The answer is A. The specific heat of water explains capillary action I just took the exam and got it correct
Answer:
(a) Ionic
(b) Nonpolar covalent
(c) Polar covalent
(d) Polar covalent
(e) Nonpolar covalent
(f) Polar covalent
<em>For those substances with polar covalent bonds, which has the least polar bond?</em> NO₂
<em>For those substances with polar covalent bonds, which has the most polar bond?</em> BF₃
Explanation:
<em>Are the bonds in each of the following substances ionic, nonpolar covalent, or polar covalent?</em>
The nature of a bond depends on the modulus of the difference of electronegativity (|ΔEN|) between the atoms that form it.
- If |ΔEN| = 0, the bond is nonpolar covalent.
- If 0 < |ΔEN| ≤ 2, the bond is polar covalent.
- If |ΔEN| > 2, the bond is ionic.
<em>(a) KCl</em> |ΔEN| = |EN(K) - EN(Cl)| = |0.8 - 3.0| = 2.2. The bond is ionic.
<em>(b) P₄</em> |ΔEN| = |EN(P) - EN(P)| = |2.1 - 2.1| = 0.0. The bond is nonpolar covalent.
<em>(c) BF₃</em> |ΔEN| = |EN(B) - EN(F)| = |2.0 - 4.0| = 2.0. The bond is polar covalent.
<em>(d) SO₂</em> |ΔEN| = |EN(S) - EN(O)| = |2.5 - 3.5| = 1.0. The bond is polar covalent.
<em>(e) Br₂</em> |ΔEN| = |EN(Br) - EN(Br)| = |2.8 - 2.8| = 0.0. The bond is nonpolar covalent.
<em>(f) NO₂</em> |ΔEN| = |EN(N) - EN(O)| = |3.0 - 3.5| = 0.5. The bond is polar covalent.
Answer: Option C - Radical
Explanation:
A radical is a chemical specie carrying a lone electron. In the halogenation of alkanes: take Methane CH4 as the alkane, and Chlorine Cl as the halogen.
The step by step halogenation process is as follows:
CH4 + Cl2 --> CH3• + HCl + Cl•
CH3• + Cl2 --> CH3Cl + HCl
CH3Cl + Cl2 --> CH2Cl2 + HCl + Cl•
CH2Cl2 + Cl2 --> CHCl3 + HCl
CHCl3 + Cl2 --> CCl4 + HCl + Cl•
Chlorine molecule attack methane knocking off an hydrogen atom from it and forming a methyl radical (CH3•), that is subsequently attack by another chlorine molecule. This cycle repeats itself, until no hydrogen atom is available for substitution by the highly reactive chlorine radical.
Note: no cation or anion is formed in the halogenation process
Ba(OH)2 can be neutralized by adding an acid to the solution. We can add HCl to produce:
Ba(OH)2 + 2HCl ===> BaCl2 + 2H2O ; producing a salt BaCl2 and water.
or we can add H2SO4 to produce:
Ba(OH)2 + H2SO4 ===> BaSO4 + 2H2O; producing a salt, BaSO4, and water. <span />
