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

it is to be noted that an alpha particle is an helium nucleus. so it contains 2 protons 2 neutrons and has a mass number of 4. when alpha decay particles are ejected from the nucleus it gets transformed into an atom which has atomic number less than 2 and a mass number less than 4.

an equation that could represent alpha decay is radium 226 decay of alpha particle forms radon 222.

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chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.