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The answer to the questions. All 15 of them.
Answer:
0.08 mL
Explanation:
The solution of the skin test has a concentration of 0.2% (w/v), which means that there are 0.2 g of the antigen per 100 mL of the solution. If a new solution will be done using it, then this solution will be diluted, and the mass of the antigen added must be the same in the volume taken and at the diluted solution.
The mass is the concentration (in g/mL) multiplied by the volume of the solution (in mL), so, if m is the mass, C the concentration, V the volume, 1 the initial solution, and 2 the diluted:
m1 = m2
C1*V1 = C2*V2
Where
C1 = 0.2 g/100 mL = 0.002 g/mL
V1 = ?
C2 = 0.04 mg/mL = 0.00004 g/mL
V2 = 4 mL
0.002*V1 = 0.00004*4
V1 = 0.08 mL
You are given the neutralization of acetic acid with sodium hydroxide. Also, you are given the k for acetic acid, which is 1.8 x 10⁻⁵. You are asked to find the<span> approximate value of the equilibrium constant, kn, for the neutralization. We will have a reaction of both acetic acid and sodium hydroxide.
CH</span>₃COOH + NaOH → CH₃COONa + H₂O
which comes from
CH₃COOH → CH₃COO⁻ + H⁺
H⁺ + OH⁻ → H₂O<span>
</span>The k for water is always 1.0 x 10¹⁴. The Ksp for the reaction will be
<span>
Ksp = [</span>CH₃COOH][H₂O]
Ksp = (1.8 x 10⁻⁵)(1.0 x 10¹⁴)
<span>Ksp = 1.8 x 10</span>⁹
