Answer: NO2, NO, and O2.
<span>Free radicals are toxic substances produced by the body. In normal circumstances,the body can neutralize but<span>
when the level of these substances is to much,they accumulate
and can generate diseases,
such as osteoporosis and cancer.</span></span>
2: <span>Volume V = a*b*c = 6.0*3.0*3.0 = 54.0 cm^3 density ρ = mass/volume = 146/54 = 2.70 g/cm^3
3: Volume = (27.8 -21.2) cm^3
mass = 22.4 g
density = 22.4/(27.8-21.2) g/cm^3
</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
Atomic mass Ni = 58.69 a.m.u
58.69 g ----------------- 6.02x10²³ atoms
?? g --------------------- 7.5x10¹⁵ atoms
58.69x (7.5x10¹⁵) / 6.02x10²³
=> 7.31x10⁻⁷ g
