This is a bit long right now.
(n l m s) these are the numbers.
n=2, second orbital level.
l is the type of orbit: l=1 elongated, l=0 spherical
m=0, magnetic number: 0
s=spin, both have spin positive
You need still to round it up. Srry!
(3.5mol)(24.106 g/1mol c6h6) =84.371 g C6H6
NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa
NaOH + CH3COOH → CH3COONa + H2O
Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH
Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH
These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L
Molarity of CH3COOH = 0.0106/0.071 = 0.1493M
CH3COONa = 0.0076 / 0.071 = 0.1070M
pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74.
pH using Henderson - Hasselbalch equation:
pH = pKa + log ([salt]/[acid])
pH = 4.74 + log ( 0.1070/0.1493)
pH = 4.74 + log 0.717
pH = 4.74 + (-0.14)
pH = 4.60.
Answer:
a) After helping our partner, we should immediately report the incident to the lab manager or any person in charge of the emergencies occurring in the lab.
b) We should have a copy of the Material Safety Data Sheet to give to the responders. This is because the responder can identify what materials were being used by the person ans what other security measures need to be taken.
Answer:
pH = 13.1
Explanation:
Hello there!
In this case, according to the given information, we can set up the following equation:
Thus, since there is 1:1 mole ratio of HCl to KOH, we can find the reacting moles as follows:
Thus, since there are less moles of HCl, we calculate the remaining moles of KOH as follows:
And the resulting concentration of KOH and OH ions as this is a strong base:
![[KOH]=[OH^-]=\frac{0.00576mol}{0.012L+0.032L}=0.131M](https://tex.z-dn.net/?f=%5BKOH%5D%3D%5BOH%5E-%5D%3D%5Cfrac%7B0.00576mol%7D%7B0.012L%2B0.032L%7D%3D0.131M)
And the resulting pH is:
Regards!
