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arsen [322]
3 years ago
15

A galvanic (voltaic) cell has the generic metals X and Y as electrodes. X is more reactive than Y , that is, X more readily reac

ts to form a cation than Y does. Classify the descriptions by whether they apply to the X or Y electrode.
Chemistry
1 answer:
Margaret [11]3 years ago
7 0

Answer:

(1) cathode: Y

(2) anode X

(3) electrons in the wire flow toward: Y

(4) electrons in the wire flow away from: X

(5) anions from the salt bridge flow toward X

(6) cations from the salt bridge flow toward Y

(7) gains mass: Y

(8) looses mass X

Explanation:

The voltaic cell uses two different metal electrodes, each in an electrolyte solution. The anode will undergo oxidation and the cathode will undergo reduction. The metal of the anode will oxidize, going from an oxidation state of 0 (in the solid form) to a positive oxidation state, and it will become an ion. At the cathode, the metal ion in the solution will accept one or more electrons from the cathode, and the ion’s oxidation state will reduce to 0. This forms a solid metal that deposits on the cathode. The two electrodes must be electrically connected to each other, allowing for a flow of electrons that leave the metal of the anode and flow through this connection to the ions at the surface of the cathode. This flow of electrons is an electrical current that can be used to do work, such as turn a motor or power a light.

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If 252 grams of iron are reacted with 321 grams of chlorine gas, what is the mass of the excess reactant leftover after the reac
mario62 [17]

Answer:

Iron is in excess.

1) The mass of the iron remaining = 83.38 grams

2) Ethane is in excess. There will remain 90.06 grams ethane

Explanation:

Step 1: Data given

Mass of iron = 252 grams

Mass of Cl2 = 321 grams

Molar mass of Fe = 55.845

Molar mass of Cl2 = 70.9 g/mol

Step 2: The balanced equation

2Fe(s)+3Cl2(g)⟶2FeCl3(s)

Step 3: Calculate moles

Moles = mass / molar mass

Moles Fe = 252.0 grams / 55.845 g/mol = 4.512 moles

Moles Cl2 = 321.0 grams / 70.90 g/mol = 4.528 moles

Step 4: Calculate the limiting reactant

For 2 moles Fe we need 3 moles Cl2 to produce 2 moles Fecl3

Cl2 is the limiting reactant. It will completely be consumed (4.528 moles).

Fe is in excess. There will 4.528 * 2/3 = 3.019 moles be consumed

There will remain 4.512 - 3.019 = 1.493 moles of Fe

The mass of the iron remaining = 1.493 * 55.845 g/mol =83.38 grams

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If 152 grams of ethane (C2H6) are reacted with 231 grams of oxygen gas, what is the excess reactant?

Step 1: Data given

Mass of ethane = 152.0 grams

mass of O2 =231.0 grams

Molar mass of ethane = 30.07 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

2C2H6(g) + 7O2(g) ⟶ 4CO2(g) +  6H2O(g)

Step 3: Calculate moles

Moles = mass / molar mass

Moles ethane = 152.0 grams / 30.07 g/mol = 5.055 moles

Moles O2 = 231.0 grams / 32.0 g/mol = 7.22 moles

Step 4: Calculate limiting reactant

For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed (7.22 moles).

Ethane is in excess. There will react 7.22 * 2/7 = 2.06 moles

There will remain 5.055 - 2.06 = 2.995 moles ethane

2.995 moles ethane = 2.995 * 30.07 g/mol = 90.06 grams ethane

8 0
4 years ago
Read 2 more answers
Draw the structure of the major organic product isolated from the reaction of 3-hexyne with hydrogen (1 mol), Lindlar palladium.
ohaa [14]

Answer:

See explanation and image attached

Explanation:

When 1 mole of hydrogen is added to 3-hexyne, the addition occurs on the same face or side of the triple bond to yield cis-2-hexene in the presence of lindlar catalyst which prevents further hydrogenation. This is known as syn addition.

