3. Sketch a simple model that shows why the constellations change over the course of a year. Then, annotate the model to explain
1 answer:
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Answer:
- i) 5.00 mL of 1.00 M NaOH: before the equivalence point
- ii) 50.0 mL of 1.00 M NaOH: before the equivalence point
- iii) 100 mL of 1.00 M NaOH: at the equivalence point
- iv) 150 mL of 1.00 M NaOH: after the equivalence point
- v) 200 mL of 1.00 M NaOH: after the equivalence point
Explanation:
1. First calculate the number of mol acid in the 100 mL of 1.00 M HCl solution.
Equation:
- Molarity = numbrer of moles of solute / volume of the solution in liters.
Thus, you need to convert each volume from mL to liters, which is done dividing by 1,000.
Naming M the molarity, n the number of moles of solute (acid or base), and V the volume in liters:
2. Now calculate the number of moles of NaOH for every condition (addition)
<u>i) 5.00 mL of 1.00 M NaOH</u>
Since the number of moles of NaOH added (0.00500mol) is less than the number of moles of acid in the solution (0.100mol), this is before equivalence point.
<u>ii) 50.0 mL of 1.00 M NaOH</u>
Since the number of moles of NaOH added (0.0500mol) is less than the number of moles of acid in the solution (0.100mol), this is before equivalence point.
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<u>iii) 100 mL of 1.00 M NaOH</u>
Since the number of moles of NaOH added (0.100mol) is equal to the number of moles of acid in the solution (0.100mol), this is at the equivalence point.
<u>iv) 150 mL of 1.00 M NaOH</u>
Since the number of moles of NaOH added (0.150mol) is greater than the number of moles of acid in the solution (0.100mol), this is after the equivalence point.
<u>v) 200 mL of 1.00 M NaOH</u>
This is more volume of NaOH, then this is also after the equivalence point.