Find all the pairs of ions in the following list which cannot exist together (in appreciable concentration) in the same aqueous solution without precipitating. That is, write any combination of the following ions which would form a precipitate, a weak acid, a weak base, or water. [Note: PbCl2 is slightly soluble.] Fe^3+ H^+ OH^- Na^+ Cl^- Pb^2+ NO3^- Mg^2+

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.

the answer is a! (2.9 x 10^-18 j) hope i could someone. : )