Answer:
Radius of the interior sphere = 3.847 nm
Explanation:
The volume of the shell (Vs) is equal to the difference of the volume of the outer sphere (Vo) and the volume of the inner sphere (Vi). Then:
If we express the radius of the outer sphere (ro) in function of the radius of the inner sphere (ri), we have (e being the shell thickness):
The first equation becomes
To find ri that satisfies this equation we have to find the roots of the polynomial.
Numerically, it could be calculated that ri=3.847 nm satisfies the equation.
So if the radius of the interior sphere is 3.847 nm, the volume of the interior sphere is equal to the volume of the shell of 1nm.
No, it won't change the amount of reactants nor the products as a catalyst will only provide an alternative path where lower activation energy is needed for the process to take place.
hope this explains it
If it does, please give it a brainliest :)))
Answer:
0.471 mol/L
Explanation:
First, we'll begin by by calculating the number of mole of KMnO4 in 26g of KMnO4.
This is illustrated below:
Molar Mass of KMnO4 = 39 + 55 + (16x4) = 39 + 55 + 64 = 158g/mol
Mass of KMnO4 from the question = 26g
Mole of KMnO4 =?
Number of mole = Mass/Molar Mass
Mole of KMnO4 = 26/158 = 0.165mole
Now we can obtain the concentration of KMnO4 in mol/L as follow:
Volume of the solution = 350mL = 350/1000 = 0.35L
Mole of KMnO4 = 0.165mole
Conc. In mol/L = mole of solute(KMnO4)/volume of solution
Conc. In mol/L = 0.165mol/0.35
conc. in mol/L = 0.471mol/L
