-λt)

N=9.632*10^20 e^(-0,005008 days^(-1) * 338.8 days) =1.765*10^20

This is the number of atoms of  Po_{210} ^{84} in the sample after 338.8 days has passed  

The number of atoms  Po_{210} ^{84} transformed is equal to the number of atoms of Pb_{206} ^{82}  produced.  

The number of atoms of Po_{210} ^{84} transformed is No - N  

9.632 *10^20 – 1.765 *10^20 = 7.866*10^20

So, 7.866*10^20 is the number of atoms of Pb_{206} ^{82} produced  

We can get the mass with the Avogadro’s number

(7.866*10^20 atoms of Pb_{206} ^{82} ) / ( 6.023*10^23 atoms of Pb_{206} ^{82} / mol of Pb_{206} ^{82} =  0.001306 moles of Pb_{206} ^{82}

This number of moles have a mass of:

(0,001306 moles of Pb_{206} ^{82} )* (205.97 g of Pb_{206} ^{82} /mol of Pb_{206} ^{82} ) = 0.269 g  

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