Explanation:
The given reaction equation will be as follows.
![[FeSCN^{2+}] \rightleftharpoons [Fe^{3+}] + [SCN^{-}]](https://tex.z-dn.net/?f=%5BFeSCN%5E%7B2%2B%7D%5D%20%5Crightleftharpoons%20%5BFe%5E%7B3%2B%7D%5D%20%2B%20%5BSCN%5E%7B-%7D%5D)
Let is assume that at equilibrium the concentrations of given species are as follows.
M
M
M
Now, first calculate the value of
as follows.
![K_{eq} = \frac{[Fe^{3+}][SCN^{-}]}{[FeSCN^{2+}]}](https://tex.z-dn.net/?f=K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BFe%5E%7B3%2B%7D%5D%5BSCN%5E%7B-%7D%5D%7D%7B%5BFeSCN%5E%7B2%2B%7D%5D%7D)
= 
= 
Now, according to the concentration values at the re-established equilibrium the value for
will be calculated as follows.
![K_{eq} = \frac{[Fe^{3+}][SCN^{-}]}{[FeSCN^{2+}]}](https://tex.z-dn.net/?f=K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BFe%5E%7B3%2B%7D%5D%5BSCN%5E%7B-%7D%5D%7D%7B%5BFeSCN%5E%7B2%2B%7D%5D%7D)
M
Thus, we can conclude that the concentration of
in the new equilibrium mixture is
M.
Answer:
Mo(CO)5 is the intermediate in this reaction mechanism.
Explanation:
The reaction mechanism describes the sequence of elementary reactions that must occur to go from reactants to products. Reaction intermediates are formed in one step and then consumed in a later step of the reaction mechanism.
In this reaction mechanism, Mo(CO)5 is the product of 1st reaction and then it is used as a reactant in 2nd reaction. So, Mo(CO)5 is the reaction intermediates.
The overall balanced equation would be,
Mo(CO)6 + P(CH3) ↔ CO + Mo(CO)5 + P(CH3)3
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: possibly diffusion
Explanation:
all particles are in motion unless at a certain degree so they'd spread throughout the room diluting as they continue to spread out.
A) -0.5(9.8)*t^2 = -25(t-2) - 0.5(9.8)(t-2)^2
-4.9t^2 = -25t + 50 - 4.9(t^2-4t+4)
0 = -25t+50+19.6t - 19.6
5.4t = 30.4
t = 5.62962963 s
b) h = -4.9(5.62962963)^2
h = -155.2943759
the building is 155.2943759 m high
c) speed 0of first stone
= at
= 9.8*5.62962963
= 55.17037037 m/s
speed of second stone
= v + at
= 25+9.8*3.62962963
= 60.57037037 m/s