Explanation:
I am not understanding your question
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.
(ANS1)— P4 + 5O2 ---> 2P2O5
(ANS2)— C3H8 + 5O2---> 3CO2 + 4H20
(ANS3)— Ca2Si + 4Cl2 ---> 2CaCl2 + SiCl4
Answer:
Kindly check the explanation section.
Explanation:
Based on the description of the reacting -OH group containing Compound, the drawing of the chemical compound is given in the attached picture.
So, without mincing words let's dive straight into the solution to the question.
The reaction between the OH containing compound and PCC is an oxidation reaction.
Looking at the carbon number 1 which the first OH group and CH3 are attached to. Oxidation can not occur here as tertiary alcohol can not be oxidize.
Hence, the second OH will be oxidized into a carbonyl group, C = O. Kindly note that when alcohol oxidizes it turns into an aldehyde.
The equation for the reaction is also given the the attached picture.
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