0.4649331785818406 is what 27.4 grams is converted to! You're welcome!! :)
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
The number of moles of NaOh that are contained in 65ml of 2.20M solution NaOh in H2o is calculated using the below formula
moles = molarity x volume /1000
that is 65 x2.20 /1000= 0.143 moles
The molar mass of NH3
N = 14
H3 = 3
total = 17
The fraction of nitrogen is 14/17
So 14/17 x 125 = 102.94g
Molar mass Na = 23g/mol
46g = 456/2 = 2mol
1mol = 6.022*10^23 atoms
2mol = 2*6.022*10623
= 1.204*10^24 atoms