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neonofarm [45]
3 years ago
7

Currently this lab is run qualitatively asking only for whether a solution is acidic, neutral, or basic. If quantitative informa

tion was to be acquired for this experiment, what range of pHs would the universal indicator described in your lab manual be unable to distinguish?
(A) 7.6-8.0
(B) 7.0
(C) 8.0-8.6
(D) 6.0-7.6
Chemistry
1 answer:
crimeas [40]3 years ago
8 0

Answer:

The only PH range which is not covered by any of the given components of the universal indicator is 7.6-8.0

Hence the PH range 7.6-8 can't be described using universal indicator.

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Explanation:

17. it goes from solid copper to aqueous copper:

Cu(s) --> Cu₂(aq) + 2e⁻

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19. net ionic, must include only reacting species, so

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Balance the following redox reaction in acidic solution. Zn(s)+MnO−4(aq)→ Zn+2(aq)+Mn+2(aq)
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Answer:

2MnO₄⁻ + 5Zn + 16H⁺ → 2Mn²⁺ + 8H₂O + 5Zn²⁺

Explanation:

To balance a redox reaction in an acidic medium, we simply follow some rules:

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  2. By inspecting, balance the half equations with respect to the charges and atoms.
  3. In acidic medium, one atom of H₂O is used to balance up each oxygen atom and one H⁺ balances up each hydrogen atom on the deficient side of the equation.
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  5. Multiply both equations with appropriate factors to balance the electrons in the two half equations.
  6. Add up the balanced half equations and cancel out any specie that occur on both sides.
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Solution

                            Zn + MnO₄⁻ → Zn²⁺ + Mn²⁺

The half equations:

                      Zn → Zn²⁺                          Oxidation half

                      MnO₄⁻ → Mn²⁺                  Reduction half

Balancing of atoms(in acidic medium)

                     Zn → Zn²⁺

                    MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O

Balancing of charge

                   Zn → Zn²⁺ + 2e⁻

                    MnO₄⁻ + 8H⁺ + 5e⁻→ Mn²⁺ + 4H₂O

Balancing of electrons

         Multiply the oxidation half by 5 and reduction half by 2:

                          5Zn → 5Zn²⁺ + 10e⁻

                        2MnO₄⁻ + 16H⁺ + 10e⁻→ 2Mn²⁺ + 8H₂O

Adding up the two equations gives:

              5Zn + 2MnO₄⁻ + 16H⁺ + 10e⁻ → 5Zn²⁺ + 10e⁻ + 2Mn²⁺ + 8H₂O

The net equation gives:

         5Zn + 2MnO₄⁻ + 16H⁺ → 5Zn²⁺ + 2Mn²⁺ + 8H₂O

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