Answer:
The reaction will be non spontaneous at these concentrations.
Explanation:
![AgBr(s)\rightarrow Ag^+(aq) + Br^- (aq)](https://tex.z-dn.net/?f=AgBr%28s%29%5Crightarrow%20Ag%5E%2B%28aq%29%20%2B%20Br%5E-%20%28aq%29)
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 R\times T\times \log K_c](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D-2.303%5Ctimes%20R%5Ctimes%20T%5Ctimes%20%5Clog%20K_c)
![\Delta G^o=-2.303\times R\times T\times \log K_{sp}](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D-2.303%5Ctimes%20R%5Ctimes%20T%5Ctimes%20%5Clog%20K_%7Bsp%7D)
![\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)
![\Delta G^o=69,117.84 J/mol=69.117 kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D69%2C117.84%20J%2Fmol%3D69.117%20kJ%2Fmol)
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=\Delta G^o+(2.303\times R\times T\times \log Q)](https://tex.z-dn.net/?f=%5CDelta%20G%3D%5CDelta%20G%5Eo%2B%282.303%5Ctimes%20R%5Ctimes%20T%5Ctimes%20%5Clog%20Q%29)
![\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)
![\Delta G=40.588 kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20G%3D40.588%20kJ%2Fmol)
- 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:
4- A material that transfers heat energy more easily than another material will experience a greater rate of thermal energy loss than an object that does not transfer heat energy easily.
Explanation:
Thermal energy loss has to do with loss of heat energy by a body to another body or its environment. The aim of the process is usually the attainment of thermal equilibrium between the body and its environment.
On a cold day, a material that transfers thermal energy more easily will loose thermal energy faster than an object that does not transfer thermal energy. The rate of heat transfer of a body determines its rate of loss of thermal energy.
Explanation:
The atomic number is equal to the number of protons in an atom's nucleus. Hydrogen's atomic number is 1 because all hydrogen atoms contain exactly one proton.
Hydrocarbons are compounds formed by only hydrogen atoms and carbon.
Answer (2)
hope this helps!
Answer:
In the presence of a base, blue litmus paper will turn red........