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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Answer:
b. 4/3
Explanation:
Given data
- Final pressure: P₂ = 3 P₁
- Final temperature: T₂ = 4 T₁
We can find by what factor will the volume of the sample change using the combined gas law.

<span>the process or action by which one thing absorbs or is absorbed by another.</span>
Answer: The value of
for this reaction is 250000.
Explanation:
The given equation is as follows.

... (1)
... (2)
To balance the atoms, multiply equation (2) by 3. Hence, the equation (2) can be re-written as follows.
... (3)
Now, subtract equation (1) from equation (3). So, the equation formed will be as follows.

This equation can also be re-written as follows.

This equation is similar to the equilibrium equation given to us.
Therefore, during this subtraction the equation constants get divided as follows.
Thus, we can conclude that the value of
for this reaction is 250000.
Answer:
b
. 0.351 L.
Explanation:
Hello!
In this case, since diluted solutions are prepared by adding an extra amount of diluent to a stock-concentrated solution, we infer that the number of moles of solute remains the same, therefore we can write:

Thus, solving for the volume of the stock solution, V1, we obtain:

Now, by plugging in the given data we obtain:

Therefore, the answer is b
. 0.351 L.
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