Answer: C. Z is an intermediate.
Explanation: The given reaction is ![X_{2} + Y + Z \rightarrow XY + XZ](https://tex.z-dn.net/?f=X_%7B2%7D%20%2B%20Y%20%2B%20Z%20%5Crightarrow%20XY%20%2B%20XZ)
And the rate equation is rate = k [X_2][Y]
From this, we can imply that either the concentration of Z is very small in comparison tot he other reactants or the reactant Z is an intermediate which is available only for the given small amount of time.
There can be various steps in the mechanism for the proposed reaction and Z can also react in any of the steps. It is not necessary that it should react in a step other than the rate determining step.
This might be possible that the activation energy for Z to react is very low But since it has made a new kind of product that is XZ, then the former cannot be true.
This is the answer for the question above:
In order to solve this, first use stoichiometry for the energy in kJ released by, converting grams of CuO2 to moles:
<span>56.96g*(1 mole/143.09 g)=0.41904 moles of Cu2O </span>
<span>2 moles of Cu2O=-292.0kJ Convert 0.41904 moles of Cu2O to kJ. </span>
<span>Finally, solve for kJ: </span>
<span>0.41904 moles of Cu2O*(-292.0kJ/2 moles of Cu2O)=-61.18 kJ of energy released</span>
Answer:
Here's what I get
Explanation:
(g) Titration curves
I can't draw two curves on the same graph, but I can draw two separate curves for you.
The graph in part (d) had an equivalence point at 20 mL.
In the second titration, the NaOH was twice as concentrated, so the volume to equivalence point would be half as much — 10 mL.
The two titration curves are below.
(h) Evidence of reaction
HCl and NaOH are both colourless.
They don't evolve a gas or form a precipitate when they react.
The student probably noticed that the Erlenmeyer flask warmed up — a sign of a chemical change.