Use the average acceleration obtained in Activity 1 to determine how far the ball should drop at 0.3 second, 0.5 second, and 0.7
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Answer:
0.3152 M
Explanation:
The reaction that takes place is:
- 2MnO₄⁻ + 5C₂O₄⁻² + 16H⁺ → 2Mn⁺² + 8H₂O + 10CO₂
First we <u>calculate the MnO₄⁻ moles used up in the titration</u>, <em>by multiplying the volume times the concentration</em>:
- 21.93 mL * 0.1725 M = 3.783 mmol MnO₄⁻
Then we <u>convert MnO₄⁻ moles to C₂O₄⁻² moles</u>:
- 3.783 mmoles MnO₄⁻ *
= 9.457 mmol C₂O₄⁻²
Finally we <u>calculate the oxalate ion concentration</u>,<em> by dividing the moles by the volume</em>:
- 9.457 mmol C₂O₄⁻² / 30.00 mL = 0.3152 M