Suppose of electrons must be transported from one side of an electrochemical cell to another in minutes. Calculate the size of electric current that must flow. Be sure your answer has the correct unit symbol and round your answer to significant digits.

I = 104.54 A

Explanation:

Suppose 650 mmol of electrons must be transported from one side of an electro chemical cell to another in 10.0 minutes Then, we have to calculate the size of electric current that must flow.

Then we know that

Current I = Charge(q) / time {t}

Charge = mols of e- × charge of electron

= (0.650 mol of e- ) × (96500 C/mol of e- ) = 62725 C

now, time in seconds t = 10 min = 10 min × 60 s / min = 600 s

therefore, I = C/t = (62725)/600

I = 104.54 C/s

Also,  1 C/s = 1 Amp

Hence, I = 104.54 A

some sources quote the lethal dose, for an average 170lbs opiate-naive individual is between 75 and 375 mg