compound formed from Pb^4+ and C2H3O^2- is?
i know the anwser is Pb(C2H3O2)2 ... can some one explain step by step what to do to get from Pb^4+ and C2H3O^2- to Pb(C2H3O2)2
..... it would be alot of help exam very soon
2007-01-27 05:35:06 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Pb4 + C2H3O2- Pb(C2H3O)2
The 4 at the Pb would trade places with the 2 at the end of O leaving the Pb with 2 and the oxygen with 4, but then since the ratio is not at its simplest form you divide both by 2 leaving the Pb with nothing and the C2H3O with 2.
2007-01-27 05:46:19 · answer #1 · answered by Anonymous · 0⤊ 2⤋
I agree with Gerald. Although being old fashioned, I call it lead acetate.
Acetate as a polyatomic ion has a charge of -1. If the Pb is indeed (IV), then the correct formula is Pb(C2H3O2)4. Therefore, the question is written incorrectly.
Unless it is really Pb(II). Then it would be OK.
2007-01-27 15:19:34 · answer #2 · answered by reb1240 7 · 0⤊ 0⤋
you get the answer Pb(C2H3O2)2 because the charges need to be balanced
the Pb has a charge of 4 so the C2H3O2 must have a charge of -4 in order to cancel out the charges
you see that Pb has the +4 so in order to get the C2H3O2 balanced with it you must multiple it by 2 so you can get -4
you put it in parenthesis because if you put it in the coefficient it would make the charge 8 and not 4
put it in parenthesis so that the charges balance out
hope this helps
2007-01-27 13:52:18 · answer #3 · answered by stephen s 1 · 0⤊ 1⤋
Based on your given data,
Please note that for a compound to be electrically neutral in your case, then the number of positive charges MUST BE equal to the number of negative charges. You have 4 positive on Pb and only 2 negative on acetate. To make them equal, you need to multiply the acetate by 2 or a total of -4. +4 -4 = 0. Thus, you need the subscript 2 on the acetate.
Your teacher probably taught you to use the CROSS method. It is essentially the same.
Consider the example Al+3 and SO4 -2. Please note that the numbers are superscripts. To find the compound formed you cross the superscipt number as SUBscripts below. Or,
(Al)2(SO4)3 . Total of six positive charges and a total of 6 negative charges.
2007-01-27 13:42:25 · answer #4 · answered by Aldo 5 · 0⤊ 0⤋
None of the above have pointed out that lead(IV) ethanoate (using Pb4+ ions) does not exist, and if it did, it would have the formula
Pb(CH3COO)4.
Lead(II) ethanoate (using Pb2+ ions) is a well-known compound, however, and has the formula
Pb(CH3COO)2.
2007-01-27 13:56:10 · answer #5 · answered by Gervald F 7 · 0⤊ 0⤋
that's difficult
2007-01-27 13:36:58 · answer #6 · answered by Anonymous · 0⤊ 1⤋