The radioactive decay obeys first order kinetics
the rate law expression for radioactive decay is
![ln\frac{[A_{0}]}{[A_{t}]}=kt](https://tex.z-dn.net/?f=ln%5Cfrac%7B%5BA_%7B0%7D%5D%7D%7B%5BA_%7Bt%7D%5D%7D%3Dkt)
Where
A0 = initial concentration
At = concentration after time "t"
t = time
k = rate constant
For first order reaction the relation between rate constant and half life is:
Let us calculate k
k = 0.693 / 72 = 0.009625 years⁻¹
Given
At = 0.25 A0
time = 144 years
So after 144 years the sample contains 25% parent isotope and 75% daughter isotopes**
Simply two half lives
D) refracted
because , the light changes direction.. causing the penny to look bigger than it is.
Answer:
sun-sun char HAHAHAHAH EVERYDAY
Explanation:
mass H2O2 = 55 mL(1.407 g/mL) = 80.85 g
molar mass H2O2 = 2(1.01 g/mol) + 2(16.00 g/mol) = 34.02 g/mol
moles H2O2 = 80.85 g/34.02 g/mol = 2.377 moles H2O2
For each mole of H2O2 you obtain 0.5 mole of O2 (see the equation).
moles O2 = 2.377 moles H2O2 (1 mole O2)/(2 moles H2O2) = 1.188 moles O2
Now, you need the temperature. If you are at STP (273 K, and 1.00 atm) then 1 mole of an ideal gas at STP has a volume of 22.4 L. Without temperature you are not really able to continue. I will assume you are at STP.
Volume O2 = 1.188 moles O2(22.4 L/mole) = 0.0530 L of O2.
which is 53 mL.
The balanced equation for the above reaction is as follows;
2S + 3O₂ --> 2SO₃
Stoichiometry of O₂ to SO₃ is 3:2
O₂ is the limiting reactant and S is provided in excess. since O₂ is the limiting reactant, the whole amount is consumed in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of O₂ moles reacted- 4 g / 32 g/mol = 0.125 mol
3 mol of O₂ forms 2 mol of SO₃
therefore when 0.125 mol of O₂ reacts number of SO₃ moles - 2/3 x 0.125 mol
Number of SO₃ moles formed - 0.0833 mol
Answer is 4) 0.08 mol
