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:
Kc = 1.54e - 31 / 2.61e - 24
Explanation:
1 ) 
; Kc = 1.54e - 31
2) 
; Kc = 2.16e - 24
upon reversing ( 2 ) equation
Kc = 1/2.16e - 24
now adding 1 and reversed equation (2)
we get ,
Kc = 1.54e-31 × 1/2.61e - 24
equilibrium constant of equation (3) is -
Kc = 1.54e - 31 / 2.61e - 24
D sublevel because the s sublevel has one orbital, the p sublevel has three orbitals, the d sublevel has five orbitals, and the f sublevel has seven orbitals. In the first period, only the 1s sublevel is being filled.
Answer:
13.5 %
Explanation:
First we<u> calculate the mass of 500 mL of water</u>, using <em>its density</em>:
- 500 mL * 1.00 g/mL = 500 g
Then we <u>calculate the mass percent of potassium sulfate</u>, using the formula:
Mass of Potassium Sulfate / Total Mass * 100%
- 78 g / (78 + 500) g * 100 % = 13.5 %
Answer:
It is more important because of the freedom.
Explanation:
While at home you can do your work of course... but you could lay down, take a nap. You could get on the game, play around. You could draw, and fiddle and dance and do WHATEVER you want with no teacher to stop you so you have to be your own motivation. You have to be your own teacher or its VERY easy to fail.
