<u>Answer:</u> The age of the mineral sample is ![1.09\times 10^9yrs](https://tex.z-dn.net/?f=1.09%5Ctimes%2010%5E9yrs)
<u>Explanation:</u>
The equation used to calculate rate constant from given half life for first order kinetics:
![t_{1/2}=\frac{0.693}{k}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B0.693%7D%7Bk%7D)
where,
= half life of the reaction = ![1.27\times 10^9yrs](https://tex.z-dn.net/?f=1.27%5Ctimes%2010%5E9yrs)
Putting values in above equation, we get:
![k=\frac{0.693}{1.27\times 10^9yrs}=5.46\times 10^{-10}yrs^{-1}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B0.693%7D%7B1.27%5Ctimes%2010%5E9yrs%7D%3D5.46%5Ctimes%2010%5E%7B-10%7Dyrs%5E%7B-1%7D)
We are given:
Mass ratio of K-40 to Ar-40 = 0.812 : 1.00
Rate law expression for first order kinetics is given by the equation:
![k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = ![5.46\times 10^{-10}yr^{-1}](https://tex.z-dn.net/?f=5.46%5Ctimes%2010%5E%7B-10%7Dyr%5E%7B-1%7D)
t = time taken for decay process = ? yr
= initial amount of the sample = [1.00 + 0.812] = 1.812 grams
[A] = amount left after decay process = 1.00 grams
Putting values in above equation, we get:
![5.46\times 10^{-10}=\frac{2.303}{t}\log\frac{1.812}{1}\\\\t=1.09\times 10^9yrs](https://tex.z-dn.net/?f=5.46%5Ctimes%2010%5E%7B-10%7D%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B1.812%7D%7B1%7D%5C%5C%5C%5Ct%3D1.09%5Ctimes%2010%5E9yrs)
Hence, the age of the mineral sample is ![1.09\times 10^9yrs](https://tex.z-dn.net/?f=1.09%5Ctimes%2010%5E9yrs)
Answer:
- .
Explanation:
Ned help po rin ako
pero once in the best way I can get a great day 2
I am not sure about my answer, but I think that it's density allows it to fall to the ground/floor.
Answer:
46.9 atm
Explanation:
Applying Van der waals equation
P = [nRT/(V-nb)]- (an²/V²)............... Equation 1
Where P = pressure, T = temperature, n = number of moles, V = volume, R = Universal constant, a and b = gas constant
Given: n = 20 moles, R = 0.08205 L.atm/mol.K, T = 273 K, V = 10.0 L, a = 0.0341 atm.L²/mol², b = 0.0237L/mol
Substitute these values into equation 1
P = [(20×0.08205×273)/(10-20×0.0237)]-(0.0341×20²/10²)
P = (447.993/9.526)-0.1364
P = 47.028-0.1364
P = 46.9 atm