All categories

2 months ago

Be sure to answer all parts.

Chemistry

2 answers:

VMariaS [2.9K]2 months ago

8 0

Response:

a. 3 Br₂(l) + 6 OH⁻(aq) → 5 Br⁻(aq) + BrO₃⁻(aq) + 3 H₂O

b. Br₂

c. Br₂

Clarification:

Balancing a redox reactionis performed using the ion-electron method.

Step 1: Identify both half-reactions.

Reduction: Br₂(l) → Br⁻(aq)

Oxidation: Br₂(l) → BrO₃⁻(aq)

Step 2: Perform mass balance. This reaction occurs in basic conditions, thus we must add OH⁻ and H₂O as needed.

0.5 Br₂(l) → Br⁻(aq)

6 OH⁻(aq) + 0.5 Br₂(l) → BrO₃⁻(aq) + 3 H₂O

Step 3: Ensure electrical balance by incorporating electrons when necessary.

1 e⁻ + 0.5 Br₂(l) → Br⁻(aq)

6 OH⁻(aq) + 0.5 Br₂(l) → BrO₃⁻(aq) + 3 H₂O + 5 e⁻

Step 4: Scale both half-reactions to ensure the electron counts balance out.

5 × (1 e⁻ + 0.5 Br₂(l) → Br⁻(aq))

1 × (6 OH⁻(aq) + 0.5 Br₂(l) → BrO₃⁻(aq) + 3 H₂O + 5 e⁻)

Step 5: Combine both half-reactions and simplify as appropriate.

5 e⁻ + 3 Br₂(l) + 6 OH⁻(aq) → 5 Br⁻(aq) + BrO₃⁻(aq) + 3 H₂O + 5 e⁻

3 Br₂(l) + 6 OH⁻(aq) → 5 Br⁻(aq) + BrO₃⁻(aq) + 3 H₂O

The species that undergoes reduction is identified as the oxidizer. The species that undergoes oxidation is termed the reducer. In this situation, Br₂ qualifies as both.

Alekssandra [3K]2 months ago

6 0

Response:

the oxidizing agent is $Br_{2}$ while the reducing agent is $Br_{2}$

Clarification:

In reduction-oxidation reactions, electrons are transferred between compounds. During reduction, the oxidation number decreases, while it increases during oxidation. A reducing agent loses electrons, increasing its oxidation number, whereas an oxidizing agent gains electrons, reducing its oxidation number.

The unbalanced equation is:

$Br_{2(l)}$ → $BrO^{-} _{3(aq)}$ + $Br_{(aq)} ^{-}$

For oxidation reactions:

$Br_{2} + 6H_{2}O$ → $2BrO^{-} _{3(aq)}$ + 12 $H^{+}$ + $10e^{-}$

For reduction reactions:

$Br_{2(l)}$ + $2e^{-}$ → $2Br^{-}$

The next step is to equalize the electrons gained and lost:

$Br_{2} + 6H_{2}O$ → $2BrO^{-} _{3(aq)}$ + 12 $H^{+}$ + $10e^{-}$

+ $5Br_{2(l)}$ → $10e^{-}$ $10Br^{-}$

Finally, combine both equations:

⇌ $3Br_{2(aq)}+3H_{2}O_{(l)}$ + $2BrO^{-} _{3(aq)}$ + $5Br_{2(l)}$ $6H_{(aq)} ^{+}$

Thus, the oxidizing agent is $Br_{2}$ and the reducing agent is $Br_{2}$

You might be interested in

For each reaction, identify the element that gets reduced and the element that gets oxidized. 2 AgCl + Zn ⟶ 2 Ag + ZnCl 2 2AgCl+

lorasvet [2795]

Answer:

Explanation:

Oxidation:

Oxidation refers to the process where electrons are lost, resulting in an increase in the oxidation state of an atom of an element.

Reduction:

Reduction is characterized by the acquisition of electrons, leading to a decrease in the oxidation number.

Oxidizing agents:

Oxidizing agents facilitate oxidation of other substances while they themselves undergo reduction.

Reducing agents:

Reducing agents cause the other element to be reduced, while they are oxidized in the process.

Examine the following reaction:

2AgCl + Zn → 2Ag + ZnCl₂

In this instance, the oxidation state of Zn on the left is 0 and rises to +2 on the right, indicating it is oxidized, while the oxidation state of Ag drops from +1 on the left to 0 on the right, showing it is reduced.

4NH₃ + 3O₂ → 2N₂ + 6H₂O

For this reaction, nitrogen's oxidation state shifts from -3 on the left to 0 on the right, signifying it is oxidized, whereas oxygen decreases from 0 to -2, indicating it gets reduced.

Fe₂O₃ + 2Al → Al₂O₃ + 2Fe

Here, the oxidation state of iron reduces from +3 on the left to 0 on the right, meaning it is reduced, while aluminum shifts from 0 to +3, indicating it is oxidized.

4 0

1 month ago

Assume the weight of an average adult is 70. kg, and that 420. kJ of heat are evolved per mole of oxygen consumed as a result of

castortr0y [3046]

The temperature difference after 3 hours is 5.16 K. Given that the moles of O₂ inhaled rate at 0.02 mole/min, which converts to 1.2 mole/hour, we know the average heat released during metabolism is 7.2 kJ/h·kg. Therefore, the amount of heat generated within 3 hours will be 7.2 kJ/h·kg multiplied by 3 hours, giving a result of 21.6 kJ/kg, or 21.6 x 10³ J/kg. Applying the formula Qp = Cp x ΔT, and taking the body's heat capacity to be 4.18 J/g·K, we find ΔT = 5.16 K.

6 0

2 months ago

Convert 3.4 x 10^23 molecules of NaCl to grams

Alekssandra [3086]

In this case, to find the grams of sodium chloride starting from its molecules, the first step is to determine the moles of sodium chloride by utilizing Avogadro's number. After that, we can obtain grams directly using the molar mass of sodium chloride (58.45 g/mol).

7 0

2 months ago