How many particles would be found in a 1.224 g sample of K2O
2 answers:
Answer:
7.8286×10²¹ particles.
Explanation:
First we need to calculate the total molar mass of the compound, in this case:
Potassium (K) = 39.1 g/mol × 2 = 78.2 +
Oxigen (O) = 16 g/mol × 1 = <u> 16 </u>
Total(K₂O) = 94.2 g/mol
Then, we calculate the number of moles of the compound in the sample, this is done dividing de mass of the sample by the molar mass:
mol = 0.013 moles in our sample.
Finally, we calculate the total number of particles. The costant known as Avogadro number (6.022×10²³) is the number of particles or atoms contained in a mole of any substance. We need to multiply the number of moles by the Avogadro number.
particles = 0.013 mol × (6.022×10²³ particles/mol) = 7.8286×10²¹ particles.
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Answer : The correct answer is 96.68 yrs
Radioactivity Decay :
it is a process in which a nucleus of unstable atom emit energy in form of radiations like alpha particle , beta particle etc .
Radioactive decay follows first order kinetics , so its rate , rate constant , amount o isotopes can be calculated using first order equations .
The first order equation for radioactive decay can be expressed as :
----------- equation (1)
Where : N = amount of radioisotope after time "t"
N₀ = Initial amount of radioisotope
k = decay constant and t = time
Following steps can be used to find time :
1) To find deacy constant :
Decay constant can be calculated using half life . Decay constant and half life can be related as :
---------equation (2)
Given : Half life of Strontium -90 = 28.8 years
Plugging value of in above formula (equation 2) :
Multiply both side by k
Dividing both side by 28.8 yrs
(ln 2 = 0.693 )
k = 0.0241 yrs⁻¹
Step 2 : To find time :
Given : N₀ = 10.3 ppm N = 1.0 ppm k = 0.0241 yrs⁻¹
Plugging these value in equation (1) as :
(ln 0.0971 = - 2.33 )
Dividing both side by - 0.0241 yrs⁻¹
t = 96.68 yrs
Hence the concentration of Strontium-90 will drop from 10.3 ppm to 1.0 ppm is 96.68 yrs