Answer:
In a neutral molecule, the sum of the bonding valance electrons must be equal. So the products of the negative element and its charges and the positive element and its charge must be equal.
Explanation:
C1×N1 = C2×N2
If we have a 3 valance electrons , the 'A' charge will be either +3 or -5 for a full octet and valance electron in 'B' atoms will mostly result in acquisition of additional electrons (2) for an octet and relative charge of -2.
Balancing the two,
3 × A = -2 × B
To be equal, A = 2 and B = 3
Therefore, A²B³
Heat & pressure. hope this helps
Answer:
The humid continental climate has hot summers, while the subarctic climate has short, cool summers.
Explanation:
I did the lesson already and got it correct lol
Answer :
- Carbonyl group : It is a functional group composed of a carbon atom that double bonded to oxygen atom. It is represented as

Carboxylic group : It is the class of organic compound in which the carboxylic (-COOH) group is attached to a hydrocarbon is known as carboxylic.
The general formula of carboxylic is,
. According to the IUPAC naming, the carboxylic are named as alkanoic acids.
Aldehyde group : It is the class of organic compound in which the (-CHO) group is attached to a hydrocarbon is known as aldehyde.
The general representation of aldehyde is,
. According to the IUPAC naming, the aldehyde are named as alkanals.
Ketone group : It is the class of organic compound in which the (-CO) group is directly attached to the two alkyl group of carbon is known as ketone.
The general representation of ketone is,
. According to the IUPAC naming, the ketone are named as alkanone.
Ester group : It is the class of organic compound in which the (-COO) group is directly attached to the two alkyl group of carbon is known as ester.
The general representation of ester is,
. According to the IUPAC naming, the ester are named as alkyl alkanoate.
Answer : The radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.
Explanation :
As we are given that the Na⁺ radius is 56.4% of the Cl⁻ radius.
Let us assume that the radius of Cl⁻ be, (x) pm
So, the radius of Na⁺ = 
In the crystal structure of NaCl, 2 Cl⁻ ions present at the corner and 1 Na⁺ ion present at the edge of lattice.
Thus, the edge length is equal to the sum of 2 radius of Cl⁻ ion and 2 radius of Na⁺ ion.
Given:
Distance between Na⁺ nuclei = 566 pm
Thus, the relation will be:





The radius of Cl⁻ ion = (x) pm = 181 pm
The radius of Na⁺ ion = (0.564x) pm = (0.564 × 181) pm =102.084 pm ≈ 102 pm
Thus, the radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.