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.
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For example, at atmospheric corrosion of copper:
2Cu + CO₂ + O₂ + H₂O = CuCO₃*Cu(OH)₂
and disolution of substances of copper in an acid condition:
CuCO₃*Cu(OH)₂(s) + 4HNO₃(aq) = 2Cu(NO₃)₂(aq) + CO₂(g) + 3H₂O(l)
CuCO₃*Cu(OH)₂(s) + 4H⁺ = 2Cu²⁺ + CO₂ + 3H₂O
Answer:
176.36g/mol
Explanation:
It was given that:
Mass of ascorbic acid=0.552 g
Volume of water=20.0 mL
Concentration of KOH=0.1103 M
Volume of KOH=28.42 mL. = 0.02842l
pH of solution at 10.0 mL= 3.72
At equivalence point, number of moles of acid is equal to the number of moles of base.
Number of moles of base= (KOH) x Volume
=0.1103 x 0.02842 L
=0.00313 moles.
Therefore number of miles of acid= 0.00313moles
No of moles= Mass/molar mass
Molar Mass= Mass of Acid/ No of moles of acid
= 0.552g/0.00312moles
= 176.36g/mol