When ketone is reacted with phosphorous pentachloride, chlorination takes place at the carbonyl carbon with substitution of the oxygen atom to give a geminal dichloride (with 2 Cl atoms on same carbon) according to the following equation:
so we can say that acetone is converted into 2,2-dichloropropane by action of PCl₅
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Answer:
- <em>The net charge of the ionic compound calcium fluoride is </em><u><em>zero (0).</em></u>
<em>Explanation:</em>
<em>Ionic compounds,</em> such as covalent ones, have zero net charge; this is, they are neutral.
Substances with net positive charge are cations and substances with net negative charge are anions.
The charges in the <em>ionic compound calcium flouride</em> are distributed in this way:
- Calcium charge: Ca²⁺: this is, each calcium ion has a 2 positive charge
- Fluoride charge: F⁻: each fluoride ion has a 1 negative charge.
- Then, the <em>net charge</em> is: 1 × (2+) + 2 × (1-) = +2 - 2 = 0.
So, a two positve charge, from one calcium ion, is equal to two negative charges, from two fluoride tions, yielding a <u>zero net charge</u>.
Answer:
Electronegativity in group 1 decreases as we go from Lithium to Francium.
Explanation:
Electronegativity is defined as the tendency of an element to attract an electron pair towards itself.
In a group generally this tendency decreases from top to bottom as the size of the atom increases and hence the positive nucleus get far from the outer orbital.
In the same way group 1 elements i.e. from Lithium to Francium electronegativity decreases.
Chromium has the electron configuration [Ar]4s13d5 and exhibits oxidation numbers 2+, 3+, and 6+. When chromium loses two electrons, it forms the Cr2+ ion and has the configuration [Ar]3d4.
The Answer is B. [Ar]3d4
Answer: A volume of 500 mL water is required to prepare 0.1 M 
from 100 ml of 0.5 M solution.
Explanation:
Given: 
= 0.1 M, 
= ?
= 0.5 M, 
= 100 mL
Formula used to calculate the volume of water is as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that a volume of 500 mL water is required to prepare 0.1 M from 100 ml of 0.5 M solution.
