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
Sorry I don’t get the question Buh I’m sure you’ll get it eventually
Answer: C, 146
Explanation:
NaCl has a molar mass of 58.44 g/mol.
To change moles to g, we do:
2.50 moles × 58.44 g/mol = 146.1 g, so C
