76 plus 54 plus 92 plus 88 plus 76 plus 88 plus 75 plus 93 plus 92 plus 68 plus 88 plus 76 plus 76 plus 88 plus 80 plus 70 plus 88plus 72 equal 1,440
It is 77
43
491
6300
oxidation
6.3 x 10-6
Yes. They react to form lead (II) oxide.
Cr(s) | Cr3+(aq) Pb2+(aq) | Pb(s)
Lead (IV) ion
The Pb K-egde Xanes data reveals that Pb is in a mixed valence state of Pb4+ and Pb2+. However in literature Pb is claimed to be in Pb2+ state. The Pb 6s2 electrons hybridize with the O 2p electrons to form strong covalent bonding which results in the relative displacement of Pb cage with respect to the O-octahedron. This results in increase ferroelectric properties of PbTiO3.However the question is that in the covalent state of the Pb2+ will it appear as Pb4+ state in the Pb K-edge? The reason argued here is that the Pb will lose the 6s2 electrons to form the bond and hence appear to be Pb4+. Hence the argument placed by this pool of thought is from the EXAFS data what appears to be Pb4+ is actually the covalently bonded Pb2+ while what appears to be Pb2+ is actually the ionic type Pb2+.What is the oxidation state of Pb and Ti in PbTiO3 ?
The spectator ions in this precipitation equation are K+ and NO3-. The non-spectator ions are Pb2+ and I-. They combine to form the precipitate PbI2.
76 plus 54 plus 92 plus 88 plus 76 plus 88 plus 75 plus 93 plus 92 plus 68 plus 88 plus 76 plus 76 plus 88 plus 80 plus 70 plus 88plus 72 equal 1,440
The answer is 672.
27x3+6=87.
132
The answer to your question is 366.