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Kaylis [27]
3 years ago
15

Assume that Aluminum and Silver Sulfide are the starting substances (reactants) in the reaction: a. Write a balanced chemical eq

uation describing the "re-creation" of silver, using the information in the case study. b. State the names of the products that are produced from this reaction. c. What type of reaction(s) is/are being represented by the chemical reaction you wrote in part (a)? d. Is the reaction in part (a) an oxidation-reduction (redox) reaction? e. If this is a redox reaction, then identify the following: What is undergoing oxidation (what is being oxidized)? Support your answer using oxidation numbers. What is undergoing reduction (what is being reduced)? Support your answer using oxidation numbers. What is the reducing agent? What is the oxidizing agent?
Chemistry
1 answer:
Naddik [55]3 years ago
6 0

Answer:

Check Explanation

Explanation:

a) The balanced chemical reaction between Aluminium and Silver Sulfide is represented below

2Al + 3Ag₂S → Al₂S₃ + 6Ag

Aluminium displaces Silver from Silver sulfide because it is higher than Silver in the electrochemical series.

b) Names of the products from this reaction

Ag is called Silver metal. Free Silver metal.

Al₂S₃ is called Aluminium Sulfide.

c) This reaction is a single-displacement reaction because an element directly displaces and replaces another element in a compound.

It is also a redox reaction (reduction-oxidation reaction) because there are species being oxidized and reduced simultaneously!

d) Yes, this reaction is an oxidation-reduction (redox) reaction because there are species being oxidized and reduced simultaneously!

e) The specie that is being oxidized is said to undergo oxidation. And oxidation is defined as the loss of electrons, thereby leading to an increase in oxidation number.

In this reaction, it is evident that Aluminium undergoes oxidation as its oxidation number increases from 0 in the free state to +3 when it displaces Silver and becomes Aluminium sulfide.

Al → Al³⁺

0 → +3 (Oxidation)

The specie that is being reduced is said to undergo reduction. And reduction is defined as the gain of electrons, thereby leading to a decrease in oxidation number.

In this reaction, it is evident that the silver ion undergoes reduction as its oxidation number decreases from +1 in the Silver Sulfide compound to 0 when it is displaced and becomes Silver in free state.

Ag⁺ → Ag

+1 → 0 (Reduction)

The reducing agent is the specie that brings about reduction. Since Silver ion in Silver sulfide is reduced, Aluminium is the reducing agent that initiates the reduction process.

The oxidation agent is the specie that brings about oxidation. Since, Aluminium is the specie that undergoes oxidation, the oxidizing agent is the Silver Sulfide that brings about the oxidation.

Hope this Helps!!!

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How many grams of Fe3O4 are required to react completely with 300 grams of H2?
vekshin1

Answer:

2023.04 g

Explanation:

Magnetite reacts with hydrogen to produce Iron metal and steam. Steam instead of water is produced as the reaction occurs at temperatures above the boiling point of water.

Fe₃O₄ + 4 H₂ → 3 Fe +4 H₂O

From the equation, 1 mole of Fe₃O₄ reacts with 4 moles of H₂.

69.76 grams of H₂ has the following number of moles.

Number of moles= mass/RAM

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The reaction ratio of Fe₃O₄ to H₂ is 1:4

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write the symbols for the ions that form when potassium and iodine react to form the ionic compound potassium iodide.
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I believe it isK+  + I-=KI
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In an experiment to study the photoelectric effect, a scientist measures the kinetic energy of ejected electrons as afunction of
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Answer:

a) v₀ = 4.41 × 10¹⁴ s⁻¹

b) W₀ = 176 KJ/mol of ejected electrons

c) From the graph, light of frequency less than v₀ will not cause electrons to break free from the surface of the metal. Electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.

d) When frequency of the light exceeds v₀, there is an increase of electron kinetic energy from zero steadily upwards with a constant slope. This is because, once light frequency exceeds, v₀, its energy too exceeds the work function of the metal and the electrons instantaneously gain the energy of incident light and convert this energy to kinetic energy by breaking free and going into motion. The energy keeps increasing as the energy and frequency of incident light increases and electrons gain more speed.

e) The slope of the line segment gives the Planck's constant. Explanation is in the section below.

Explanation:

The plot for this question which is attached to this solution has Electron kinetic energy on the y-axis and frequency of incident light on the x-axis.

a) Wavelength, λ = 680 nm = 680 × 10⁻⁹ m

Speed of light = 3 × 10⁸ m/s

The frequency of the light, v₀ = ?

Frequency = speed of light/wavelength

v₀ = (3 × 10⁸)/(680 × 10⁻⁹) = 4.41 × 10¹⁴ s⁻¹

b) Work function, W₀ = energy of the light photons with the wavelength of v₀ = E = hv₀

h = Planck's constant = 6.63 × 10⁻³⁴ J.s

E = 6.63 × 10⁻³⁴ × 4.41 × 10¹⁴ = 2.92 × 10⁻¹⁹J

E in J/mol of ejected electrons

Ecalculated × Avogadros constant

= 2.92 × 10⁻¹⁹ × 6.023 × 10²³

= 1.76 × 10⁵ J/mol of ejected electrons = 176 KJ/mol of ejected electrons

c) Light of frequency less than v₀ does not possess enough energy to cause electrons to break free from the metal surface. The energy of light with frequency less than v₀ is less than the work function of the metal (which is the minimum amount of energy of light required to excite electrons on metal surface enough to break free).

As evident from the graph, electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.

d) When frequency of the light exceeds v₀, there is an increase of electron kinetic energy from zero steadily upwards with a constant slope. This is because, once light frequency exceeds, v₀, its energy too exceeds the work function of the metal and the electrons instantaneously gain the energy of incident light and convert this energy to kinetic energy by breaking free and going into motion. The energy keeps increasing as the energy and frequency of incident light increases and electrons gain more speed.

e) The slope of the line segment gives the Planck's constant. From the mathematical relationship, E = hv₀,

And the slope of the line segment is Energy of ejected electrons/frequency of incident light, E/v₀, which adequately matches the Planck's constant, h = 6.63 × 10⁻³⁴ J.s

Hope this Helps!!!

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