The reaction of hydrogen peroxide with iodine, h2o2(aq)+i2(aq)⇌oh−(aq)+hio(aq) is first order in h2o2 and first order in i2. if the concentration of h2o2 was increased by half and the concentration of i2 was quadrupled, by what factor would the reaction rate increase?

When H2O2 & i2 is a first order and the rate can be as the following:
rate = K [H2O2]^1 [i2]^1
and by assuming that [H2O2] & [i2] = 1 so by substitution 
the rate = K (1) (1) = 1 
and when H2O2 increase by half it becomes 1.5 and when i2 increase by quadruple it becomes 4.
by substitution:
the rate = K (1.5)(4) = 6 
So the rate will increase by factor 6

You know how the first one if filled up like the objects do the same but with the other categories

If a hydrogen atom is neutral, it must also have  one electron. an oxygen atom (o) has an atomic number of eight (8). this means a neutral oxygen atom has eight protons and eight electrons. the element actium (ac) has an atomic number of 89, so it has 89 protons and 89 electrons in a neutral atom