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4 years ago

12

Round the number 0.00234 to 3 significant figures

1 answer:

Sergeu [11.5K]4 years ago

6 0

Answer: 0.002

Explanation:

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The amount of I−3(aq) in a solution can be determined by titration with a solution containing a known concentration of S2O2−3(aq

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Answer : The molarity of $I_3^-$ in the solution is, 0.128 M

Explanation :

The given balanced chemical reaction is,

$2S_2O_2^{-3}(aq)+I_3(aq)\rightarrow S_4O_2^{-6}(aq)+3I^-(aq)$

First we have to calculate the moles of $Na_2S_2O_3$ .

$\text{Moles of }Na_2S_2O_3=\text{Molarity of }Na_2S_2O_3\times \text{Volume of solution}$

$\text{Moles of }Na_2S_2O_3=0.260mole/L\times 0.0296L=0.007696mole$

Conversion used : (1 L = 1000 ml)

Now we have to calculate the moles of $I_3^-$ .

From the balanced chemical reaction, we conclude that

As, 2 moles of $S_2O_2^{-3}$ react with 1 mole of $I_3^-$

So, 0.007696 moles of $S_2O_2^{-3}$ react with $\frac{0.007696}{2}=0.003848$ mole of $I_3^-$

The moles of $I_3^-$ = 0.003848 mole

Now we have to calculate the molarity of $I_3^-$ .

$\text{Molarity of }I_3^-=\frac{\text{Moles of }I_3^-}{\text{Volume of solution}}$

Now put all the given values in this formula, we get:

$\text{Molarity of }I_3^-=\frac{0.003848mole}{0.03L}=0.128mole/L=0.128M$

Therefore, the molarity of $I_3^-$ in the solution is, 0.128 M

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