R    CH3OH(g)+      HCl(g)⇌       CH3Cl(g)+  H2O(g)       

Initial  0.250 atm   0.600 atm             0   0
Change     -x   -x                  +x   +x
Equilibrium 0.250−x   0.600−x                  x   x


   Kp=P_CH3OHP_HCl /P_CH3ClP_H2O

x2 /(0.250−x) (0.600−x) =4.7×103
        x=0.249962 atm
P(HCl,eq)=0.600-x=0.600-0.249962=0.350"atm"

You can do this question without RICE table also.
Since K_pis very large, reaction goes almost to completion (methanol is limiting), so x≈0.250.
Here temperature and volume are same for both gases so pressures are directly proportional to moles.
HCl is in excess left unused.
Finally, methanol is zero, and convert pressure of methanol to HCl pressure used taking them as mole ratio.
0.250 atm of methanol reacts with 0.250 atm of HCl .Therefore,
final partial pressure of HCl left: 0.600- 0.250 = 0.350 atm

Hi Mam (This is Mahee)!

Attached is the question I made mistakes on. (The second screenshot is the question)  

NaF is completely soluble in water, so it dissociates to give Na⁺(aq) and F⁻(aq). The fluoride ion, F⁻, is the conjugate base of the weak acid HF, so it behaves as a Brønsted–Lowry base (proton acceptor) in water. Water, in this reaction, acts as the Brønsted–Lowry acid (proton donor).

Hydrolysis equation (definition-based):

F-(aq) + H2O(l) <--> HF(aq) + OH-(aq)

F⁻ accepts H⁺ → forms HF

H₂O donates H⁺ → forms OH⁻

Because HF is weak, the equilibrium lies mostly to the left, but enough OH⁻ forms to make the solution basic.

Hi Mam!

I am confused about how to do this problem.

Hello! I have a few questions related to NMR that I would love clarification on, thank you so much!

Hello! I have a few questions related to NMR that I would love clarification on, thank you so much!

Hello! I have a few questions related to NMR that I would love clarification on, thank you so much!


   

Hello! I have a few questions related to NMR that I would love clarification on, thank you so much!
 

Hello! I have a few questions related to NMR that I would love clarification on, thank you so much!

 Substances that donate H⁺ in water are Brønsted–Lowry acids, and substances that accept H⁺ are Brønsted–Lowry bases.
H₃PO₄ and H₂PO₄⁻ can both act as Brønsted–Lowry acids because each can donate a proton. However, H₃PO₄ is a stronger acid than H₂PO₄⁻ since it carries no negative charge and more readily donates H⁺.
I⁻ and Ba(OH)₂ act as bases.
I⁻ can accept a proton to form HI.
Ba(OH)₂ provides OH⁻ ions, which readily accept H⁺ to form water.
H₂PO₄⁻ is amphiprotic—it can also act as a base by accepting a proton to form H₃PO₄.