Answer:
(a) ₁₉K: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹
(b) ₁₀Ne: 1s² 2s² 2p⁶
---
(a) 3
(b) 6
(c) 7
Explanation:
We can state the ground-state electron configuration for each element following Aufbau's principle.
(a) ₁₉K: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹
(b) ₁₀Ne: 1s² 2s² 2p⁶
Second part
(a) Al belongs to Group 13 in the Periodic Table. It has 13-10=3 electrons in the valence shell.
(b) O belongs to Group 16 in the Periodic Table. It has 16-10=6 electrons in the valence shell.
(c) F belongs to Group 17 in the Periodic Table. It has 17-10=7 electrons in the valence shell.
Answer: CF4
Explanation:
Calculate the molar mass of each compound. Divide the molar mass of Carbon by the molar mass of each compound, then multiply the answer by 100 to get the percentage.
CF4= 12+(19X 4)
=12+76= 88 g/mol
%C= 12/88 x 100= 13.64%
CO2= 12+(16 X 2)
12+32= 44 G/MOL
%C= 12/44 x 100= 27.3%
CH4= 12+ (1 X4)
=12+4
=16 G/MOL
%C= 12/16 X 100= 75%
C204
(12X2) + (16X4)
24+64
= 88 g/mol
%C= 24/88 x 100
= 27.3%
<span>0.5035 so go with .50 moles.
</span><span>
</span>
I believe that The answer is
Answer:
The answer to your question is d. 0.5 M
Explanation:
Data
[A] = 1M
K = 0.5
Concentration of B and C at equilibrium = x
Concentration of A at equilibrium = 1 - x
Equation of equilibrium
k = ![\frac{[B][C]}{A}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BB%5D%5BC%5D%7D%7BA%7D)
Substitution
![0.5 = \frac{[x][x]}{1 - x}](https://tex.z-dn.net/?f=0.5%20%3D%20%5Cfrac%7B%5Bx%5D%5Bx%5D%7D%7B1%20-%20x%7D)
Simplification
0.5 = 
Solve for x
0.5(1 - x) = x²
0.5 - 0.5x = x²
x² + 0.5x - 0.5 = 0
Find the roots x₁ = 0.5 x₂ = -1
There are no negative concentrations so the concentration of A at equilibrium is
[A] = 1 - 0.5
= 0.5 M
