212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.
Answer:
Option A.
Explanation:
Similar to Avagadro's law, there is another law termed as dilution law. As the product of volume and normality of the reactant is equal to the product of volume and normality of the product from the Avagadro's law. In dilution law, it will be as product of volume and concentration of the solute of the reactant is equal to the product of volume and concentration of solution.

So, as per the given question C1 = 5.45 M of lead nitrate and V1 has to be found. While C2 is 1.41 M of lead nitrate and V2 is 820.7 ml.
Then, 

So nearly 212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.
Answer:they will out live us himans
Explanation:
You can use the mass of neuron divided by the mass of conversion factor: 1.67*10^(-25)/(1.66054*10^(-24))≈1 amu. So the answer is 1 amu.
1s2 2s2 2p6 3s2 3p6 4s1
s orbital can hold 2 electron
p orbitals can hold 6 electron
All organic compounds have at least 1 carbon and 2 hydrogen atoms.