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: A. dipole
Explanation:
Dipole can be defined as the separation of the equal and opposite charges present in the molecule via ionic bonding. The water molecule comprising of two molecules of hydrogen being positively charged and one molecule of oxygen being negatively charged are connected via ionic bonding thus water molecule is an example of dipole molecule. There is unequal distribution of the electrical density throughout the molecule.
Answer:
B. ADDITION OF TWO GROUPS ACROSS A DOUBLE BOND
Explanation:
Addition reaction of alkenes involves the conversion of the double bond in alkenes Inyo single bonds by the addition of two groups of atoms or radicals.
During this addition reaction, two substances, an unsaturated compound(e.g. ethane) and an attacking reagent (hydrogen, halogens, hydrogen halides, chlorine and bromine water) combines to form a single new compound without forming any other products. So a saturated product or one in which is an increase in degree of saturation is formed.
Answer:
0.0900 mol/L
Explanation:
<em>A chemist makes 330. mL of nickel(II) chloride working solution by adding distilled water to 220. mL of a 0.135 mol/L stock solution of nickel(II) chloride in water. Calculate the concentration of the chemist's working solution. Round your answer to significant digits.</em>
Step 1: Given data
- Initial concentration (C₁): 0.135 mol/L
- Initial volume (V₁): 220. mL
- Final concentration (C₂): ?
- Final volume (V₂): 330. mL
Step 2: Calculate the concentration of the final solution
We prepare a dilute solution from a concentrated one. We can calculate the concentration of the working solution using the dilution rule.
C₁ × V₁ = C₂ × V₂
C₂ = C₁ × V₁/V₂
C₂ = 0.135 mol/L × 220. mL/330. mL = 0.0900 mol/L
Chemical Formula of Sodium Sulfate is Na₂SO₄.
As we know, 1 mole of any substance contains 6.022 × 10²³ particles. So, 1 mole of Na₂SO₄ also contains 6.022 × 10²³ formula units of Na₂SO₄.
Also,
1 mole of Na₂SO₄ contains = 6.022 × 10²³ Atoms of S
So,
4.5 moles of Na₂SO₄ will contain = X atoms of S
Solving for X,
X = (4.5 moles × 6.022 × 10²³ Atoms) ÷ 1 mole
X = 2.70 × 10²⁴ Atoms of S
Result:
4.5 Moles of Na₂SO₄ contains 2.70 × 10²⁴ Atoms of Sulfur.
