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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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Answer:

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Explanation:

You are most aware when the vehicle is accelerating. At constant velocity you would not be aware of the motion. Only if the system is accelerated the dynamics must be solved considering a pseudo-force (of inertial origin) acting.

It's because of this that:

(A) False. The acceleration can be detected from the inside of a closed car.

(B) False. You would be aware of the motion, but not because humans can sense speed but acceleration.

(C) False. Constant velocity cannot be felt in a closed car.

(D) False. Again, you can't feel constant speed.

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A Smart Car, which has a mass of 1000 kg, is going 20 m/s. When it hits the barrier, it stops with a time of 0.5 seconds. What i
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Explanation:

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Suppose you increase your walking speed from 4 m/s to 13 m/s in a period of 3 s. What is your acceleration?
iogann1982 [59]

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3 years ago
What is the volume of a cone with a height of 27 cm
JulijaS [17]

Explanation:

→ Volume of cone = πr² × h/3

Here,

  • Radius (r) = 13 cm
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→ Volume of cone = π(13)² × 27/3 cm³

→ Volume of cone = 169π × 9 cm³

→ Volume of cone = 1521π cm³

→ Volume of cone = 1521 × 22/7 cm³

→ Volume of cone = 33462/7 cm³

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A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit
Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

P = \frac{1}{2} \mu \omega^2 A^2 v

Here,

\mu = Linear mass density of the string

\omega =  Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

v = \sqrt{\frac{T}{\mu}} \rightarrow Here T is the Period

For the linear mass density we have that

\mu = \frac{m}{l}

And the angular frequency can be written as

\omega = 2\pi f

Replacing this terms and the first equation we have that

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})

P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})

PART A ) Replacing our values here we have that

P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

P = 0.2320W

PART B) The new amplitude A' that is half ot the wavelength of the wave is

A' = \frac{1.8*10^{-3}}{2}

A' = 0.9*10^{-3}

Replacing at the equation of power we have that

P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

P = 0.058W

8 0
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