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

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.