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WINSTONCH [101]
1 year ago
10

Please Help Me!!

Chemistry
1 answer:
Cerrena [4.2K]1 year ago
3 0

Answer:

1. Galvanic oxidation. Example is the corrosion of aluminium wires when in contact with copper wires under wet conditions.

2. Rainwater or Damp/moist air

3. Chromium-plated steel screws or stainless steel screws or galvanized steel screws

Explanation:

1. Galvanic oxidation or corrosion occurs when two different metals with different electrode potentials are brought into contact with each other by means of an electrolyte (usually a aqueous solution), such that a redox reaction occurs leading to one metal with the more negative electrode potential (the anode) becoming oxidized, while the other less negative potential (the cathode) is reduced.

In order for galvanic corrosion to occur, three elements are required.

i. Two metals with different corrosion potentials (anode and cathode)

ii. Direct metal-to-metal electrical contact

iii. A conductive electrolyte solution (e.g. water) must connect the two metals on a regular basis.

For example oxidation (corrosion) of aluminium wires when in contact with copper wire under wet conditions.

2. The most likely electrolyte will be rainwater containing dissoved solutes (if the panel is in an exposed part of the house) or damp/moist air.

3. From the table, the most likely screw will be chromium-plated steel screws or stainless steel (made of iron and nickel) screws or galvanized steel (zinc-plated) screws.

All these possible screw components have a more negative electrode potential than copper. Thus they will serve as the anode in a galvanic oxidation with copper.

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The answer:
<span>The equation of its dissolution in water is: AgNO3 → Ag + (aq)  +  NO3- (aq)

and     </span>AgNO3 →   Ag + (aq)  +  NO3- (aq)
          1 mol          1mol                1mol 
             ?  --------   0.854mo
so for finding the value, it is sufficients to complute 1 x 0.854 mol =0.854 mol
so,  0.854 mol is required for the reaction to form 0.854 mol of Ag

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A 3.5 kg iron shovel is left outside through the winter. The shovel, now orange with rust, is rediscovered in the spring. Its ma
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Answer:

a

Explanation:

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1 year ago
The solubility of acetanilide in hot water (5.5 g/100 ml at 100 C) is not very great, and it has some solubility in cold water (
sashaice [31]

To answer the problem given:

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= 3.47 / 4.0 * 100%

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The maximum theoretical percent recovery from the crystallization of 4.0 g of acetanilide from 100 ml of water is 87%. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

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A 12.0 g sample of a metal is heated to 90.0 ◦C. It is then dropped into 25.0 g of water. The temperature of the water rises fro
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Answer:

The specific heat of the metal is 0.335 J/g°C

Explanation:

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

Mass of the metal = 12.0 grams

Initial temperature of the metal = 90.0 °C

Mass of the water = 25.0 grams

Initial temperature of water = 22.5 °C

Final temperature of water (and metal) = 25.0 °C

The specific heat of water = 4.18 J/g°C

<u>Step 2:</u> Calculate the specific heat of the metal

Qgained  = -Qlost

Qwater = -Qmetal

Q= m*c*ΔT

m(metal) *c(metal)*ΔT(metal) = -m(water)*c(water)*ΔT(water)

⇒ mass of the metal = 12.0 grams

⇒ c(metal) = TO BE DETERMINED

⇒ ΔT(metal) = T2 - T1 = 25.0 - 90.0 °C = -65.0

⇒ mass of the water = 25.0 grams

⇒ c(water) = the specific heat of water = 4.18 J/g°C

⇒ ΔT(water) = T2 - T1 = 25.0 - 22.5 = 2.5°C

12.0 * c(metal) * -65.0 = -25.0 * 4.18 * 2.5

c(metal) = 0.335 J/g°C

The specific heat of the metal is 0.335 J/g°C

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