Answer:
0.27 atm
Explanation:
<em>At 25ºC, Kp = 2.9 x 10⁻³ for the reaction NH₄OCONH₂(s) ⇌ 2 NH₃(g) + CO₂(g). In an experiment carried out at 25ºC, a certain amount of NH₄OCONH₂ is placed in an evacuated rigid container and allowed to come to equilibrium. Calculate the total pressure in the container at equilibrium.</em>
Step 1: Make an ICE chart
Solid and liquids are ignored in ICE charts.
NH₄OCONH₂(s) ⇌ 2 NH₃(g) + CO₂(g)
I 0 0
C +2x +x
E 2x x
Step 2: Write the pressure equilibrium constant expression (Kp)
Kp = [NH₃]² × [CO₂]
Kp = (2x)² × x
2.9 × 10⁻³ = 4 x³
x = 0.090 atm
Step 3: Calculate the pressures at equilbrium
pNH₃ = 2x = 2(0.090 atm) = 0.18 atm
pCO₂ = x = 0.090 atm
The total pressure is:
P = 0.18 atm + 0.090 atm = 0.27 atm
Because some atoms<span> are more stable when they </span>gain or lose<span> an </span>electron<span> and </span>form ions<span>.</span>
Answer:
37 mmol of acetate need to add to this solution.
Explanation:
Acetic acid is an weak acid. According to Henderson-Hasselbalch equation for a buffer consist of weak acid (acetic acid) and its conjugate base (acetate)-
![pH=pK_{a}(acetic acid)+log[\frac{mmol of CH_{3}COO^{-}}{mmol of CH_{3}COOH }]](https://tex.z-dn.net/?f=pH%3DpK_%7Ba%7D%28acetic%20acid%29%2Blog%5B%5Cfrac%7Bmmol%20of%20CH_%7B3%7DCOO%5E%7B-%7D%7D%7Bmmol%20of%20CH_%7B3%7DCOOH%20%7D%5D)
Here pH is 5.31, 
(acetic acid) is 4.74 and number of mmol of acetic acid is 10 mmol.
Plug in all the values in the above equation:
![5.31=4.74+log[\frac{mmol of CH_{3}COO^{-}}{10}]](https://tex.z-dn.net/?f=5.31%3D4.74%2Blog%5B%5Cfrac%7Bmmol%20of%20CH_%7B3%7DCOO%5E%7B-%7D%7D%7B10%7D%5D)
or, mmol of 
= 37
So 37 mmol of acetate need to add to this solution.
8.................................
What your question for number 3
