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)
Answers:
A) Li
B) Ar
C) Br
D) Ne
E) B
Answer:
<em>first </em><em>example</em><em> </em><em>is</em><em>. </em><em>Air </em><em> </em><em>it </em><em>is </em><em>mixture </em><em>of </em><em>many </em>
<em>gases </em><em> </em>
<em>it's </em><em>us </em><em> </em><em>it </em><em>keeps </em><em>us </em><em>alive </em><em> </em><em>and </em><em>many </em><em>more </em><em>uses</em>
<em>second </em><em>example</em><em> </em><em>is </em><em>water </em><em>is </em><em>also </em><em>a </em><em>mixture</em><em> </em>
<em>made </em><em>up </em><em>of </em><em>carbon </em><em>and </em><em>hydrogen</em><em> </em>
<em>and </em><em>its uses </em><em>is </em><em>to </em><em>fulfill</em><em> </em><em>our </em><em>needs </em>
<em>hope </em><em>it </em><em>helps</em>
The starting substances in a chemical reaction are called reactants - they are written on the left side of a chemical equation.
Answer:
Reduction reaction
This is so because considering the reaction Cu 2+ has been reduced to Cu(s) by losing two electrons , and any other element that undergoes this stage is termed as a reduction reaction.
Hope this helps