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S₂O₈²⁻
(aq) + 2I⁻
(aq) → I₂(aq) + 2SO₄
²⁻(aq)
2S₂O₃²⁻
(aq) + I₂(aq) → S₄O₆²⁻
(aq) + 2I⁻
(aq)
<u>Explanation:</u>
S₂O₈²⁻
(aq) + 2I⁻
(aq) → I₂(aq) + 2SO₄
²⁻(aq)
To measure the rate of this reaction we must measure the rate of concentration change of one of the reactants or products. To do this, we will include (to the reacting S₂O₈
²⁻ and I⁻
i) a small amount of sodium thiosulfate, Na₂S₂O₃,
ii) some starch indicator.
The added Na₂S₂O₃ does not interfere with the rate of above reaction, but it does consume the I₂ as soon as it is formed.
2S₂O₃²⁻
(aq) + I₂(aq) → S₄O₆²⁻
(aq) + 2I⁻
(aq)
This reaction is much faster than the previous, so the conversion of I2 back to I⁻ is essentially instantaneous.
![rate = \frac{dI2}{dt} = \frac{1/2 [S2O3^2^-]}{t}](https://tex.z-dn.net/?f=rate%20%3D%20%5Cfrac%7BdI2%7D%7Bdt%7D%20%3D%20%5Cfrac%7B1%2F2%20%5BS2O3%5E2%5E-%5D%7D%7Bt%7D)
Answer:
Here's what I get
Explanation:
1. Balanced equation
HQ⁻ + CH₃-Br ⟶ HQ-CH₃ + Br⁻
(I must use HQ because the Brainly Editor thinks the O makes a forbidden word)
2. Mechanism
HQ⁻ + CH₃-Br ⟶[HQ···CH₃···Br]⁻⟶ HQ-CH₃ + Br⁻
A C B
The hydroxide ion attacks the back side of the carbon atom in the bromomethane (A).
At the same time as the Q-H bond starts to form, the C-Br bond starts to break.
At the half-way point, we have a high-energy intermediate (C) with partially formed C-O and C-Br bonds.
As the reaction proceeds further, the Br atom drops off to form the products — methanol and bromide ion (B).
3. Energy diagram
See the diagram below.
Answer:
Led is a light-emitting diode, a semiconductor diode that glows when a voltage is applied. So, there is always a sure result of whether a substance conducts electricity or not as the light glows. That is why it is a better option for testing conduction of electricity.
Explanation:
<h3>
Answer:</h3>
14 milliliters
<h3>
Explanation:</h3>
We are given;
Prepared solution;
- Volume of solution as 0.350 L
- Molarity as 0.40 M
We are required to determine the initial volume of HNO₃
- We are going to use the dilution formula;
- The dilution formula is;
M₁V₁ = M₂V₂
Rearranging the formula;
V₁ = M₂V₂ ÷ M₁
=(0.40 M × 0.350 L) ÷ 10.0 M
= 0.014 L
But, 1 L = 1000 mL
Therefore,
Volume = 14 mL
Thus, the volume of 10.0 M HNO₃ is 14 mL