Answer:
1) 1,... 2
2) 18
3) n= 3 and I=1
Explanation:
1) when l= 0, its an s-sub-level, and only 1 orbital is possible which can carry only 2-electrons
2) the maximum number of electron is given by 2n^2= 2×3^2= 18
3) in 3p, the coefficient of p is the value of n= 3 and l-value of P is 1
<span>NO2 weighs 46.005 grams per mol. There are 6.02x10^23 molecules in a mol. In the given sample of 189.5 grams, there are 4.12 mols. This means that there are 2.48x10^24 molecules of NO2 in the given sample.</span>
In NaMnO₄, Mn has the highest oxidation number.
The question is incomplete, the complete question is;
Which of the following species contains manganese with the highest oxidation number?
A) Mn
B) MnF₂
C) Mn₃(PO₄)₂
D) MnCl₄
E) NaMnO₄
In order to ascertain the specie that contains manganese with the highest oxidation number, we must calculate the oxidation number of manganese in each of the species one after the other.
1) For Mn, the oxidation number of Mn is zero because the atom is uncombined.
2) For MnF₂;
Mn has an oxidation number of +2
3) For Mn₃(PO₄)₂
Mn has an oxidation number of +2
4) For MnCl₄
Mn has an oxidation number of +4
5) For NaMnO₄
Mn has an oxidation number of +7
Hence in NaMnO₄, Mn has the highest oxidation number.
Learn more: brainly.com/question/10079361
Answer:
What is the total number of valence electrons in the Lewis structure of nh4+?
Total valence electrons pairs
Total electron pairs are determined by dividing the number total valence electrons by two. For, NH4+, Total pairs of electrons are 4.
Explanation:
Hope it helps! Correct me if I am wrong :>
Im sure about my answer :>
Can you please mark me as brainlest?
