Answer:
6.22 × 10⁻⁵
Explanation:
Step 1: Write the dissociation reaction
HC₆H₅COO ⇄ C₆H₅COO⁻ + H⁺
Step 2: Calculate the concentration of H⁺
The pH of the solution is 2.78.
pH = -log [H⁺]
[H⁺] = antilog -pH = antilog -2.78 = 1.66 × 10⁻³ M
Step 3: Calculate the molar concentration of the benzoic acid
We will use the following expression.
Ca = mass HC₆H₅COO/molar mass HC₆H₅COO × liters of solution
Ca = 0.541 g/(122.12 g/mol) × 0.100 L = 0.0443 M
Step 4: Calculate the acid dissociation constant (Ka) for benzoic acid
We will use the following expression.
Ka = [H⁺]²/Ca
Ka = (1.66 × 10⁻³)²/0.0443 = 6.22 × 10⁻⁵
0.250 L*3M=0.250 L*3mol/L= 0.750 mol
Answer:
see calculations in explanation
Explanation:
percent = part/total x 100%
part = ∑ atomic mass of element
- hydrogen = 1.008 amu (atomic mass units)
- carbon = 12.011 amu
- nitrogen = 14.007 amu
total = ∑ molecular mass of compound
= H amu + C amu + Namu
= 1.008 amu + 12.011 amu + 14.007 amu
= 27.026 amu
%H = (1.008amu/27.026amu)100% = 3.730%
%C = (12.011amu/27.026amu)100% = 44.442%
%N = (14.007amu/27.026amu)100% = 51.827%
Check results ∑%values = 100%
3.730% + 44.442% + 51.827% = 99.999% ≅ 100%
Answer:
12.7m
Explanation:
FORMULA
for an approximate result, divide the length value by 39.370078
I know they are fishing and there is an alligator in the lake. The alligator might grab the bait and attack the fishers.
Hope I helped!
Let me know if you need anything else!
~ Zoe
