The electronic configuration for vanadium (V) in the periodic table is as follows: 1s2 2s2 2p6 3s2 3p6 4s2 3d3 (option D).
<h3>What is electronic configuration?</h3>
Electronic configuration is the the arrangement of electrons in an atom, molecule, or other physical structure like a crystal.
Vanadium is the 23rd element on the periodic table and has chemical symbol V with atomic number 23. It is a transition metal, used in the production of special steels.
This suggests that the electronic configuration of Vanadium will be written as follows: 1s2 2s2 2p6 3s2 3p6 4s2 3d3
Therefore, the electronic configuration for vanadium (V) in the periodic table is as follows: 1s2 2s2 2p6 3s2 3p6 4s2 3d3.
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Answer:
The ratio of HC2H3O2(aq) in the flask after the addition of 5.0mL of NaOH(aq) to HC2H3O2(aq) in the flask after the addition of 1.0mL of NaOH(aq) is 15 : 19 .
Explanation:
HC2H3O2 is CH₃⁻ COOH, which is also known as Acetic acid.
IUPAC name of this compound is Ethanoic acid.
Acetic acid has a basicity of 1. so there is one acidic hydrogen is acetic acid.
Given that, equivalence point was reached when 20.0mL of NaOH is added.
let the normality of acetic acid is N₁ and that of NaOH is N₂.
volume of acetic acid is V₁ and that of NaOH is V₂.
Equivalence point occurs when, N₁ × V₁ = N₂ × V₂.
⇒ N₁ × V₁ = N₂ × 20.
after the addition of 5.0mL of NaOH(aq), remaining N₁ × V° = N₂ × (20 - 5).
= N₂ × 15.
after the addition of 1.0mL of NaOH(aq), remaining N₁ × Vˣ = N₂ × (20 - 1).
= N₂ × 19.
⇒ V° : Vˣ = 15 : 19 .
⇒
For the compound B the following statement is correct-
B. It is an ether because it is unable to form a hydrogen bond, so it is less soluble in water.
The solubility of alcohol in water depends upon the capability of formation of hydrogen bond in the solute. Now in alcohol the -OH group is polar in nature which enhance the possibility of hydrogen bond formation and it is more soluble in water.
On the other hand although there presence a -O- functional group in ether. It is less soluble in water due to non polarity of the functional group.
From the given data it is seen that compound A is more soluble in water than compound B. Thus it may be predicted that compound A is alcohol and B is ether.
Henceforth, for the compound B the following statement is correct-
B. It is an ether because it is unable to form a hydrogen bond, so it is less soluble in water.
The reason of incorrect options:
A. compound B cannot be an alcohol as it is less soluble in water.
C. In ether the functional group is -O-, thus electronegative atom (O) is present.
D. As both the compound (alcohol and ether) has equal molecular mass thus the organic chain will be same in alcohol and the hydrogen bond interaction will be more prominent than the dispersion force between the -OH group.
The question is incomplete, the complete question is;
Chlorine monoxide accumulates in the stratosphere above Antarctica each winter 3nd plays a key role the formation of the ozone hole above the South Pole each spring Eventually. CIO decomposes acco to the equation: 2CIO(g) rightarrow CL2(g) + O2(g) The second-order rate constant for the decomposition of CIO is 6950000000 M-1 s-1 at a particular temperature Determine the half-life of CIO when its initial concentration is .0000000185 M
Answer:
7.8 * 10^-3 s
Explanation:
Given that the half life of a second order reaction is obtained from the formula;
t1/2 = k-1[A]o-1
t1/2 = 1/k[A]o
second order rate constant (k) = 6950000000 M-1 s-1
initial concentration ([A]o) =0.0000000185 M
t1/2 = 1/6950000000 * 0.0000000185
t1/2 = 7.8 * 10^-3 s
Answer:
32
Step-by-step explanation:
There are two ways you can count the valence electrons.
A. From the Periodic Table
1 × P (Group 15) = 5
4 × O (Group 16) = 4 × 6 = 24
+3 e⁻ (for the charges) = <u> 3</u>
Total = 32
B. From the Lewis structure
In the <em>Lewis structure</em> (below), each line (bond) represents a pair of bonding electrons, and each dot represents an unbound electron (half a lone pair).
5 lines (bonds) = 5 × 2 = 10
3 single-bonded O atoms = 3 × 6 = 18
1 double-bonded O atom = <u> 4</u>
Total = 32
