Answer:
The rate of the reaction will increase by a factor of 9.
Explanation:
Hello,
In this case, considering the given second-order reaction, whose rate law results:
![r=k[A] [B]^2](https://tex.z-dn.net/?f=r%3Dk%5BA%5D%20%5BB%5D%5E2)
We easily infer that at constant concentration of A but tripling the concentration of B, we are going to obtain the following increasing factor while holding the remaining variables constant:
![Increase\ factor=\frac{r_{final}}{r_{initial}} =\frac{k[A][3*B]^2}{k[A][B]^2} =\frac{3^2}{1} \\Increase\ factor=9](https://tex.z-dn.net/?f=Increase%5C%20factor%3D%5Cfrac%7Br_%7Bfinal%7D%7D%7Br_%7Binitial%7D%7D%20%3D%5Cfrac%7Bk%5BA%5D%5B3%2AB%5D%5E2%7D%7Bk%5BA%5D%5BB%5D%5E2%7D%20%3D%5Cfrac%7B3%5E2%7D%7B1%7D%20%5C%5CIncrease%5C%20factor%3D9)
Best regards.
This problem is simply converting the concentration from molality to molarity. Molality has units of mol solute/kg solvent, while molarity has units of mol solute/L solution.
2.24 mol H2SO4/kg H2O * (0.25806 kg H2SO4/mol H2SO4) = 0.578 kg H2SO4/kg H2O
That means the solution weighs a total of 1 kg + 0.578 kg = 1.578 kg. Then, convert it to liters using the density data:
1.578 kg * (1000g / 1kg) * (1 mL/1.135 g) = 1390 mL or 1.39 L.
Hence, the molarity is
2.24/1.39 = 1.61 M
The empirical formula of metal iodide : CoI₃(Cobalt(III) Iodide)
<h3>Further explanation</h3>
13.02 g sample of Cobalt , then mol Co(MW=58.933 g/mol) :

Mass of metal iodide formed : 97.12 g, so mass of Iodine :

Then mol iodine (MW=126.9045 g/mol) :

mol ratio of Cobalt and Iodine in the compound :