They would repel one another because they they have opposing charges
<span>1. The correct option is A, THE OXYGEN ATOM HAS A NEGATIVE CHARGE. A water molecule is made up of two atoms of hydrogen and one atom of oxygen. The molecules of water are arranged in such a way that, the negative hydrogen atom is attracted to the positive hydrogen atom and the overall structure is bent. Because oxygen is more electronegative than hydrogen, oxygen atom draws the shared electrons toward itself , this gives the oxygen end of the molecule a partial negative charge and the hydrogen end, a partial positive charge.
2. The correct option is: WITHOUT THE PROPER BALANCE OF WATER, CHEMICAL REACTIONS WILL NOT TAKE PLACE. Metabolic chemical reactions take place in living cells all the time because water, which is a necessary condition for the reaction is in place. Th fluids find in the cells are mostly water; water creates suitable conditions for biochemical reactions to take place. Without water, the cells will not be able to carry out any chemical reaction.
3. The correct option is A. Hydrogen bond allows oxygen and water molecules to be bonded together. Hydrogen bonds are weak interactions that formed between the hydrogen with a patial positive end and oxygen with a patial negative end. The hydrogen bond in water is responsible for the unique properties associated with water.</span><span />
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.
