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FromTheMoon [43]
1 year ago
15

Balance the following skeleton reactions and identify the oxidizing and reducing agents:(b) Mn²⁺(aq) + BiO₃⁻(aq) → MnO₄⁻(aq) + B

i³⁺(aq) [acidic]
Chemistry
1 answer:
amm18121 year ago
5 0

The balanced skeleton reactions are given below

2Mn²⁺(aq) +5BiO₃⁻(aq) +  14H^{+}(aq)  →  2MnO₄⁻(aq) + 5Bi³⁺(aq) + 7H_{2}O(l)

The oxidising agent and  reducing agents are Bi and Mn.

There are following steps involve in balance redox reaction in acidic medium .

  • First divide the reaction into two half reactions .
  • Balance the elements other than oxygen and hydrogen atom.
  • Balance the oxygen atom by adding H_{2}O on that side where number of oxygen  atom is less .
  • Balance the hydrogen atom by adding H^{+} on another side of in which water attached .
  • Balance the charge by adding e^{-} .
  • Add the two half reaction  .

The species which reduces the other species in the redox reaction is called oxidizing agent . Similarly , the species which oxidize the other species in the redox reaction  is called reducing agent .

To learn more about reducing agents please click here ,

brainly.com/question/2890416?

#SPJ4

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2 years ago
What is the final pH of a solution obtained by mixing 300 ml of 0.4 M NH3 with 175 ml of 0.3 M HCl? (Kb = 1.8 x 10-5) Show all o
MariettaO [177]

Answer:

pH of the final solution = 9.15

Explanation:

Equation of the reaction: HCl + NH₃ ----> NH₄Cl

Number of moles of  NH₃ = molarity * volume (L)

= 0.4 M * (300/1000) * 1 L =  0.12 moles

Number of moles of HCl =  molarity * volume (L)

= 0.3 M * (175/1000) * 1 L = 0.0525 moles

Since all he acid is used up in the reaction, number of moles of acid used up equals number of moles of NH₄Cl produced

Number moles of NH₄Cl produced =  0.0525 moles

Number of moles of base left unreacted =  0.12 - 0.0525 = 0.0675

pOH = pKb + log([salt]/[base])

pKb = -logKb

pOH = -log (1.8 * 10⁻⁵) + log (0.0525/0.06755)

pOh = 4.744 + 0.109

pOH = 4.853

pH = 14 - pOH

pH = 14 - 4.853

pH = 9.15

Therefore, pH of the final solution = 9.15

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Write the empirical formula for at least four ionic compounds that could be formed from the following ions:
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Answer:

FeSO_{4},FeCH_{3}CO,Fe_{2}(SO_{4})_{3},Fe_{2}(CH_{3}CO)_{3}

Explanation:

Empirical formula of ionic compound formed by two ions A^{x+} and B^{y-} is A_{y}B_{x} (for x\neq y) of AB (for x = y)

The above empirical formula is in accordance with charge neutrality principle

Here each cation (Fe^{3+} and Fe^{2+}) can form two ionic compounds by combining with two given anions (SO_{4}^{2-} and CH_{3}CO^{2-}).

So the four ionic compounds are: FeSO_{4},FeCH_{3}CO,Fe_{2}(SO_{4})_{3},Fe_{2}(CH_{3}CO)_{3}

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4 years ago
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