Answer:
The given statement - The main criterion for sigma bond formation is that the two bonded atoms have valence orbitals with lobes that point directly at each other along the line between the two nuclei , is <u>True.</u>
Explanation:
The above statement is correct , because the sigma bond is produced by the head on overlapping, the orbitals should all point in the same direction.
<u>SIGMA BONDS -</u> Sigma bonds (bonds) are the strongest type of covalent chemical bond in chemistry. They're made up of atomic orbitals that collide head-on. For diatomic molecules, sigma bonding is best characterized using the language and tools of symmetry groups.
Head-on overlapping of atomic orbitals produces sigma bonds. The concept of sigma bonding is expanded to include bonding interactions where a single lobe of one orbital overlaps with a single lobe of another. Propane, for example, is made up of ten sigma bonds, one for each of the two CC bonds and one for each of the eight CH bonds.
Hence , the answer is true .
Answer:175⋅mL of the given sulfuric acid
Explanation:
Answer:
The phrase "opposites attract" is true for atoms. The positively charged ion (cation) and the negatively charged ion (anion) are attracted to each other. It is this attraction, created by the transfer of electrons, that forms the ionic bond. The transfer of one electron creates a single bond.
Explanation:The phrase "opposites attract" is true for atoms. The positively charged ion (cation) and the negatively charged ion (anion) are attracted to each other. It is this attraction, created by the transfer of electrons, that forms the ionic bond. The transfer of one electron creates a single bond.
Answer:
a) Kb = 10^-9
b) pH = 3.02
Explanation:
a) pH 5.0 titration with a 100 mL sample containing 500 mL of 0.10 M HCl, or 0.05 moles of HCl. Therefore we have the following:
[NaA] and [A-] = 0.05/0.6 = 0.083 M
Kb = Kw/Ka = 10^-14/[H+] = 10^-14/10^-5 = 10^-9
b) For the stoichiometric point in the titration, 0.100 moles of NaA have to be found in a 1.1L solution, and this is equal to:
[A-] = [H+] = (0.1 L)*(1 M)/1.1 L = 0.091 M
pKb = 10^-9
Ka = 10^-5
HA = H+ + A-
Ka = 10^-5 = ([H+]*[A-])/[HA] = [H+]^2/(0.091 - [H+])
[H+]^2 + 10^5 * [H+] - 10^-5 * 0.091 = 0
Clearing [H+]:
[H+] = 0.00095 M
pH = -log([H+]) = -log(0.00095) = 3.02
