A fact can be disapproved with further evidence different details. So basically evidence can change a fact!! Hope this helps
Answer:

Explanation:
Hello,
In this case, we could considering this as a redox titration:

Thus, the balance turns out (by adding both hydrogen ions and water):

Thus, by stoichiometry, the grams of Fe+2 ions result:

Finally, the mass percent is:

Best regards.
Answer:
The formula of the compound is:
N2H2
Explanation:
Data obtained from the question:
Nitrogen (N) = 93.28%
Hydrogen (H) = 6.72%
Next, we shall determine the empirical formula for the unknown compound. This is illustrated below:
N = 93.28%
H = 6.72%
Divide by their molar mass
N = 93.28 /14 = 6.663
H = 6.72 /1 = 6.7
Divide by the smallest
N = 6.663 / 6.663 = 1
H = 6.72 /6.663 = 1
Therefore, the empirical formula is NH.
Now, we can obtain the formula of the compound as follow:
The formula of a compound is simply a multiple of the empirical formula.
[NH]n = 30.04
[14 + 1]n = 30.04
15n = 30.04
Divide both side by 15
n = 30.04/15
n = 2
Therefore, the formula of the compound is:
[NH]n => [NH]2 => N2H2
<u>Given:</u>
Concentration of Ba(OH)2 = 0.348 M
<u>To determine:</u>
pOH of the above solution
<u>Explanation:</u>
Based on the stoichiometry-
1 mole of Ba(OH)2 is composed of 1 mole of Ba2+ ion and 2 moles of OH- ion
Therefore, concentration of OH- ion = 2*0.348 = 0.696 M
pOH = -log[OH-] = - log[0.696] = 0.157
Ans: pOH of 0.348M Ba(OH)2 is 0.157