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Savatey [412]
3 years ago
8

Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring

in an electrochemical cell at 2 5°C. (The equation is balanced.) Pb(s) + Br2(l) → Pb2+(aq) + 2 Br-(aq) Pb2+(aq) + 2 e- → Pb(s) E° = -0.13 V Br2(l) + 2 e- → 2 Br-(aq) E° = +1.07 V
Chemistry
1 answer:
iren2701 [21]3 years ago
4 0

Answer:

1.20 V

Explanation:

The standard cell potential is calculated from the expression

ε⁰ cell = ε⁰ oxidation + ε⁰  reduction

The species that will be reduced is the one with the higher standard reduction potential and the species that will be oxidized will be the one with the more negative reduction potential.

Thus for our question we will have

oxidation:

Pb(s)  →   Pb2+(aq) + 2 e-       ε⁰ oxidation       =  -   ε⁰  reduction

                                                                          =   - ( - 0.13 V ) = + 0.13 V

reduction    

Br2(l) + 2 e- → 2 Br-(aq)           ε⁰  reduction     = +1.07 V

ε⁰ cell = ε⁰ oxidation + ε⁰  reduction = + 0.13 V + 1.07 V  = 1.20 V

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Answer:

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Explanation:

5 0
2 years ago
Chromium is dissolved in sulfuric acid according to the following equation: Cr + H2SO4 ⇒ Cr2 (SO4) 3 + H2
Usimov [2.4K]

Answer:

\large \boxed{\text{a)188.4 g; b) 98.67 $\, \%$}}

Explanation:

We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:                      98.08           392.18

             2Cr + 3H₂SO₄ ⟶ Cr₂(SO₄)₃ + 3H₂

To solve the stoichiometry problem, you must

  • Use the molar mass of H₂SO₄ to convert  the mass of H₂SO₄ to moles of H₂SO₄
  • Use the molar ratio to convert moles of H₂SO₄ to moles of Cr₂(SO₄)₃
  • Use the molar mass of Cr₂(SO₄)₃ to convert moles of Cr₂(SO₄)₃ to mass of Cr₂(SO₄)₃

a) Mass of Cr₂(SO₄)₃

(i) Mass of pure H₂SO₄

\text{Mass of pure} = \text{165 g impure} \times \dfrac{\text{85.67 g pure} }{\text{100 g impure}} = \text{141.36 g pure}

(ii) Moles of H₂SO₄

\text{Moles of H$_{2}$SO}_{4} = \text{141.36 g H$_{2}$SO}_{4} \times \dfrac{\text{1 mol H$_{2}$SO}_{4}}{\text{98.08 g H$_{2}$SO}_{4}} = \text{1.441 mol H$_{2}$SO}_{4}

(iii) Moles of Cr₂(SO₄)₃

The molar ratio is 1 mol Cr₂(SO₄)₃:3 mol H₂SO₄ \text{Moles of Cr$_{2}$(SO$_{4}$)}_{3} = \text{1.441 mol H$_{2}$SO}_{4} \times \dfrac{\text{1 mol Cr$_{2}$(SO$_{4}$)}_{3}}{\text{3 mol H$_{2}$SO}_{4}} = \text{0.4804 mol Cr$_{2}$(SO$_{4}$)}_{3}

(iv) Mass of Cr₂(SO₄)₃ \text{Mass of Cr$_{2}$(SO$_{4}$)}_{3} = \text{0.4804 mol Cr$_{2}$(SO$_{4}$)}_{3} \times \dfrac{\text{392.18 g Cr$_{2}$(SO$_{4}$)}_{3}}{\text{1 mol Cr$_{2}$(SO$_{4}$)}_{3}} = \textbf{188.4 g Cr$_{2}$(SO$_{4}$)}_{3}\\\text{The mass of Cr$_{2}$(SO$_{4}$)$_{3}$ formed is $\large \boxed{\textbf{188.4 g}}$}

b) Percentage yield

It is impossible to get a yield of 485.9 g. I will assume you meant 185.9 g.

\text{Percentage yield} = \dfrac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 \, \% = \dfrac{\text{185.9 g}}{\text{188.4 g}} \times 100 \, \% = \mathbf{98.67 \, \%}\\\\\text{The percentage yield is $\large \boxed{\mathbf{98.67 \, \%}}$}

7 0
4 years ago
The yearly amounts of carbon emissions from cars in Belgium are normally distributed with a mean of 13.9 gigagrams per year and
zavuch27 [327]

Answer:

The correct option is;

c. 0.167

Explanation:

The parameters given are;

The mean, μ = 13.9 Gigagrams/year

The standard deviation, σ = 5.8 Gigagrams/year

The z-score formula is given as follows;

z = \dfrac{x - \mu }{\sigma }

Where:

x = Observed score = 11.5 Gigagrams/year

We have;

z = \dfrac{11.5 - 13.9 }{5.8 } =  \dfrac{-2.4 }{5.8 } = 0.4138

From the z-score table relations/computation, the probability (p-value) = 0.6605

Where:

x = Observed score = 14 Gigagrams/year

We have;

z = \dfrac{14- 13.9 }{5.8 } =  \dfrac{0.1}{5.8 } = 0.01724

From the z-score table relations/computation, the probability (p-value) = 0.4931

Therefore, the probability, p_{ca}, that the amount of carbon emissions from cars in Belgium for a randomly selected year are between 15.5 Gigagrams/year and 14 Gigagrams/year = The area under the normal curve bounded by the p-values for the two amounts of carbon emission

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p_{ca} = 0.6605 - 0.4931 = 0.1674 ≈ 0.167

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4 years ago
How many milligrams of magnesium should you take a day?.
nevsk [136]

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Explanation:

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3 years ago
_____ is a mixture of dead organic material that can be used as fertilizer
AleksAgata [21]

The answer is compost. It is an organic matter that has been disintegrated and cast-off as fertilizer and soil alteration. This is a significant component in organic farming. The procedure of composting necessitates creating a mound of wet organic substance identified as green wastes – these are leaves, food waste and waiting for the resources to collapse into humus after weeks or months.

7 0
4 years ago
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