Answer:
B
Explanation:
<em>Both plain and plateau have flat surfaces. However, a plain is located in a low-lying area while a plateau is located on an elevated area. In essence, a plateau can be viewed as an elevated plain or a plain that is bordered by cliffs.</em>
The correct option is B.
<u>Given information:</u>
Concentration of NaF = 0.10 M
Ka of HF = 6.8*10⁻⁴
<u>To determine:</u>
pH of 0.1 M NaF
<u>Explanation:</u>
NaF (aq) ↔ Na+ (aq) + F-(aq)
[Na+] = [F-] = 0.10 M
F- will then react with water in the solution as follows:
F- + H2O ↔ HF + OH-
Kb = [OH-][HF]/[F-]
Kw/Ka = [OH-][HF]/[F-]
At equilibrium: [OH-]=[HF] = x and [F-] = 0.1 - x
10⁻¹⁴/6.8*10⁻⁴ = x²/0.1-x
x = [OH-] = 1.21*10⁻⁶ M
pOH = -log[OH-] = -log[1.21*10⁻⁶] = 5.92
pH = 14 - pOH = 14-5.92 = 8.08
Ans: (b)
pH of 0.10 M NaF is 8.08
Answer:
The answer to your question is Ferns
Explanation:
Ferns are called nonflowering plants and produce spores instead of seeds.
Explanation:
The shapes and relative energies of the orbitals s,p,d and f orbitals are given by the principal quantum number and the azimuthal quantum number.
The principal quantum number gives the main energy level and the azimuthal quantum number denotes the shape of the orbitals.
- For the principal quantum number, they represent the energy levels in which the orbital is located or the average distance of the orbital from the nucleus. It takes the number n = 1,2,3,4,5,6,7......
- The azimuthal quantum number(L) shows the shape of the orbitals in subshells accommodating electrons. The number of possible shapes is limited by the the principal quantum number.
L Name of orbital shape of orbital
0 s spherical
1 p dumb-bell
2 d double dumb-bell
3 f complex
Principal Azimuthal Orbital
Quantum Quantum Designation of
Number (N) Number(l) Sublevel
1 0 1s
2 0 2s
1 2p
3 0 3s
1 3p
2 3d
4 0 4s
1 4p
2 4d
3 4f
Learn more:
Atomic orbitals brainly.com/question/9514863
#learnwithBrainly
Answer:
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
Explanation:
The reaction performed in the experiment is;
2 Cu(NO3)2 + 4 KI → 2 CuI (s) + 4 KNO3 + I2
The iodide ions reduces Cu^2+ to Cu^+ which is insoluble in water hence the precipitate. This is so because iodine is a good oxidizing agent seeing that it requires one electron to fill its outermost shell. Potassium on the other hand is a good reducing agent since it easily looses its one electron.
The oxidation - reduction equation is as follows;
2Cu2^+ + 2e ----> 2Cu^+ reduction half equation
2I^- ----> I2 + 2e. Oxidation half equation
Balanced redox reaction equation;
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
