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Paraphin [41]
3 years ago
14

What mass of silver can be plated onto an object in 33.5 minutes at 8.70 a of current? ag (aq e- ? ag(s what mass of silver can

be plated onto an object in 33.5 minutes at 8.70 a of current? ag (aq e- ? ag(s 9.78 g 0.102 g 3.07 g 0.326 g 19.6 g?
Physics
2 answers:
Allisa [31]3 years ago
5 0

Answer:

Last option 19.6 g

Explanation:

In order to do this, we need to apply the Faraday law which is the following:

q = I*t (1)

Where:

I: current in Amperius

t: time in seconds

q: charge in Coulombs

the time in seconds is:

t = 33.5 min * 60 s/min = 2,010 s

Replacing in (1):

q = 8.7 * 2,010 = 17,487 C

With this value, we need to calculate the moles of Ag. This can be done dividing this value with the respective charge in one mole:

moles = q/n

n: value of 1 mole of electron / C = 96500 C/mol e

Replacing we have:

moles = 17.487 / 96,500

moles = 0,1812 moles e

Remember now that each mole of Ag+ can only catch one electron at a time, so, 0.1812 moles e, would be 0.1812 moles.

Finally, we need the molecular mass of Ag, which is 107.87 g/mol so:

m = 0.1812 * 107.87

m = 19.55 g or 19.6 g

Schach [20]3 years ago
4 0
To determine the mass plated, we use Faraday's Law of Electrolysis. We calculate as follows:

q = It
q = 8.70 (33.5) (60)
q = 17487 C

mass = 17487 C ( 1 mol e- / 96500 C) ( 1 mol / 2 mol e-) (107.9 g /mol)
mass = 9.78 g

Hope this helps.

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well it would be A because 55 degrees is going strait well 75 is going literally straight up

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3 years ago
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A flat uniform circular disk (r= 2.00m,
dusya [7]

Incomplete question.The Complete question is here

A flat uniform circular disk (radius = 2.00 m, mass = 1.00 ✕ 102 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a friction less axis perpendicular to the center of the disk. A 40.0-kg person, standing 1.25 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 2.00 m/s relative to the ground.

a.) Find the resulting angular speed of the disk (in rad/s) and describe the direction of the rotation.

b.) Determine the time it takes for a spot marking the starting point to pass again beneath the runner's feet.

Answer:

(a)ω = 1 rad/s

(b)t = 2.41 s

Explanation:

(a) initial angular momentum = final angular momentum  

0 = L for disk + L............... for runner

0 = Iω² - mv²r ...................they're opposite in direction

0 = (MR²/2)(ω²) - mv²r ................where is ω is angular speed which is required in part (a) of question

0 = [(1.00×10²kg)(2.00 m)² / 2](ω²) - (40.0 kg)(2.00 m/s)²(1.25 m)

0=200ω²-200

200=200ω²

ω = 1 rad/s

b.)

lets assume the "starting point" is a point marked on the disk.

The person's angular speed is  

v/r = (2.00 m/s) / (1.25 m) = 1.6 rad/s

As the person and the disk are moving in opposite directions, the person will run part of a revolution and the turning disk would complete the whole revolution.

(angle) + (angle disk turns) = 2π

(1.6 rad/s)(t) + ωt = 2π

t[1.6 rad/s + 1 rad/s] = 2π

t = 2.41 s

6 0
3 years ago
Which of the following has the greatest kinetic energy? A. a mass of 4m at velocity v. B. a mass of 3m at velocity 2v. C. a mass
Artist 52 [7]

Answer:

Option C

Explanation:

Kinetic energy is the energy that the body possesses by virtue of its motion.

The formula for Kinetic energy is given by \frac{1}{2} mv^2

Using this formula let us find kinetic energy for the bodies given and find out which is the greatest

A) KE = \frac{1}{2} (4m)(v^2) = 2mv^2

B) KE =\frac{1}{2} (3m)(2v)^2 = 6mv^2

C) KE = \frac{1}{2} (2m)(3v)^2 = 9mv^2

D) KE = \frac{1}{2} (3)(4v)^2 = 8mv^2

Comparing these we find that 9mv^2 is the highest.

Hence option C is the answer.

4 0
3 years ago
8. What is the frequency of green light waves that have a wavelength of 5.2 x 10-7 m.? The speed of light is 3.0 x 108 m/s
o-na [289]

Answer:

f=5.76\times 10^{14}\ Hz

Explanation:

We need to find the frequency of green light having wavelength o5.2\times 10^{-7}\ m. It can be calculated as follows :

c=f\lambda\\\\f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{5.2\times 10^{-7}}\\\\f=5.76\times 10^{14}\ Hz

So, the required frequency of green light is equal to 5.76\times 10^{14}\ Hz.

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madreJ [45]
Gamma rays have the highest energies and the shortest wavelengths.
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