Explanation:
Hybridization of O in 
So, water molecule has four hybrid orbitals.
Two hybrid orbitals form 2 sigma bond with two H atoms.
Remaining two hybrid orbitals are occupied by two lone pairs.
Because of lone pair-lone pair repulsion, shape of 
becomes bent.
Water molecule is polar because of difference in eletronegativities of O and H.
O is more electronegative as comapared to hydrogen. So bonding electrons get attracted towards O atom which results in the development of partial negative charge on O atom and partial positive charge on H atoms.
Because of development of partial negative and partial positive charge, water molecule becomes polar.
The balanced chemical reaction would be
<span>fecl2 + 2naoh = fe(oh)2(s) + 2nacl
Initial amounts of the reactants are given, so, we need to determine which of the reactants is the limiting reactant and use this amount to determine what is asked. However, what is being asked is how many of the FeCl2 is used in the reaction, showing that it is NaOH that is the limiting reactants. Thus, we just use the initial amount of NaOH and relate the substances by the chemical reaction as follows:
6 mol NaOH ( 1 mol FeCl2 / 2 mol NaOH ) = 3 mol FeCl2
Therefore, 3 moles of FeCl2 is used up and 3 moles of FeCl2 is also left after the reaction.</span>
Answer:
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Explanation:
The answer is A. Water
Bronsted-Lowry base compounds are those that can accept protons
Bronsted-Lowry Acid Compounds are those that can recieve one
Water / H2O is an Amphoteric compund which mean that its molecul can act as a Base and Acid compound, so the answer is A.
