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.
Answer:
C
Explanation:
because valence electrons are located at the last energy level
Explanation:
Half life is simply the amount of time it takes for half of a substance to decompose.
Options;
- Carbon-14 has a half-life of 5,730 years. A 30 gram sample will be 10 grams after 5,730 years. This is incorrect. After 5730 years, 15g of the sample ought to remain.
- Nickel-59 has a half-life of 76,000 years. A sample would go through 3 half-lives in 228,000 years. This is correct. 3 * 76000 = 228,000
- Hafnium-182 has a half-life of 9 million years. A 38 gram sample would be 4.75 grams in 27 million years. This is incorrect. Mass after 3 half lives (27/9) = 9.5 (38 / 2 / 2)
- Iron-60 has a half-life of 1.5 million years. In 6 million years a 40 gram sample would be reduced to 10 grams. This is incorrect. Mass after 4 half lives (6 / 1.5) = 2.5 gram (40 / 2 / 2 /2 / 2)
- Lead-202 has a half-life of 52,500 years. The original sample must have been 120 grams if you have a 60 gram sample after 105,000 years. This is incorrect. Original sampe = 240 gram. So after 2 half lives (105,000/52500), mass left = 60 (240 / 2 /2)
d. When aluminum-28 undergoes beta decay, silicon-28 is produced.
Explanation:
When the nuclei of aluminium-28 decays, it produces silicon- 28:
Aluminium ²⁸₁₃Al
Silicon 28 ²⁸₁₄Si
beta particle ⁰₋₁
²⁸₁₃Al → ²⁸₁₄Si + ⁰₋₁
This way, the mass and atomic number are conserved.
Conservation of mass number:
28 = 28 + 0, 28 = 28
13 = 14 -1 , 13 = 13
Learn more:
Balancing nuclear equations brainly.com/question/10094982
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