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.
It would be most similar to neon. it wouldn’t be sulfur because that’s in the same group as oxygen and has the same number of electrons. and carbon has less than that so the only one that makes sense is neon
Answer:
pH= 0.92
Explanation:
HNO3-> H^+ +NO3^-
HNO3 is a strong acid, so it fully dissociates
[HNO3] = 0.12M [H^+] = 0.12M
pH= -log[H^+]
pH=-log[.12] = 0.92
pH = 0.92
Answer:
Nitrate is a polyatomic ion with the molecular formula NO⁻₃ and a molecular mass of 62.0049 u. Organic compounds that contain the nitrate ester as a functional group are also called nitrates.