Answer:
The situation given here is imaginary such that the life of Rock has to be found using the half-life of the element lokium that has been found inside the rock.
Half-life of any material is the amount of time taken by that particular material to decay. Now the amount of lokium found in rock can show after how many half-lives this amount has been left out.
The time elapsed will be log (L) atoms X half-life.
Explanation:
Question #1
Potasium hydroxide (known)
volume used is 25 ml
Molarity (concentration) = 0.150 M
Moles of KOH used
0.150 × 25/1000 = 0.00375 moles
Sulfuric acid (H2SO4)
volume used = 15.0 ml
unknown concentration
The equation for the reaction is
2KOH (aq)+ H2SO4(aq) = K2SO4(aq) + 2H2O(l)
Thus, the Mole ratio of KOH to H2SO4 is 2:1
Therefore, moles of H2SO4 used will be;
0.00375 × 1/2 = 0.001875 moles
Acid (sulfuric acid) concentration
0.001875 moles × 1000/15
= 0.125 M
Question #2
Hydrogen bromide (acid)
Volume used = 30 ml
Concentration is 0.250 M
Moles of HBr used;
0.25 × 30/1000
= 0.0075 moles
Sodium Hydroxide (base)
Volume used 20 ml
Concentration (unknown)
The equation for the reaction is
NaOH + HBr = NaBr + H2O
The mole ratio of NaOH : HBr is 1 : 1
Therefore, moles of NaOH used;
= 0.0075 moles
NaOH concentration will be
= 0.0075 moles × 1000/20
= 0.375 M
Answer: -
1 mol
Explanation: -
Number of moles of Sulphur S = 7
Number of moles of O2 = 9
The balanced chemical equation for the reaction is
2S (s)+3 O2 (g)→2SO3(g)
From the above reaction we can see that
3 mol of O2 react with 2 mol of S
9 mol of O2 will react with
= 6 mol of S
Unreacted S = 7 - = 1 mol.
If a reaction vessel initially contains 7 mol S and 9 mol O2
1 mole of s will be in the reaction vessel once the reactants have reacted as much as possible
You slap some dough on to a beach and wait for it to become bread
