Answer:
C.)One electron in each p orbital
Explanation:
In a P-sublevel with 3 electrons, they should be arranged with one electron going into each p-orbitals.
This is in accordance with the Hund's rule of maximum multiplicity.
The rule states that "electrons go into degenerate orbitals or sub-levels(p,d and f) singly before paring up".
Since the p-orbital is 3-fold degenerate with a capacity to accommodate a maximum number of 6 electrons, given 3 electrons, they will follow the Hund's rule in order to fill the orbitals.
So one electron will go in each p - orbitals easily.
Answer:
1.Metals
These are very hard except sodium
These are malleable and ductile pieces
These are shiny
Electropositive in nature
Non-metals
These are soft except diamond
These are brittle and can break down into pieces
These are non-lustrous except iodine
Electronegative in nature
2. The electrochemical series helps to pick out substances that are good oxidizing agents and those which are good reducing agents.In an electrochemical series the species which are placed above hydrogen are more difficult to be reduced and their standard reduction potential values are negative.
3. Arrhenius theory, theory, introduced in 1887 by the Swedish scientist Svante Arrhenius, that acids are substances that dissociate in water to yield electrically charged atoms or molecules, called ions, one of which is a hydrogen ion (H+), and that bases ionize in water to yield hydroxide ions (OH−).
4. The common application of indicators is the detection of end points of titrations. The colour of an indicator alters when the acidity or the oxidizing strength of the solution, or the concentration of a certain chemical species, reaches a critical range of values.
Answer:
True
Explanation:
Your welcome! :) Good luck!
Answer:
Option D. 5.5
Explanation:
The equation is this:
2A + 6B ⇒ 3C
With the amounts that we were given, let's determine which is the <em>limting reactant</em>
2 A reacts with 6 B
4 A will react with ( 4 .6)/2 = 12B
I have 11 B, so the limiting is B
6 B react with 2 A
11 B will react with (11 .2 )/6 =3.66 A
I have 4 A, so A is the excess.
6 B produce 3 C
11 B will produce ( 11 .3)/6 = 5.5C
The question is incomplete, the complete question is;
1.000 atm of Oxygen gas, placed in a container having apinhole opening in its side. leaks from the container 2.14 timesfaster thatn 1.000 atm of an unknown gas placed in this sameapparatus. Which of the following species could be theunknown gas?
A. CL2
B. SF6
C. Kr
D. UF6
E. Xe
Answer:
SF6
Explanation:
From Graham's law;
Let the rate of diffusion of oxygen be R1
Let the rate of diffusion of unknown base be R2
Let the molar mass of oxygen by M1
Let the molar mass of unknown gas be M2
Hence;
R1/R2 = √M2/M1
So;
2.14/1 = √M2/32
(2.14/1)^2 = M/32
M= (2.14/1)^2 × 32
M= 146.6
This is the molar mass of SF6 hence the answer above.
