Answer is: C) the fact that the number of lone pairs of electrons on the central atom is greater in the case of water.
Carbon(IV) oxide is nonpolar because CO₂ is linear molecule and the oxygen atoms are symmetrical (bond angles 180°).
Water is polar because of the bent shape of the molecule.
Oxygen atom in water molecule has sp3 hybridization. The bond angle between the two hydrogen atoms is approximately 104.45°.
Oxygen atom has atomic number 8, it means it has eight protons and eight electrons, so atom has neutral charge. Oxygen is a nonmetal.
Electron configuration of oxygen atom: ₈O 1s² 2s² 2p⁴.
Oxygen atom has six valence electrons
, two lone pairs and two electrons that form two sigma bonds with hydrogen atoms.
Carbon is a chemical element with symbol C and atomic number 6, which means it has 6 protons and six electrons. Four valence electrons are in 2s and 2p orbitals.
Electron configuration of carbon atom: ₆C 1s² 2s² 2p².
In carbon dioxide, carban has sp hybridization with no lone pairs.
Answer:
b) +2 and +3.
Explanation:
Hello,
In this case, given the molecular formulas:

And:

We can relate the subscripts with the oxidation states by knowing that they are crossed when the compound is formed, for that reason, we notice that oxygen oxidation state should be -2 for both cases and the oxidation state of X in the first formula must be +2 since both X and O has one as their subscript as they were simplified:

Moreover, for the second case the oxidation state of X should be +3 in order to obtain 3 as the subscript of oxygen:

Thus, answer is b)+2 and +3
Best regards.
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.