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)
Weast to east , east to west, nort to south, south to nort.
Answer:
Sucrose to is the substrate of the reaction because is a simple sugar( disaccharide) which produce glucose and fructose in the presence of sucrase or when it is hydrolysed.
Answer:
0.071 is the correct answer using the formula d=m/v
Answer:
301.8 g
Explanation:
We prepare a solution with 200.4 g of water (solvent) and 101.42 g of salt (solute). The mass of the solution is equal to the sum of the mass of the solvent and the mass of the solute.
m(solution) = m(solute) + m(solvent)
m(solution) = 200.4 g + 101.42 g
m(solution) = 301.8 g (we round-off to one decimal according to the significant figures rules)