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
Lithium
Li → Li⁺ + e⁻
Li 1s²2s¹ → Li⁺ 1s² + e⁻
Answer:
1.135 M.
Explanation:
- For the reaction: <em>2HI → H₂ + I₂,</em>
The reaction is a second order reaction of HI,so the rate law of the reaction is: Rate = k[HI]².
- To solve this problem, we can use the integral law of second-order reactions:
<em>1/[A] = kt + 1/[A₀],</em>
where, k is the reate constant of the reaction (k = 1.57 x 10⁻⁵ M⁻¹s⁻¹),
t is the time of the reaction (t = 8 hours x 60 x 60 = 28800 s),
[A₀] is the initial concentration of HI ([A₀] = ?? M).
[A] is the remaining concentration of HI after hours ([A₀] = 0.75 M).
∵ 1/[A] = kt + 1/[A₀],
∴ 1/[A₀] = 1/[A] - kt
∴ 1/[A₀] = [1/(0.75 M)] - (1.57 x 10⁻⁵ M⁻¹s⁻¹)(28800 s) = 1.333 M⁻¹ - 0.4522 M⁻¹ = 0.8808 M⁻¹.
∴ [A₀] = 1/(0.0.8808 M⁻¹) = 1.135 M.
<em>So, the concentration of HI 8 hours earlier = 1.135 M.</em>
I think the answer is a and c
I
