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Chemistry 77 Online
OpenStudy (anonymous):

c. If a piece of jewelry were electroplated with gold for 25 seconds at 1.5 A, how many grams of gold would be plated? (Note: 1 A = 6.241 1018 e–/s. The reaction for the reduction of gold ions to gold metal is Au+ + 1e– Au(s).) (1 point)

OpenStudy (australopithecus):

Reaction is Au+ + 1e– -> Au(s) Figure out how many electrons are put into the reaction in 25 seconds at 1.5 A, this will equal the amount of Au(s) atoms you produce from atoms using avagadros number you can find moles from moles you can find grams

OpenStudy (australopithecus):

Any questions?

OpenStudy (australopithecus):

If you are having trouble with the math let me know I can help you through it

OpenStudy (anonymous):

here's what i have so far.. 1 A = 6.241 1018 e–/s so…. 1.5(6.241 x 1018 e-/s)25s = number of atoms reduced number of atoms reduced = 2.34 E 20

OpenStudy (anonymous):

so do i multiply that by avogrado's number(6.0221413e+23)

OpenStudy (anonymous):

so 2.34 E 20*6.0221413e+23)

OpenStudy (anonymous):

1.4091811e+24 moles right?

OpenStudy (anonymous):

no wait i did this wrong

OpenStudy (australopithecus):

http://www.wolframalpha.com/input/?i=1.5%286.241+x+10^18+e-%2Fs%2925 Yeah you messed up the first calculation

OpenStudy (anonymous):

i did plug that into my calculator but i must've misread something

OpenStudy (australopithecus):

number of atoms reduced 6.362*10^20

OpenStudy (anonymous):

yep i got that

OpenStudy (anonymous):

and then multiply by avogadro's #?

OpenStudy (anonymous):

dont i need to find the mass of gold?

OpenStudy (australopithecus):

Yes but if you multiply you are going to get an absurd number, Avogadro's number has units, 6.022141*10^23 atom/mol

OpenStudy (australopithecus):

you will end up with atom^2/mol which is not useful

OpenStudy (australopithecus):

as units

OpenStudy (australopithecus):

1.4091811e+24 moles is 3.8*mass of the moon lol that is a lot of gold

OpenStudy (anonymous):

oh wow yea haha

OpenStudy (australopithecus):

always pay attention to number sizes in chemistry and really any science

OpenStudy (anonymous):

i realized i multiplied by 1018 instead of 10^18 it was a typo

OpenStudy (anonymous):

Au+ + 1e– has to be used next...?

OpenStudy (australopithecus):

not really a big deal you can tell the number was in scientific notation

OpenStudy (australopithecus):

well it is a 1 to 1 reaction thus the number of electrons reacted = number of Au(s) produced

OpenStudy (australopithecus):

so you can just say that the number of electrons = number of atoms of Au(s) produced

OpenStudy (anonymous):

i think my teacher said we have to convert this to moles and then grams.

OpenStudy (australopithecus):

yes you convert to moles using avagadros number

OpenStudy (anonymous):

i know that for grams it's : mol of substance x (Atomic mass of substance/ 1 mol)

OpenStudy (anonymous):

oh so i have to use avogadros #

OpenStudy (anonymous):

so from above, we know that there are 6.362x10^20 atoms reduced. so i multiply this by avogadro's number right?

OpenStudy (australopithecus):

well look at the units of your two numbers avagadros number: 6.022141*10^23 atom/mol number of Au(s) atoms: 6.362*10^20 atom you need to set these numbers up so you end up with just the units mole, remember just like numbers units can cancel out when divded and can be raised to powers when multiplied together

OpenStudy (anonymous):

oh okay i got it!

OpenStudy (anonymous):

so its 1.056434912x10^-3

OpenStudy (anonymous):

moles*

OpenStudy (anonymous):

|dw:1407964662954:dw|

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