Complete ionic:
Cu(aq) + 2Cl(aq) + 8O(aq) + 2Na(aq) + C(aq) + 3O(aq) = CaCO3(s) + 2Na(aq) + Cl(aq) + 4O(aq)
Net ionic:
Cu(aq) + Cl(aq) + 4O(aq) + 2Na(aq) + C(aq) + 3O(aq) = CaCO3(s)
So write everything out as IF it will dissociate in water. So everything that is aq splits but solid just floats to the bottom of the mixture. Cancel what you can (in this case the two from the ClO4 on the left of the equation cancels with the ClO4 from the right) and the 2Na cancels. Then, write out the whole solution and you are done!
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:
Gas, has no shape but does take up volume
Experiments test the scientists' ideas.