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natima [27]
1 year ago
12

What mass of cu(s)cu(s) is electroplated by running 14. 5 aa of current through a cu2 (aq)cu2 (aq) solution for 4. 00 hh?

Chemistry
2 answers:
AfilCa [17]1 year ago
7 0

Mass of copper would be=92.46 g

I = Current = 19.5 A

t = 4 hours =4×60×60=14400 s

F = Faraday constant = 96485.33 C/mol

Molar mass of copper = 63.546 g/mol

A charge is given by

Q=19.5×14400=280880 C

Moles of electrons are given by Q/F=280880/96485.33=2.91 mol

Moles of copper is=1/2×2.91=1.455 mol

Mass of copper would be=1.455×63.546=92.46 g

Learn more about mass here brainly.com/question/19600926

#4206.

galina1969 [7]1 year ago
4 0

The mass of Copper electroplated is 68.76 g

<h3>What is electroplating?</h3>

The process of plating a metal onto another is known as electroplating.

It is often used to prevent corrosion of metal or for the decorative purposes

In this process, electric current is passed through an aqueous solution containing dissolved cations.

The dissolved cations are reduced developing a thin metal coating on the electrode.

At cathode,

Cu^{2+}(aq) + 2e^-\rightarrow      Cu(s)

Current, I = 14.5 A

Time, t = 4 hrs = 4×60×60 = 14400 sec

Charge, q = It = 14.5×14400= 208800 C

Copper metal deposited by 2×96487 C = 63.55 g

Copper metal deposited by 208800 C = \frac{63.55 \times208800}{2\times96487}

                                                         = 68.76g

Hence, The mass of Copper electroplated is 68.76 g

Learn more about electroplating:

brainly.com/question/22104403

#SPJ4

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A 25.0 L tank of nitrogen gas is at 25 oC and 2.05 atm . If the temperature stays at 25 oC and the volume is decreased to 14.5 L
NeTakaya

Answer:

\boxed {\boxed {\sf P_2 \approx 3.53 \ atm}}

Explanation:

In this problem, the temperature stays constant. The volume and pressure change, so we use Boyle's Law. This states that the pressure of a gas is inversely proportional to the volume. The formula is:

P_1V_1=P_2V_2

Now we can substitute any known values into the formula.

Originally, the gas has a volume of 25.0 liters and a pressure of 2.05 atmospheres.

25.0 \ L * 2.05 \ atm = P_2V_2

The volume is decreased to 14.5 liters, but the pressure is unknown.

25.0 \ L * 2.05 \ atm = P_2 * 14.5 \ L

Since we are solving for the new pressure, or P₂, we must isolate the variable. It is being multiplied by 14.5 liters and the inverse of multiplication is division. Divide both sides by 14.5 L .

\frac {25.0 \ L * 2.05 \ atm }{14.5 \ L}=\frac{P_2 *14.5 \ L}{14.5 \ L}

\frac {25.0 \ L * 2.05 \ atm }{14.5 \ L}= P_2

The units of liters cancel.

\frac {25.0  * 2.05 \ atm }{14.5 }=P_2

\frac {50.25\  atm }{14.5 }=P_2

3.53448276 \ atm = P_2

The original values of volume and pressure have 3 significant figures, so our answer must have the same.

For the number we found, that is the hundredth place.

  • 3.53448276

The 4 in the thousandth place (in bold above) tells us to leave the 3 in the hundredth place.

3.53 \ atm \approx P_2

The new pressure is approximately <u>3.53 atmospheres.</u>

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