Oxidation half reaction is written as follows when using using reduction potential chart
example when using copper it is written as follows
CU2+ +2e- --> c(s) +0.34v
oxidasation is the loos of electron hence copper oxidation potential is as follows
cu (s) --> CU2+ +2e -0.34v
Because H is a non-metal, it has the highest electronegativity. Option (1) H is correct.
Hope this would help~
The equilibrium constant, k of the reaction in which case, the concentrations of the given reactants and products are as indicated is; Choice A; K = 3.1 x 10⁵
<h3>What is the equilibrium constant , k of the reaction as described in the task content?</h3>
It follows from above that the concentrations of the reactants and products are as follows; [H2] = 0.10 M, [N2] = 0.10 M, and [NH3] = 5.6 M at equilibrium.
Hence, the equilibrium constant of the reaction in discuss is;
K = [5.6]²/[0.10]³[0.10]
k = 5.6² × 10⁴
k = 3.136 × 10⁵
K = 3.1 × 10⁵.
Read more on equilibrium constant;
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Answer:
(a) 5s. n = 5. Sublevel s, l = 0. Number of orbitals = 1
(b) 3p. n = 3. Sublevel p, l = 1. Number of orbitals = 3
(c) 4f. n =4. Sublevel f, l = 3. Number of orbitals = 7
Explanation:
The rules for electron quantum numbers are:
1. Shell number, 1 ≤ n
2. Sublevel number, 0 ≤ l ≤ n − 1
So,
(a) 5s. n = 5, shell number 5. Sublevel s, l = 0. Number of orbitals = 2l +1 = 1
(b) 3p. n = 3, shell number 3. Sublevel p, l = 1. Number of orbitals = 2l +1 = 3
(c) 4f. n =4, shell number 4. Sublevel f, l = 3. Number of orbitals = 2l +1 = 7
Answer:
There are four stages in the metamorphosis of butterflies and moths: egg, larva, pupa, and adult.
Hope this helps :)
