Carbon have 4 valence electron
The electron configuration of C is : 1s2 2s2 2ps
the outer most s and p orbital contains 4 electrons and these are the valence electron of C atom
Key: to determine the valence electron of an element, you have to refer to the group it is located in, in the periodic table. (C is in group 4 (or 14) so VE = 4)
hope its clear enough
5.7 gallons
if you divide 279 / 49 it = 5.69, which you would just round up to 5.7
Answer:
Kb = 1.6 × 10⁻⁵
Explanation:
Step 1: Given data
Acid dissociation constant of hydrocyanic acid (Ka): 6.2 × 10⁻¹⁰
Concentration of cyanide ion (Cb): 0.1 M
Step 2: Calculate the basic dissociation constant (Kb) of cyanide ion
We have the Ka of HCN. We can calculate the Kb of its conjugate base using the following expression.
Ka × Kb = Kw = 1.0 × 10⁻¹⁴
Kb = 1.0 × 10⁻¹⁴/Ka
Kb = 1.0 × 10⁻¹⁴/6.2 × 10⁻¹⁰
Kb = 1.6 × 10⁻⁵
Consider the halogenation of ethene is as follows:
CH₂=CH₂(g) + X₂(g) → H₂CX-CH₂X(g)
We can expect that this reaction occurring by breaking of a C=C bond and forming of two C-X bonds.
When bond break it is endothermic and when bond is formed it is exothermic.
So we can calculate the overall enthalpy change as a sum of the required bonds in the products:
Part a)
C=C break = +611 kJ
2 C-F formed = (2 * - 552) = -1104 kJ
Δ H = + 611 - 1104 = - 493 kJ
2C-Cl formed = (2 * -339) = - 678 kJ
ΔH = + 611 - 678 = -67 kJ
2 C-Br formed = (2 * -280) = -560 kJ
ΔH = + 611 - 560 = + 51 kJ
2 C-I Formed = (2 * -209) = -418 kJ
ΔH = + 611 - 418 = + 193 kJ
Part b)
As we can see that the highest exothermic bond formed is C-F bond so from bond energies we can found that addition of fluoride is the most exothermic reaction