Answer:
The answer to your question is 2 molecules, or B(edg 2021).
Explanation:
edg 2021
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:
when an atom, typically a metal, loses an electron or electrons, and becomes a positive ion, or cation. Another atom, typically a non-metal, is able to acquire the electron(s) to become a negative ion, or anion.
Explanation:
The least common multiples of 4, 10, and 16 is 80.
Answer:
27.64 liters
Explanation:
From the balanced equation, 2 moles of K2Cr2O7 requires 3 moles of CH3OH.
Mole of CH3OH = 1.9/32.04 = 0.0593 mole
Mole of K2Cr2O7 that will require 0.0593 mole of CH3OH:
2 x 0.0593/3 = 0.0395 mole
mole = molarity x volume
Volume of K2Cr2O7 needed = 0.0395/0.00143
= 27.64 Liter
<em>Hence, 27.64 liters of 0.00143 M K2Cr2O7 will be required to titrate 1.90 g of CH3OH dissolved in 50.0 mL of solution</em>
