need help figuring out number 5 and also the second trial for number 4. if you c
ID: 1045959 • Letter: N
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need help figuring out number 5 and also the second trial for number 4. if you can't see the number in the first box in trial one it's 5.8532. please help with those I haven't done and also make sure the rest is done right
DATA TABLE Trial 1 Trial 2 Initial mass of copper electrodes(g) ( Final mass of copper electrodes (g) 6,822 Average current (A) Time of current application (s) 5.8220e 5.793% 0.4995 A 0, 4992 A 1so seconds SO Seconds DATA ANALYSIS 1 Caiculate the tota? charge, in-coulombs (C) that passed through the electrolytic cell- for each-trial. 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 101 coulombs per electron. 2. Trial 2-sul 2 x ID-18 3. 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) Thal 12.301 x Io0 ?? Tnal 2Q.soe x IO 4. 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) Tnal a- Calculate the number of copper atoms in a mole of copper for each trial. Compare this value to Avogadro's number 5 48 Advanced Chemistry with VerniecExplanation / Answer
Q4) trial 2)
mass of Cu electrode lost=5.8226g-5.7938g=0.0288g
Cu2+ +2e --->Cu(s)
number of Cu in the electrolysis=2*number of electrons in the electrolysis=2*Charge/charge of 1 electron
Also charge=current*time =0.4995 C/s*180s=89.91 C
number of Cu in the electrolysis=2*89.91 C/(1.602*10^-19C/e)=1.122*10^21 Cu atoms
number of Cu atoms per gram=( 1.122*10^21atoms )/0.0288g)=3.896*10^22 Cu atoms /gram lost
5)trial 1) mass of Cu lost =5.8532g-5.8226g=0.0306g
mol of Cu atoms lost=mass/molar mass of Cu=0.0306g/(63.546g/mol)=0.000481 mol
number of Cu atoms lost=Avogadro's number*mol of Cu=NA*0.000481 mol
But acording to calculations,
number of Cu in the electrolysis=1.122*10^21 Cu atoms
So,NA*0.000481 mol=1.122*10^21 Cu atoms
NA=1.122*10^21 Cu atoms/0.000481mol=2.332*10^24 >>Actual Avogadro's number=1.066*10^23 atoms/mol
) trial 2)
mass of Cu electrode lost=5.8226g-5.7938g=0.0288g
mol of Cu atoms lost=mass/molar mass of Cu=0.0288g/(63.546g/mol)=0.000453 mol
number of Cu atoms lost=Avogadro's number*mol of Cu=NA*0.000453 mol
But acording to calculations,
number of Cu in the electrolysis=2* number of electrons=2*5.612*10^18 Cu atoms
So,NA*0.000453 mol=2*5.612*10^18 Cu atoms
NA=2*5.612*10^18 Cu atoms/0.000453mol=2.478*10^22<<Actual Avogadro's number=1.066*10^23 atoms/mol
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