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Anna35 [415]
3 years ago
9

Write empirical formula

Chemistry
1 answer:
AURORKA [14]3 years ago
6 0

Answer:

Fe(MnO_{4})_{2}\\\\ Fe(IO_{3})_{2}\\\\NH_{4}MnO_{4}\\\\NH_{4}IO_{3}

Explanation:

Fe^{2+}(MnO_{4}^{-})_{2}--->Fe(MnO_{4})_{2}\\\\Fe^{2+}(IO_{3}^{-})_{2}---> Fe(IO_{3})_{2}\\\\NH_{4}^{+}MnO_{4}^{-}--->NH_{4}MnO_{4}\\\\NH_{4}^{+}IO_{3}^{-}--->NH_{4}IO_{3}

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Nitrogen dioxide is a red-brown gas responsible for the brown color of smog. A container of nitrogen dioxide that is at low pres
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Answer:

Initially 1.51\times 10^{-2} moles of nitrogen dioxide were in the container .

Explanation:

Volume of the container at low pressure and at room temperature =V_1=3.4 L

Number of moles in the container = n_1

After more addition of nitrogen gas at the same pressure and temperature.

Volume of the container after addition = V_2=5.11 L

Number of moles in the container after addition=n_2=2.28\times 10^{-2} mol

Applying Avogadro's law:

\frac{Volume}{Moles}=constant (at constant pressure and temperature)

\frac{V_1}{n_1}=\frac{V_2}{n_2}

n_1=\frac{V_1\times n_2}{V_2}=\frac{3.4 L\times 2.28\times 10^{-2} mol}{5.11 L}

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3 years ago
Which of the following has the strongest buffering capacity? A. H2O B. 0.1 M HCl C. 0.1 M carbonic/bicarbonate (H2CO3/HCO3-) at
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Explanation:

(A)   As we know that carbonic acid (H_{2}CO_{3}) and Sodium bicarbonate (NaHCO_{3}) forms an acidic buffer.

Therefore, pH of an acidic buffer is given by Hendeerson-Hasselbalch equation as follows.

               pH = pK_{a} + log(\frac{[Salt]}{[Acid]}) ........... (1)

So mathematically,  if [Salt] = [Acid]  then \frac{[Salt]}{[Acid]} = 1 .

And,  log (\frac{[Salt]}{[Acid]}) = 0

Therefore, equation (1) gives us the following.

         pH = pK_{a} (when acid and salt are equal in concentration)

Hence, pK_{a} of H_{2}CO_{3} (carbonic acid) is 6.35.

And, with this we have following results.

In (A) and (D) we have the case \frac{[NaHCO_{3}]}{[H_{2}CO_{3}]}[/tex] i.e. [Salt] = [Acid].

Hence, for the cases pH = pK_{a} = 6.35.

(B)    [NaHCO_{3}] = 0.045 M and,  [H_{2}CO_{3}] = 0.45 M

Hence,   pH = 6.35 + log([NaHCO_{3}][[H_{2}CO_{3}])

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Therefore, it means that this buffer will be most suitable buffer as it has pH on acidic side and addition of slight excess base will not affect much of its pH value.

(C)    [NaHCO_{3}] = 0.45 M [H_{2}CO_{3}]

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So,       pH = 6.35 + log(\frac{[NaHCO_{3}]}{[H_{2}CO_{3}]})

                  = 6.35 + log(\frac{0.45}{0.045})

                  = 6.35 + (+1)

                 = 7.35

This means that pH on Basic side makes it no more acidic buffer.

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