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TEA [102]
3 years ago
8

Phosphorus-32 is a commonly used radioactive nuclide in biochemical research, particularly in studies of nucleic acids. the half

-life of phosphorus-32 is 14.3 days. what mass of phosphorus-32 is left from an original sample of 157 mg of na332po4 after 42.0 days? assume that the atomic mass of 32p is 32.0.
Chemistry
1 answer:
lora16 [44]3 years ago
3 0

given days = 42 days 
 So, number of half life, n = 42/14.3 = 2.93 
 So, after 42 days, the mass of sodium phosphate sample left = original mass
x (1/2)^n, 
 = 157 mg x (1/2)^2.93 = 20.60 g of Na3PO4 after 42 days  
 Formula mass of Sodium phosphate = 3(23) + 32 + 4(16) amu = 165 amu  
 165 g of sodium phosphate contains 32 g of Phosphorus 
 20.60 g ------------------------------? = 20.60 x 32/ 165 =3.99g of P32
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Anna007 [38]

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Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

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According to stoichiometry :

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Thus 0.571 moles of NH_4NO_3 will require=\frac{1}{1}\times 0.571=0.571moles  of N_2O  

Mass of N_2O=moles\times {\text {Molar mass}}=0.571moles\times 44.01g/mol=25.13g

1 mole of NH_4NO_3 produce = 2 moles of H_2O

Thus 0.571 moles of NH_4NO_3 will require=\frac{2}{1}\times 0.571=1.142moles  of H_2O  

Mass of H_2O=moles\times {\text {Molar mass}}=1.142moles\times 18g/mol=20.56g

Thus 25.13 g of N_2O  and 20.56 g of H_2O will be produced from 45.70 g of NH_4NO_3

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