the number in trial 1 for initial mass is 5.8532 g in case it\'s hard to read. n

Question

the number in trial 1 for initial mass is 5.8532 g in case it's hard to read. need all questions answered clearly and numbered please thank you

DATA TABLE Trial 1 Trial 2 Initial mass of copper electrodes(g) Final mass of copper electrodes (g) Average current (A) 5.8220e 5.793% 0.4995 | 6 . 8 2210 o.4992 A time of curent application ts)130 seconds3o Secouds SO SecoNdS DATA ANALYSIS 1. Caiculate the total charge, in coulombs (C) that passed through the electrolytic cell-for each-trial. 2. Use your answers from 1 above to calculate the number of electrons in the electrolysis for each trial. Recall from the famous Millikan oil-drop experiment that the charge of an electron is 1.602 x 10-19 coulombs per electron. Determine the number of copper atoms lost from the anode in each trial. Remember that the electrolysis process uses two electrons to produce one copper ion (Cu") 3 Calculate the number of copper atoms per gram of copper lost at the anode for each trial. The mass lost at the anode is equal to both the mass of copper atoms lost and the mass of copper ions produced (the mass of the electrons is negligible) 4. 5 Calculate the number of copper atoms in a mole of copper for each trial Compare this value to Avogadro's number 48 Advanced Chemistry with Vernier

Explanation / Answer

For trial 1

1-

Total charge passed = 0.4992 A x 180 s

= 89.856 Coulomb

2 -

Number of electrons

=( 89.856 Coulomb) /(1.602*10^-19 Coulomb /electrons)

= 5.608 x 10^20 electrons

3-

2 electrons produces = 1 copper ion

5.608 x 10^20 electrons produces = 5.608 x 10^20 /2

= 2.804 x 10^20 copper ions

Number of copper atoms = 2.804 x 10^20

4-

Number of copper atoms/g of Cu lost =

= 2.804 x 10^20/(6.8226 - 5.8532)

= 2.893 x 10^20

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