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
thanks for the answers ッ. (btw they’re on the bottom of the question if anyone doesn’t see it.
Br2 == 2Br
24% dissociated => n total moles, 0.24 mol*n of Br, and 0.76*n mol of Br2
=> partial pressure of Br, P Br = 0.24 bar, and
partical pressure of Br2, P Br2 = 0.76 bar
kp = (P Br)^2 / P Br2 = (0.24)^2 / 0.76 = 0.0758
is this for a test or are you genuinely interested? molality = mols sugar/kg solvent
Solve for molality
delta T = Kf*m
Solve for delta T and subtract from zero C to find the new freezing point.
or
-5.58
Answer:- Mass of copper piece is 290 gram.
Solution:- We know that, mass = density * volume
density of copper is given as 8.96 gram per mL.
Volume of copper piece is the rise change in volume.
Volume of copper piece = 137 mL - 105 mL = 32 mL
Let's multiply the volume by density to calculate the mass of copper:
mass of copper = 
mass of copper = 286.72 g
Volume has two significant figures, so if we round the mass to two significant figures then it becomes 290 g.