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:
One chemical change- It was when he toasted the bread. The heat changed the bread so that it has the crust on the outside.
Explanation:
Pressure is 5.7 atm
<u>Explanation:</u>
P1 = Standard pressure = 1 atm
P2 = ?
V1 = Volume = 10L
V2= 2.4L
T1 = 0°C + 273 K = 273 K
T2 = 100°C + 273 K = 373 K
We have to find the pressure of the gas, by using the gas formula as,

P2 can be found by rewriting the above expression as,

Plugin the above values as,

Answer:
A) ∆Suniv >0, ∆G<0, T∆Suniv >0.
Explanation:
The connection between entropy and the spontaneity of a reaction is expressed by the <u>second law of thermodynamics</u><u>: The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process</u>.
Mathematically, we can express the second law of thermodynamics as follows:
For a spontaneous process: ΔSuniv = ΔSsys + ΔSsurr > 0
Therefore, the second law of thermodynamics tells us that a spontaneous reaction increases the entropy of the universe; that is, ΔSuniv > 0.
If we want spontaneity expressed only in terms of the properties of the system (ΔHsys and ΔSsys), we use the following equation:
-TΔSuniv = ΔHsys - TΔSsys < 0
That means that T∆Suniv >0.
This equation says that for a process carried out at constant pressure and temperature T, if the changes in enthalpy and entropy of the system are such that <u>ΔHsys - TΔSsys is less than zero, the process must be spontaneous.</u>
Finally, if the change in free energy is less than zero (ΔG<0), the reaction is spontaneous in the forward direction.
Answer:
2.65 M
Explanation:
Convert grams of K₂CO₃ to moles. The molar mass is 138.205 g/mol.
(110 g)/(138.205 g/mol) = 0.796 mol
Convert milliliters of solution to liters.
300 mL = 0.300 L
Divide moles of K₂CO₃ by liters of solution.
0.796 mol/0.300 L = 2.65 mol/L = 0.265 M