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)
Finding percent composition is fairly easy. You only need to divide the mass of an element by the total mass of the compound. We can do this one element at a time.
First, let's find the total mass by using the masses of the elements given on the periodic table.
7 x 12.011 (mass of Carbon) = 84.077
5 x 1.008 (mass of Hydrogen) = 5.04
3 x 14.007 (mass of Nitrogen) = 42.021
6 x 15.999 (mass of Oxygen) = 95.994
Add all of those pieces together.
84.077 + 5.04 + 42.021 + 95.994 = 227.132 g/mol is your total. Since we also just found the mass of each individual element, the next step will be very easy.
Carbon: 84.077 / 227.132 = 0.37016 ≈ 37.01 %
Hydrogen: 5.04 / 227.132 = 0.022189 ≈ 2.22 %
Nitrogen: 42.021 / 227.132 = 0.185 ≈ 18.5 %
Oxygen: 95.994 / 227.132 = 0.42263 ≈ 42.26 %
You can check your work by making sure they add up to 100%. The ones I just found add up to 99.99, which is close enough. A small difference (no more than 0.03 in my experience) is just a matter of where you rounded your numbers.
Spectroscopy because it talks about the study of spectrum of light.
Answer: 5
Explanation: you divide the distance by the time and get the speed.
