Radio active decay reactions follow first order rate kinetics.
a) The half life and decay constant for radio active decay reactions are related by the equation:
Where k is the decay constant
b) Finding out the decay constant for the decay of C-14 isotope:
c) Finding the age of the sample :
35 % of the radiocarbon is present currently.
The first order rate equation is,
![[A] = [A_{0}]e^{-kt}](https://tex.z-dn.net/?f=%20%5BA%5D%20%3D%20%5BA_%7B0%7D%5De%5E%7B-kt%7D%20%20%20)
![\frac{[A]}{[A_{0}]} = e^{-kt}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BA%5D%7D%7B%5BA_%7B0%7D%5D%7D%20%3D%20e%5E%7B-kt%7D%20%20)
t = 7923 years
Therefore, age of the sample is 7923 years.
Let's go over the given information. We have the volume, temperature and pressure. From the ideal gas equation, that's 4 out of 5 knowns. So, we actually don't need Pvap of water anymore. Assuming ideal gas, the solution is as follows:
PV=nRT
Solving for n,
n = PV/RT = (753 torr)(1 atm/760 torr)(195 mL)(1 L/1000 mL)/(0.0821 L·atm/mol·K)(25+273 K)
n = 7.897×10⁻³ mol H₂
The molar mass of H₂ is 2 g/mol.
Mass of H₂ = 7.897×10⁻³ mol * 2 g/mol = <em>0.016 g H₂</em>
Answer:0.45L
Explanation:
molarity=0.15M
Mass=5g
No of moles=mass ➗ molecular mass
Molecular mass of KCL=39.0983x1+35.453x1
Molecular mass of KCL=74.5513
No of moles=5 ➗ 74.5513
No of moles=0.067
Volume in liters=No of moles ➗ molarity
Volume in liters=0.067 ➗ 0.15
Volume in liters=0.45L
The visual criterion which has to be assessed is to ensure it's not oily or sticky.
<h3>What is Distillation?</h3>
This is a separation technique that involves the use selective boiling and condensation to separate a liquid mixture into different components.
In the case of Crude reaction product , it is best to ensure that it is not oily or sticky before semimicro distillation is done.
Read more about Distillation here brainly.com/question/552187
Explanation:
Scandium has atomic number of 21. This means that in it's neutral state its going to have 21 electrons.
a) The full electronic configuration is given as;
1s2 2s2 2p6 3s2 3p6 3d1 4s2
The final electron is placed in the d orbital. The shell is 3d
(b) When scandium has a charge if +1, it has lost an electron. The total number of electrons would now be 21-1 = 20
The electronic configuration would be given as;
1s2 2s2 2p6 3s2 3p6 4s2
The electron in the 3d orbital would be removed.
