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
the numbers are going to be small so like a power but its at the bottom
NH3, H2O2, NHO2
The balanced chemical reaction is:
<span>2 I2 + KIO3 + 6 HCl ---------> 5 ICl + KCl + 3 H2O
</span>
We are given the amount of the product to be produced from the reaction. This will be the starting point of our calculations.
28.6 g ICl (1 mol / 162.35 g ICl ) ( 2 mol I2 / 5 mol ICl ) ( 253.81 g I2 / 1 mol I2 ) = 17.88 g I2
For an approximate result, multiply the volume value by 3.785
Answer ≈ 56.7812
It is a trial aimed to reduce bias during an experiment. An example would be a sugar pill, something that has no real effect so that the results of the true trial can accurately be compared. Its like a control group.
