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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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3 years ago
A 15 kg box is sliding down an incline of 35 degrees. The incline has a coefficient of friction of 0.25. If the box starts at re
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The box has 3 forces acting on it:

• its own weight (magnitude <em>w</em>, pointing downward)

• the normal force of the incline on the box (mag. <em>n</em>, pointing upward perpendicular to the incline)

• friction (mag. <em>f</em>, opposing the box's slide down the incline and parallel to the incline)

Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have

• net parallel force:

∑ <em>F</em> = -<em>f</em> + <em>w</em> sin(35°) = <em>m a</em>

• net perpendicular force:

∑<em> F</em> = <em>n</em> - <em>w</em> cos(35°) = 0

Solve the net perpendicular force equation for the normal force:

<em>n</em> = <em>w</em> cos(35°)

<em>n</em> = (15 kg) (9.8 m/s²) cos(35°)

<em>n</em> ≈ 120 N

Solve for the mag. of friction:

<em>f</em> = <em>µ</em> <em>n</em>

<em>f</em> = 0.25 (120 N)

<em>f</em> ≈ 30 N

Solve the net parallel force equation for the acceleration:

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<em>a</em> ≈ 3.6 m/s²

Now solve for the block's speed <em>v</em> given that it starts at rest, with <em>v</em>₀ = 0, and slides down the incline a distance of ∆<em>x</em> = 3 m:

<em>v</em>² - <em>v</em>₀² = 2 <em>a</em> ∆<em>x</em>

<em>v</em>² = 2 (3.6 m/s²) (3 m)

<em>v</em> = √(21.7263 m²/s²)

<em>v</em> ≈ 4.7 m/s

4 0
2 years ago
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expeople1 [14]
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<h2>Answer:</h2>

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<h2>Explanation:</h2>

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