Suppose of electrons must be transported from one side of an electrochemical cell to another in seconds. 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.

18.03A

Explanation:

(The question is lacking some important details such as quantity of electrons transported, specific time etc

However here's a more detailed question ).

Suppose 550.mmol of electrons must be transported from one side of an electrochemical cell to another in 49.0 minutes. Calculate the size of electric current that must flow.

Be sure your answer has the correct unit symbol and round your answer to 3 significant digits.

Solution

From the above question,

Number of moles = 550mmol = 0.55mol

Time taken = 49 minutes = 2940sec

Q = I*T

Q = charge passing through the body

I = current passing through the body

T = time take for the charge to pass through.

Number of electrons passing through = 0.55mol.

But charge on 1ē = 96500C

Q = 0.55 * 96500

Q = 53075C

Since we already know that Q = I*T

T = 2940sec

I = Q / T

I = 53075 / 2940

I = 18.03A

The current passing through is 18.03A

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :