Elements with three p-electrons....
That would be N, P, As, Sb, and Bi -- elements in group 15
For example, energy diagram showing "empty" orbitals up through the 3p.
.....3p __ __ __
3s __
.....2p __ __ __
2s __
1s __
Energy diagram of phosphorous showing three unpaired electrons in 3p-sublevel
.....3p ↑_ ↑_ ↑_
3s ↑↓
.....2p ↑↓ ↑↓ ↑↓
2s ↑↓
1s ↑↓
According to Hund's rule, the electrons singly occupy the p-orbitals, and all have the same spin.
<span>Gather your materials. Gather the sand, salt and iron filing mixture; a magnet; a paper towel; a cup of warm water; an empty cup; and a piece of filter paper.Use the magnet to remove the iron. ...Use warm water to dissolve the salt. ...<span>Remove the sand from the salt water.</span></span>
Answer:
there are 3 different elements in that equation
Explanation:
Fe is the element Iron, Cr is the element Chromium, and O is oxygen
Answer 1) : To rank the species according to the decreasing energy needed to raise the temperature of 10 g of the substance by 25 °C, the oder of energy needed is given below as per their specific heat.
Which requires more energy as per their specific heats;
- water 4.184
- aluminum 0.897
- cast iron 0.460
- Copper 0.385
- Silver 0.235
Answer 2)To rank items as equivalent, overlap them.
The average kinetic energy of the molecules after point D,
The average kinetic energy of the molecules from point B to point C,
The average kinetic energy of the molecules before point A.
Answer:
pH after the addition of 10 ml NaOH = 4.81
pH after the addition of 20.1 ml NaOH = 8.76
pH after the addition of 25 ml NaOH = 8.78
Explanation:
(1)
Moles of butanoic acid initially present = 0.1 x 20 = 2 m moles = 2 x 10⁻³ moles,
Moles of NaOH added = 10 x 0.1 = 1 x 10⁻³ moles
CH₃CH₂CH₂COOH + NaOH ⇄ CH₃CH₂CH₂COONa + H₂O
Initial conc. 2 x 10⁻³ 1 x 10⁻³ 0
Equilibrium 1 x 10⁻³ 0 1 x 10⁻³
Final volume = 20 + 10 = 30 ml = 0.03 lit
So final concentration of Acid = 
Final concentration of conjugate base [CH₃CH₂CH₂COONa]
Since a buffer solution is formed which contains the weak butanoic acid and conjugate base of that acid .
Using Henderson Hasselbalch equation to find the pH
![pH=pK_{a}+log\frac{[conjugate base]}{[acid]} \\\\=-log(1.54X10^{-5} )+log\frac{0.03}{0.03} \\\\=4.81](https://tex.z-dn.net/?f=pH%3DpK_%7Ba%7D%2Blog%5Cfrac%7B%5Bconjugate%20base%5D%7D%7B%5Bacid%5D%7D%20%20%5C%5C%5C%5C%3D-log%281.54X10%5E%7B-5%7D%20%29%2Blog%5Cfrac%7B0.03%7D%7B0.03%7D%20%5C%5C%5C%5C%3D4.81)
