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.
Answer:
0.058M is molarity of the solution
Explanation:
Molarity is defined as the ratio between the moles of solute (In this case, MgSO₄) and the liters of solution.
In the problem:
<em>Moles MgSO₄: 0.32 moles</em>
<em>Liters solution: 5.5L</em>
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Molarity is:
0.32 moles / 5.5L =
0.058M is molarity of the solution
Answer:
The electrons are found on the atomic nucleus.
Explanation:
To determine the volume you would need to find the density and mass of each.
Answer:
Linear molecule with two domains
Explanation:
