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inn [45]
3 years ago
8

Write a balanced half-reaction for the reduction of bismuth oxide ion to bismuth ion in acidic aqueous solution. be sure to add

physical state symbols where appropriate.
Chemistry
1 answer:
gavmur [86]3 years ago
8 0
BiO₃⁻ → Bi³⁺
+5           +3
Balance oxidation state:
BiO₃⁻ + 2 e⁻ → Bi³⁺ 
Balancing oxygen by adding water and balance H
BiO₃⁻ + 6 H⁺ + 2 e⁻ → Bi³⁺ + 3 H₂O
balance charge:
BiO₃⁻ + 6 H⁺ + 2 e⁻ → Bi³⁺ + 3 H₂O
      + 3                               + 3
Final:
BiO₃⁻ + 6 H⁺ + 2 e⁻ → Bi³⁺ + 3 H₂O
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Name the compound containing only fluorine and hydrogen.
Sonbull [250]
Hydrogen has only one electron. It has one proton of nucleus and one outter electron.
7 0
3 years ago
You have a solution of calcium chloride dissolved in water. You add some more solid to the solution and stir, and the solid diss
IgorC [24]

Answer:

It changed from unsaturated to saturated solution

Explanation:

Before adding the spatula tip full amount, the solution was unsaturated, but as soon as the spatula tip full amount was added and couldn't dissolve, it became unsaturated because it could not dissolve any more solute having already dissolved it maximum number of solute

5 0
3 years ago
A given hydrocarbon is converted completely to carbon dioxide and water, and equal numbers of moles of co2 and h2o are produced.
Sauron [17]
X moles of CO₂, and x moles of H₂O

CxH2x + 3x/2 O₂ -------> xCO₂ + xH₂O

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4 0
4 years ago
How can you model the cycling of matter in the Earth system?
Rus_ich [418]

Answer:

The cycling of matter is important to many Earth processes and to the survival of organisms the existing matter must cycle continuously for this planet to support life Water, carbon, nitrogen, phosphorus, and even rocks move through cycles If these materials did not cycle, Earth could not support life.

Explanation:

5 0
3 years ago
Suppose 0.795 g of sodium iodide is dissolved in 100. mL of a 39.0 m M aqueous solution of silver nitrate.
lubasha [3.4K]

Answer:

The final molarity of iodide anion is 0.053 M

Explanation:

<u>Step 1</u>: Data given

Mass of sodium iodide (NaI) = 0.795 grams

Volume of the solution = 100 mL = 0.1 L

Molarity of aqueous solution of silver nitrate (AgNO3) = 39 mM = 0.039M

The molecular mass of sodium iodide is 149.89 g/mol.

<u>Step 2:</u> The balanced equation

AgNO3(aq) + NaI(aq) → AgI(s) + NaNO3(aq)

<u>Step 3: </u>Calculate number of moles of sodium iodide

Moles NaI = mass NaI / Molar mass NaI

Moles NaI = 0.795 grams / 149.89 g/mol

Moles NaI = 0.0053 moles

For 1 mole AgNO3 consumed, we need 1 mole NaI to produce 1 mole AgI and 1 mole NaNO3

The sodium iodide will dissociate as followed:

NaI(aq) → Na+(aq) +  I-(aq)

<u>Step 4</u>: Calculate iodide ions

For 1 mole NaI, we have 1 mole of I-

For 0.0053 moles of NaI we'll have 0.0053 moles I-

<u>Step 5:</u> Calculate molarity of iodide ion

Molarity = moles I- / volume

Molarity I- = 0.0053 moles / 0.1 L

Molarity I- = 0.053 M

The final molarity of iodide anion is 0.053 M

5 0
3 years ago
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