Idk 0? I thought when an object is at potential energy it at its lowest point. Sorry guess this doesn't help If I just had the lesson I may be able 2 help.

7 0

3 years ago

How long would it take a current of 10A to deposit 6.36g of copper during the electrolysis of copper (ll) tetraoxosulphate VI so

zhuklara [117]

It take 0.54 hours to deposit 6.36g of copper

<h3>Further explanation</h3>

Faraday's Law I

"The mass of the substance formed at each electrode is proportional to the electric current flowing in the electrolysis

W = e.i.t / 96500

$\tt \large {\boxed {\bold {W \: = \: \dfrac {e \times i \: x \: t} {96500}}}}$

e = equivalent = Ar / valence

i = current, A

t = time, s

W=6.36 g

e = 63.5 : 2 =31.75

i = 10 A

$\tt t=\dfrac{W\times 96500}{e.i}\\\\t=\dfrac{6.36\times 96500}{31.75\times 10 }\\\\t=1933.04~s\approx 0.54~hours$

8 0

2 years ago

Newton’s second law of motion is F=ma A net force of 60 N north acts on an object with a of 30 kg. Use Newton’s second law of mo

horsena [70]

I’m pretty sure the immediate answer is 2. Using the equation 60N=(30kg)x2(whatever unit of measure) also it will change because if you change the mass or the Net Force, your going to have to redo the equation based on the new information.

8 0

3 years ago

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