Explanation:
when the catalyst is added the reaction with the lower activation energy occur
Answer:
<h2>The answer is 3.2 g/L</h2>
Explanation:
The density of a substance can be found by using the formula

From the question
mass of cat = 400 g
volume = final volume of water - initial volume of water
volume = 125 - 100 = 25 L
The density of the cat is

We have the final answer as
<h3>3.2 g/L</h3>
Hope this helps you
Answer:
The reaction will be non spontaneous at these concentrations.
Explanation:

Expression for an equilibrium constant
:
![K_c=\frac{[Ag^+][Br^-]}{[AgCl]}=\frac{[Ag^+][Br^-]}{1}=[Ag^+][Br^-]](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BAg%5E%2B%5D%5BBr%5E-%5D%7D%7B%5BAgCl%5D%7D%3D%5Cfrac%7B%5BAg%5E%2B%5D%5BBr%5E-%5D%7D%7B1%7D%3D%5BAg%5E%2B%5D%5BBr%5E-%5D)
Solubility product of the reaction:
![K_{sp}=[Ag^+][Br^-]=K_c=7.7\times 10^{-13}](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BAg%5E%2B%5D%5BBr%5E-%5D%3DK_c%3D7.7%5Ctimes%2010%5E%7B-13%7D%20)
Reaction between Gibb's free energy and equilibrium constant if given as:


![\Delta G^o=-2.303\times 8.314 J/K mol\times 298 K\times \log[7.7\times 10^{-13}]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D-2.303%5Ctimes%208.314%20J%2FK%20mol%5Ctimes%20298%20K%5Ctimes%20%5Clog%5B7.7%5Ctimes%2010%5E%7B-13%7D%5D)

Gibb's free energy when concentration
and ![[Br^-] = 1.0\times 10^{-3} M](https://tex.z-dn.net/?f=%5BBr%5E-%5D%20%3D%201.0%5Ctimes%2010%5E%7B-3%7D%20M)
Reaction quotient of an equilibrium = Q
![Q=[Ag^+][Br^-]=1.0\times 10^{-2} M\times 1.0\times 10^{-3} M=1.0\times 10^{-5}](https://tex.z-dn.net/?f=Q%3D%5BAg%5E%2B%5D%5BBr%5E-%5D%3D1.0%5Ctimes%2010%5E%7B-2%7D%20M%5Ctimes%201.0%5Ctimes%2010%5E%7B-3%7D%20M%3D1.0%5Ctimes%2010%5E%7B-5%7D)

![\Delta G=69.117 kJ/mol+(2.303\times 8.314 Joule/mol K\times 298 K\times \log[1.0\times 10^{-5}])](https://tex.z-dn.net/?f=%5CDelta%20G%3D69.117%20kJ%2Fmol%2B%282.303%5Ctimes%208.314%20Joule%2Fmol%20K%5Ctimes%20298%20K%5Ctimes%20%5Clog%5B1.0%5Ctimes%2010%5E%7B-5%7D%5D%29)

- For reaction to spontaneous reaction:
. - For reaction to non spontaneous reaction:
.
Since ,the value of Gibbs free energy is greater than zero which means reaction will be non spontaneous at these concentrations
Answer : The molarity after a reaction time of 5.00 days is, 0.109 M
Explanation :
The integrated rate law equation for second order reaction follows:
![k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%5Cright%29)
where,
k = rate constant = 
t = time taken = 5.00 days
[A] = concentration of substance after time 't' = ?
= Initial concentration = 0.110 M
Now put all the given values in above equation, we get:
![9.7\times 10^{-6}=\frac{1}{5.00}\left (\frac{1}{[A]}-\frac{1}{(0.110)}\right)](https://tex.z-dn.net/?f=9.7%5Ctimes%2010%5E%7B-6%7D%3D%5Cfrac%7B1%7D%7B5.00%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%280.110%29%7D%5Cright%29)
![[A]=0.109M](https://tex.z-dn.net/?f=%5BA%5D%3D0.109M)
Hence, the molarity after a reaction time of 5.00 days is, 0.109 M
Answer:
(E) changing temperature
Explanation:
Consider the following reversible balanced reaction:
aA+bB⇋cC+dD
If we know the molar concentrations of each of the reaction species, we can find the value of Kc using the relationship:
Kc = ([C]^c * [D]^d) / ([A]^a * [B]^b)
where:
[C] and [D] are the concentrations of the products in the equilibrium; [A] and [B] reagent concentrations in equilibrium; already; b; c and d are the stoichiometric coefficients of the balanced equation. Concentrations are commonly expressed in molarity, which has units of moles / 1
There are some important things to remember when calculating Kc:
- <em>Kc is a constant for a specific reaction at a specific temperature</em>. If you change the reaction temperature, then Kc also changes
- Pure solids and liquids, including solvents, are not considered for equilibrium expression.
- The reaction must be balanced with the written coefficients as the minimum possible integer value in order to obtain the correct value of Kc