Hello!
A) At pH=1
This pH is lower than the value for the pKa, so Acetic acid wouldn't be ionized, but the equilibrium would be displaced to
CH₃COOHCH₃COOH ⇄ CH₃COO⁻ + H₃O⁺ (equilibrium displaced to the
left)
The chemical structure for CH₃COOH is the first one in the attached images.
B) At pH=7
This pH is higher than the value for the pKa, so Acetic acid would be ionized, and the equilibrium would be displaced to
CH₃COO⁻
CH₃COOH ⇄ CH₃COO⁻ + H₃O⁺ (equilibrium displaced to the
right)
The chemical structure for CH₃COO⁻ is the second one in the attached images.
Have a nice day!
Hey there!:
K = Ka * Kb / Kw
Ka = 1.8*10⁻⁴
Kb = 10⁻¹⁴ / 6.8*10⁻⁴
K = 1.8*10⁻⁴ * ( 10⁻¹⁴/ 6.8*10⁻⁴ ) * ( 1 / 10⁻¹⁴ )
K = = 1.8 / 6.8
K = 0.265
Answer A
Therefore:
K is less than on the forward reaction is not favorable .
Hope That helps!
Explanation:
pH indicators are weak acids that exist as natural dyes and indicate the concentration of H+ (H3O+) ions in a solution via color change. A pH value is determined from the negative logarithm of this concentration and is used to indicate the acidic, basic, or neutral character of the substance you are testing.
Answer:
13.2 g Na2CO3
Explanation:
Convert 10.0 g NaOH to mol.
10.0 g x 1 mol/39.997 g = 0.250 mol
Use mol ration given by the equation: 2 mol NaOH to 1 mol Na2CO3
0.250 mol NaOH x 1 mol Na2CO3/2 mol NaOH = 0.125 mol Na2CO3
Finally, convert the moles of Na2CO3 to grams.
0.125 mol Na2CO3 x 105.99 g/1 mol = 13.2 g
Answer:
There is 61.538% oxygen in Al2(SO4)3.
Explanation:
Wt Of oxygen in the compound = 12*16 = 192 amu.
Total Wt. Of the compound = 2*12+3*32+12*16 = 312 amu.
Thus, percent of oxygen = Wt of oxygen/total Wt. Of compound *100
= 192/312 * 100=61.538 %