Answer:
The P-H bonds are more polar than the N-H bonds.
Explanation:
Phosphine is a polar molecule with non-polar bonds. The phosphorus atom is bonded to three hydrogen atoms and the phosphorus atom has a lone pair of electrons. Since hydrogen and phosphorus are equal in electronegativity, it implies that they attract the shared pairs of electrons the same amount,hence bonding electrons are shared equally making the covalent bonds non-polar.
The lone pair of electrons on phosphorus causes the molecule to be asymmetrical with respect to charge distribution this is why the molecule is polar even though the are non-polar bonds in the molecule.
Looking at the values of electro negativity stated in the question, one can easily see that the difference in electro negativity between nitrogen and hydrogen is 0.9 while the difference in electro negativity between phosphorus and hydrogen is zero. It is clear that NH3 is naturally more polar than PH3 since each individual N-H bond in NH3 is a polar bond while the individual P-H bonds in PH3 are nonpolar.
Answer: Two effects may occur during high current flow: 1) the wire may become overheated to the point that surface oxidation or even evaporation may take place, 2) at the connection points at each end of the wire, especially if the terminations are of a different type of metal than the wire, some atoms may migrate into or out of the wire.
Answer:
The difference between the energy of the reactants and the energy of the products is called the enthalpy change (∆H) of the reaction. For an exothermic reaction, the enthalpy change is always negative. In an endothermic reaction, the products are at a higher energy than the reactants.
Explanation:
I’m really sorry I need points I hope you find an answer
Answer:
(a) Iron is being oxidized.
(d) Sulfur is being reduced.
Explanation:
Let's consider the following redox reaction.
8 Fe(s) + S₈(s) → 8 FeS(s)
Iron is being oxidized according to the following oxidation half-reaction:
Fe(s) → Fe²⁺(s) + 2 e⁻
Sulfur is being reduced according to the following reduction half-reaction:
S₈(s) + 16 e⁻ → 8 S²⁻(s)
