Answer:
the answer would be Halogens
Answer:
32 mL
Explanation:
<em>A chemist must prepare 500.0mL of hydrobromic acid solution with a pH of 0.50 at 25°C. He will do this in three steps: Fill a 500.0mL volumetric flask about halfway with distilled water. Measure out a small volume of concentrated (5.0M) stock hydrobromic acid solution and add it to the flask. Fill the flask to the mark with distilled water. Calculate the volume of concentrated hydrobromic acid that the chemist must measure out in the second step. Round your answer to 2 significant digits.</em>
<em />
Step 1: Calculate [H⁺] of the dilute solution
pH = -log [H⁺]
[H⁺] = antilog -pH = antilog -0.50 = 0.32 M
Step 2: Calculate [HBr] of the dilute solution
HBr is a strong acid that dissociates according to the following equation.
HBr ⇒ H⁺ + Br⁻
The molar ratio of HBr to H⁺ is 1:1. The concentration of HBr is 1/1 × 0.32 M = 0.32 M.
Step 3: Calculate the volume of the concentrated HBr solution
We will use the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂ / C₁
V₁ = 0.32 M × 500.0 mL / 5.0 M
V₁ = 32 mL
Answer:
2. d
3. i
4. e
5. j
6. g
7. a
8. f
9. b
Explanation:
do a few searches when you have time to deepen understanding!
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 ![[Ag^+] = 1.0\times 10^{-2} M](https://tex.z-dn.net/?f=%5BAg%5E%2B%5D%20%3D%201.0%5Ctimes%2010%5E%7B-2%7D%20M)
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
