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WINSTONCH [101]
2 years ago
10

Please Help Me!!

Chemistry
1 answer:
Cerrena [4.2K]2 years 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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A 2.550 x 10^−2 M solution of glycerol (C3H8O3) in water is at 20.0°C. The sample was created by dissolving a sample of C3H8O3 i
BARSIC [14]

Answer:

2.557\times 10^{-2}\ molal.

Explanation:

Given , molarity of glycerol= 2.55\times 10^-^2\ M.

Volume= 1 L.

Therefore, No of moles of glycerol= molarity\times volume\ in\ liters=2.55\times 10^-^2\ moles.

Now, volume of water needed, V=998.8 mL.

Density is given as= 0.9982 g/mL.

Therefore, mass of water = volume\times density=998.8\times 0.9982=997\ g=0.997\ kg

Now, molality=\dfrac{no\ of\ moles}{mass\ of\ solvent\ in\ kg}=\dfrac{2.55\times 10^-^2\ }{0.997}=0.02557\ molal=2.557\times 10^{-2}\ molal.

Hence, this is the required solution.

3 0
1 year ago
How quickly can bacterial contamination occur?
NeX [460]
5-10 seconds xxx :) hope this helps
5 0
2 years ago
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The specific heat of aluminum is 0.214 cal/g.oC. Determine the energy, in calories, necessary to raise the temperature of a 55.5
Natasha2012 [34]
For this problem, we use the formula for sensible heat which is written below:

Q= mCpΔT
where Q is the energy
Cp is the specific heat capacity
ΔT is the temperature difference

Q = (55.5 g)(<span>0.214 cal/g</span>·°C)(48.6°C- 23°C)
<em>Q = 304.05 cal</em>
4 0
2 years ago
A different student is given a 10.0g sample labeled CaBr2 that may contain an inert (nonreacting) impurity. Identify a quantity
grandymaker [24]

Answer:

Mass of Ca in sample, Mass of Br in sample, Number of moles of Ca in sample, Number of moles of Br in sample, Mass or moles of element other than Ca or Br in sample

Explanation:

The AP Classroom will not count your answer to this question as correct unless it includes at least one of the answers listed above. If you say that theanswer to this question is density, it will be marked as incorrect, I found that out the hard way when I used the answers that brainly gave me.

Good luck,

I applaud you for using the sources avalible to you, which is /definetly not/ cheeting.

8 0
2 years ago
How much heat must be removed from 25.0g of steam at 118.0C in order to form ice at 15C
NemiM [27]

Answer:

-10778.95 J heat must be removed in order to form the ice at 15 °C.

Explanation:

Given data:

mass of steam = 25 g

Initial temperature = 118 °C

Final temperature = 15 °C

Heat released = ?

Solution:

Formula:

q = m . c . ΔT

we know that specific heat of water is 4.186 J/g.°C

ΔT = final temperature - initial temperature

ΔT = 15 °C - 118 °C

ΔT = -103 °C

now we will put the values in formula

q = m . c . ΔT

q = 25 g × 4.186 J/g.°C × -103 °C

q = -10778.95 J

so, -10778.95 J heat must be removed in order to form the ice at 15 °C.

3 0
2 years ago
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