The way you want to find the percent composition would be by breaking down the problem like so:
K= atomic mass of K which is 39.098
Mn = atomic mass of Mn which is 54.938
O= atomic mass of o which is 15.999
Then you want to add 39.098+ 54.938+ 15.999 and you get 110.035 which is the molar mass for KMnO
Then you want to take each molar mass and then divide it 110.035 and multiply by 100
Ex. K = 39.098/ 110.035 and the multiply what you get by a 100
You do this for the other elements as well good luck!
Answer:
Cu2+(aq) + 2OH-(aq) => Cu(OH)2(s)
Explanation:
Use of aqueous sodium hydroxide is a precipitation reaction to test for anions or cations. Aqueous sodium hydroxide in a precipitate test forms a insoluble precipitates along with some colors characteristics.
Aqueous sodium hydroxide (NaOH) when mixed with copper(II) (Cu2+) forms a blue precipitate. The formula is as follows:
Cu2+(aq) + 2OH-(aq) => Cu(OH)2(s)
The chemical could have more or less of a reaction to the other chemicals in the experiment