The given question is incomplete. The complete question is:
Suppose a current of 0.920 A is passed through an electroplating cell with an aqueous solution of agno3 in the cathode compartment for 47.0 seconds. Calculate the mass of pure silver deposited on a metal object made into the cathode of the cell.
Answer: 0.0484 g
Explanation:
where Q= quantity of electricity in coloumbs
I = current in amperes = 0.920 A
t= time in seconds = 47.0 sec
96500 Coloumb of electricity electrolyzes 1 mole of Ag
43.24 C of electricity deposits =
of Ag
Thus the mass of pure silver deposited on a metal object made into the cathode of the cell is 0.0484 g
The answer is- Part washers use cleaning solutions that eventually become spent and must be disposed of properly.
Cleaning solutions: Cleaning solution is a liquid solvent or solution used to clean the working surfaces and the parts of a machine.
What type of cleaning solution is Part washers?
- Parts washers are mechanical devices that are designed to be cleaned to remove debris, dirt, oil, paint, and other substances that could potentially contaminate parts to prepare them for assembly, packaging, or even painting.
- They are basically used to clean parts to make them ready for functional use.
- Cleaning solutions for washing used parts are considered a special waste because they can be hazardous waste and are waste from an industrial process.
- Used solvents are almost always hazardous waste. Both solvent and aqueous parts washers produce a sludge that is usually hazardous because it contains toxic metals and solvents from the parts being cleaned.
- Absorbents used to wipe parts after washing are also dangerous if they contain toxic metals in concentrations exceeding regulatory limits or listed hazardous solvents, and used oil may also contain hazardous waste.
- Thus, these cleaning solutions must be disposed of properly.
To learn more about the cleaning solutions, visit:
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We can calculate years by using the half-life equation. It is expressed as:
A = Ao e^-kt
<span>where A is the amount left at t years, Ao is the initial concentration, and k is a constant.
</span>From the half-life data, we can calculate for k.
1/2(Ao) = Ao e^-k(1620)
<span>k = 4.28 x 10^-4
</span>
0.125 = 1 e^-<span>4.28 x 10^-4 (</span>t)
t = 4259 years
Yes the answer is 1s22s22p63s23p64s23d5
