Answer:
Kc = 0.5951 (4 sig. figs.)
Explanation:
For A + B ⇄ C + D at standard thermodynamic conditions (298K, 1atm)
ΔG = ΔG° + R·T·lnQ => 0 = ΔG° + R·T·lnKc => ΔG° = - R·T·lnKc
=> lnKc = - ΔG°/R·T
ΔG° = +12.86 Kj/mol
R = 8.314 Kj/mol·K
T = 298K
lnKc = - (+12.86Kj) / (8.314Kj/mol·K)(298K) = - 0.519 mol⁻¹
Kc = e⁻⁰°⁵¹⁹ mol⁻¹ = 0.5957 mol⁻¹ (4 sig. figs.)
Answer:
APROXIMENTLY 17
Explanation:
lol i remebered from science class
Answer:
a) 52.8
Explanation:
M1V1 = M2V2
(7.91 M)(x ml) = (2.13 M)
(196.1 ml)
(7.91M) (xml) = 417.693 M.ml
x ml = 417.693/ 7.91
x = 52.8
Answer:
Explanation:
Hello,
In this case, given the acid, we can suppose a simple dissociation as:
Which occurs in aqueous phase, therefore, the law of mass action is written by:
![Ka=\frac{[H^+][A^-]}{[HA]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
That in terms of the change 
due to the reaction's extent we can write:
But we prefer to compute the Kb due to its exceptional weakness:
Next, the acid dissociation in the presence of the base we have:
![Kb=\frac{[OH^-][HA]}{[A^-]}=1x10^{6}=\frac{x*x}{0.1-x}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BOH%5E-%5D%5BHA%5D%7D%7B%5BA%5E-%5D%7D%3D1x10%5E%7B6%7D%3D%5Cfrac%7Bx%2Ax%7D%7B0.1-x%7D)
Whose solution is 
which equals the concentration of hydroxyl in the solution, thus we compute the pOH:
![pOH=-log([OH^-])=-log(0.0999)=1](https://tex.z-dn.net/?f=pOH%3D-log%28%5BOH%5E-%5D%29%3D-log%280.0999%29%3D1)
Finally, since the maximum scale is 14, we can compute the pH by knowing the pOH:
Regards.
