I think 3 of them are its been 1 half years since ive done this i dont take chemistry anymore
Using the Henderson-Hasselbalch equation on the solution before HCl addition: pH = pKa + log([A-]/[HA]) 8.0 = 7.4 + log([A-]/[HA]); [A-]/[HA] = 4.0. (equation 1) Also, 0.1 L * 1.0 mol/L = 0.1 moles total of the compound. Therefore, [A-] + [HA] = 0.1 (equation 2) Solving the simultaneous equations 1 and 2 gives: A- = 0.08 moles AH = 0.02 moles Adding strong acid reduces A- and increases AH by the same amount. 0.03 L * 1 mol/L = 0.03 moles HCl will be added, soA- = 0.08 - 0.03 = 0.05 moles AH = 0.02 + 0.03 = 0.05 moles Therefore, after HCl addition, [A-]/[HA] = 0.05 / 0.05 = 1.0 Resubstituting into the Henderson-Hasselbalch equation: pH = 7.4 + log(1.0) = 7.4, the final pH.
A control group is the group in an experiment that does not receive any sort of change, to then be compared to the other treated objects at the end of the study.
Answer:
The partial pressure of BrCl at equilibrium is 0.08 atm.
Explanation:
The equilibrium constant of the reaction is given by =
initial
0 0 0.500 atm
At equilbrium
p p (0.500-2p)
The equilibrium constant's expression of the reaction is given by ;
![K_p=\frac{[BrCl]^2}{[Br_2][Cl_2]}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B%5BBrCl%5D%5E2%7D%7B%5BBr_2%5D%5BCl_2%5D%7D)
Solving for p:
p = 0.21 atm
The partial pressure of BrCl at equilibrium is:
(0.500-2p) = (0.500 - 2 × 0.21 )atm = 0.08 atm[/tex]