The syn addition is achieved because the hydrogen molecule is first adsorbed on the surface of the palladium metal and is subsequently attached to the same face of the triple bond in 3-hexyne as shown in the image attached to this answer.

6 0
3 years ago
This graph represents a population of molecules in a gas versus the distribution of the average velocity(speed) of its molecules
pshichka [43]

Answer:

Part A

Given that the graph is symmetrical and bell shaped, the average kinetic energy is given by the midline of graph, which corresponds to the common speed of the highest number of the population

Part B

The formula for the average kinetic energy, K.E. = (3/2)·(R/NA)·T

Therefore, the part of the graph that indicates the temperature of the sample is the average kinetic energy. K.E.

Part C

At a lower temperature, the heat is less evenly distributed and we have the distribution T2 higher than T1

Please see the attached graph created with MS Visio

Explanation:

3 0
3 years ago
Based on the reactivities of the elements involved, which reactions will form products that are more stable than the reactants?
myrzilka [38]

2 LiI + Cl₂ → 2 LiCl + I₂

2 LiBr + F₂ → 2 LiF + Br₂

<h3>Explanation</h3>

Each of the five reactions involve one halogen molecule (F₂, Cl₂, Br₂, and I₂) substituting the ion of another halogen (F⁻, Cl⁻, Br⁻, and I⁻).

Halogen atoms are found in group 17 of the periodic table. They are all non-metal elements. Each of the halogen atom will gain one electron to form an ion of charge -1. However, the tendency to do so decreases down the group.

  • F is the first halogen in group 17. It has only two shells of electrons.
  • Cl is right under F. Its electrons occupy three main energy shells.
  • Br follows with four main energy shells.
  • I is under Br and has five main energy shells.

Atoms of all four elements have the same effective nuclear charge of +7. However, F has the smallest radius. As a result, it has the strongest hold on electrons around it. Its ion F⁻ is more stable than ions of Cl, Br, or I. Similarly, its molecule F₂ is more reactive than Cl₂, Br₂, and I₂.

As a result, the stability of halogen molecules increases down the group:

  • Stability: F₂ < Cl₂ < Br₂ < I₂.

The stability of halogen ions decreases down the group:

  • Stability: F⁻ > Cl⁻ > Br⁻ > I⁻.

Cl₂ repaces F⁻ (from LiF) in first reaction. F₂ and Cl⁻ are produced. F₂ is less stable than Cl₂. Cl⁻ is less stable than F⁻.

Cl₂ replaces I⁻ (from LiI) in the second reaction. I₂ and Cl⁻ are produced. I₂ is more stable than Cl₂. Cl⁻ is more stable than I⁻.

Br₂ replaces Cl⁻ (from LiCl) in the third reaction. Cl₂ and Br⁻ are produced. Cl₂ is less stable than Br₂. Br⁻ is less stable than Cl⁻.

F₂ replaces Br⁻ (from LiBr) in the fourth reaction. Br₂ and F⁻ are produced. Br₂ is more stable than F₂. F⁻ is more stable than Br⁻.

I₂ replaces Br⁻ (from LiBr) in the fifth reaction. Br₂ and I⁻ are produced. Br₂ is less stable than I₂. I⁻ is less stable than Br⁻.

5 0
3 years ago
Indicate how the concentration of each species in the chemical equation will change to reestablish equilibrium after changing th
Viktor [21]

Explanation:

Since this is an equilibrium problem, we apply le chatelier principle. This principle states that whenever a system at equilibrium is disturbed due to change in several factors, it would move in a way to annul such change.

C2H4(g) + Cl2 ⇔ 2C2H4Cl2(g)

When the concentration of C2H4 is increased, there is more reactant sin the system. In order to annul this change, the equilibrium position will shift to the right favoring product formation.

When the concentration of C2H4Cl2 is increased, there is more product in the system. To annul this change, the equilibrium position will shift to the left, favoring reactant formation.

7 0
4 years ago
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