The oxidation state of bromine in the bromide reactant is –1, and the oxidation state of the
bromine in the bromate is +5. The fact that the bromine of one reactant reduces the bromine
of the other should not bother you, since it is immaterial. The important thing in redox
reactions is the number of electrons. Since bromide is acting as a reducing agent, going from
the –1 oxidation state to the 0 oxidation state, it transfers one electron per bromide. Bromate,
however, is going from the +5 oxidation state to the 0 oxidation state, requiring 5 electrons.
Bromate gets these electrons from bromide, requiring 5 of them. The ratio of bromate to
bromide is therefore 1 to 5.43 . C
To figure out how many moles of Na 2 CO 3 are in 120 mL of a 1.5 M solution, perform the
following calculation:
(1.5 mol/L)(0.12 L) = 0.18 mol
To figure out how many grams are in 0.18 mol, you need to multiply the number of moles by
the formula weight:(106 g/mol)(0.18 mol) = 19 grams44 . D
Thomas Graham stated in 1832 that the rates for two gases diffusing is inversely proportional
to the square root of the molar mass. Therefore, H 2 :O 2 = [32/2]1/2. Simplify this equation and
you have a ratio of diffusion rates of 4:1 for H 2 :O 2.
45 . D
Questions of nuclear chemistry can, as in this case, often be translated into basic arithmetic
problems. Based on the law of conservation of mass, the sum of the mass numbers
(superscripted) must be the same on each side of the arrow. Similarly, conservation of charge
mandates that the sum of the nuclear charges (subscripted) be the same on each side of the
arrow. We can thus translate the nuclear question posed into two elementary arithmetic
questions:
27 + 4 = 30 + ? and 13 + 2 = 15 + ?