Question 1:
when an acid name ends in -ic acid, there is often a related acid whose name ends in -ous acid. the formula of the -ous acid has one less oxygen atom than that of the -ic acid. with this information, write formulas for:
(a) sulfurous acid and sulfuric acid
(b) nitrous acid and nitric acid.
question 2:
what is the ph of a solution that has a hydrogen ion concentration of 1.0 * 10 -9 m? show your calculation.

Question 1:

(a) Sulfurous acid: H2SO3

Sulfuric acid: H2SO4

(b) Nitrous acid: H2NO2

Nitric acid: H2NO3

Question 2:

To calculate the pH, based on concentration of H+ ions, there is one formula:

So the pH of this solution is

(the solution is basic).

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

a  solution  is always transparent, light passes through with no scattering from solute particles which are molecule in size. the  solution  is homogeneous and does not settle out. a colloid is intermediate between asolution  and a  suspension. while asuspension  will separate out a colloid will not.