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3 years ago

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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krek1111 [17]

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4 years ago

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Dafna1 [17]

Answer:

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3 years ago

Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find

FromTheMoon [43]

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Explanation:

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

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We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

$H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)$

0.025 0.025 0.025

Equation for the second dissociation of sulfuric acid:

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At eqllm: 0.025-x 0.025+x x

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$Ka_2\text{ for }H_2SO_4=0.01$

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Lady bird [3.3K]

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