They did not repeat their tests multiple times.
Reason being is using the same stopwatch doesn’t matter and changing the reactants won’t help, and you without a doubt need more than one person collecting the data so,
They did not repeat their tests multiple times is the answer.
Answer:
i. Keq=4157.99.
ii. More hydrogen sulfide will be produced.
Explanation:
Hello,
i. In this case, for the concentrations at equilibrium on the given chemical reaction, the equilibrium constant results:
![Keq=\frac{[H_2S]^2}{[H_2]^2[S_2]} =\frac{(0.97M)^2}{(0.051M)^2(0.087)} =4157.99](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BH_2S%5D%5E2%7D%7B%5BH_2%5D%5E2%5BS_2%5D%7D%20%3D%5Cfrac%7B%280.97M%29%5E2%7D%7B%280.051M%29%5E2%280.087%29%7D%20%3D4157.99)
ii. Now, by means of the Le Chatelier's principle, the addition of a reactant shifts the reaction towards products, it means that more hydrogen sulfide will be produced in order to reach equilibrium.
Best regards.
Answer:
2.11 g hydrobromic acid (correct to 3SF)
Explanation:
Molecular formula of hydrobromic acid = C2H5BrO2
mass of C2H5BrO2 = 140.96g
Beginning with what we're given, 9.03*10^21 we then make a conversion by using Avegadro's number which is 6.02*10^23 per mole (Oct. 23 at 6:02 am is national mole day :) Then, we need to convert out of moles, 140.96g hydrombromic acid per mole.
It looks like this:
9.03*10^21 molecules • (1 mol C2H5BrO2 / 6.02*10^23 molecules) • (140g C2H5BrO2 / 1 mol) = 2.1144 g C2H5BrO2
The answer is C because the weight is going down so they form a gas.
