Answer:
Types of atomic orbitals present in the third principal energy are <u>s, p and d only .</u>
Explanation:
- <u>OPTION A-: s and p atomic orbitals -</u> these two orbitals are present in second principal energy level. Therefore , the option is incorrect.
- <u> OPTION B-: p and d only -</u> This option is wrong as there is no such principal level energy where , s atomic orbital is absent .
- <u>OPTION C-: s , p and d only -</u>these orbitals are present in<u> third principal energy level</u>. The third major level of energy has one orbital, three orbitals of p, and five orbitals of d, each of which can contain up to 10 electrons. The third stage thus holds a maximum of 18 electrons. This option is correct .
- <u>OPTION D-: s , p, d and f only -</u>There is also a f sublevel at the <u>fourth and higher stages,</u> containing seven f orbitals, which can accommodate up to 14 electrons at most. Therefore, up to 32 electrons will hold the fourth level: 2 in the s orbital, 6 in the three p orbitals, 10 in the five d orbitals, and 14 in the seven f orbitals. This option is incorrect .
<u>Thus , the correct option is C (s , p and d only .)</u>
Answer:
c
Explanation:
there are billions of cells in living things. We are made out of millions of cells
The attraction of metal ions for mobile electrons
I’m pretty sure the answer you’re looking for is science...but I’m not quite sure because your question is kinda vague
Answer:
Kc → 5.58×10⁻⁴
Explanation:
Equilibrium reaction is:
2NOCl (g) ⇄ 2NO (g) + Cl₂(g)
Initially we have 1.25 moles of NOCl
After the equilibrium, we have 1.10 moles. So, during the process:
(1.25 mol - 1.1 mol) = 0.15 moles have reacted.
As ratio are 2:2, and 2:1, 0.15 moles of NO and (0.15 /2) = 0.075 moles of chlorine, were produced in the equilibrium.
Finally in equilibrium we have: 1.10 moles of NOCl, 0.15 moles of NO and 0.075 moles of Cl₂. But these amount are not molar, so we need molar concentration in order to determine Kc:
1.10 mol /2.50L = 0.44 M
0.15 mol / /2.50L = 0.06 M
0.075 mol /2.50L = 0.03 M
Let's make expression for Kc → [Cl₂] . [NO]² / [NOCl]²
Kc = (0.03 . 0.06²) / 0.44² → 5.58×10⁻⁴