I think 5.50 M x 35.0 mL x molar mass of RbOH = mass (g)
Answer:
The amount of HC₂H3₃2(aq) in the flask after the addition of 5.0mL of NaOH(aq) compared to the amount of HC₂H₃O₂(aq) in the flask after the addition of 1.0mL is much smaller because more HC₂H₃O₂(aq) is required to react with 5.0 mL NaOH than with 1.0 mL NaOH.
Explanation:
Equation of the reaction between acetic acid, HC₂H₃O₂(aq) and sodium hydroxide, NaOH(aq) is given below:
CH₃COOH (aq) + NaOH (aq) ----> CH₃COONa (aq) + H₂O
The equation of the reaction shows that acetic acid andsodium hydroxide will react in a 1:1 ratio
Since the concentration of NaOH was not given, we can assume that the concentration is 0.01 M
Moles of NaOH in 5.0 mL of 0.01 M NaOH = 0.01 × 5/1000 = 0.00005 moles
Moles of NaOH in 1.0 mL of 0.01 M NaOH = 0.01 ×1/1000 = 0.0001 moles
Ratio of moles of NaOH in 5.0 mL to 1.0 mL = 0.00005/0.00001 = 5
There are five times more moles of NaOH in 5.0 mL than in 1.0 mL and this means that 5 times more the quantity of HC₂H₃O2(aq) required to react with 1.0 mL NaoH is needed to react with 5.0 mL NaOH.
Therefore, the amount of HC₂H₃O2(aq) in the flask after the addition of 5.0mL of NaOH(aq) compared to the amount of HC₂H₃O₂(aq) in the flask after the addition of 1.0mL is much smaller because more HC₂H₃O₂(aq) is required to react with 5.0 mL NaOH than with 1.0 mL NaOH.
Henri Becquerel discovered radioactivity and it was the Curies who made experiments about the concept and became known for their d
Lithium i think!!
hope this helps
