Sandy would reach an incorrect outcome.that's because she will repeat this mistake in the future too.
You must know the concentration of the acetic acid. Suppose the concentration is 0.1 M. The solution is as follows:
CH₃COOH → CH₃COO⁻ + H⁺
I 0.1 0 0
C -x +x +x
E 0.1 - x x x
Ka = (x)(x)/(0.1 - x)
1.8×10⁻⁵ = x²/(0.1 - x)
Solving for x,
x = 1.333×10⁻³ = H⁺
pH = -log[H⁺] = -log(1.333×10⁻³)
pH = 2.88
There will be 7.5 g of Be-11 remaining after 28 s.
If 14 s = 1 half-life, 28 s = 2 half-lives.
After the first half-life, ½ of the Be-11 (15 g) will disappear, and 15 g will remain.
After the second half-life, ½ of the 15 g (7.5 g) will disappear, and 7.5 g will remain.
In symbols,
<em>N</em> = <em>N</em>₀(½)^<em>n</em>
where
<em>n</em> = the number of half-lives
<em>N</em>₀ = the original amount
<em>N</em> = the amount remaining after <em>n</em> half-lives
(D) At equilibrium, the concentration of the products will be much higher than the concentration of the reactants.
Answer:
0.048 M
Explanation:
Note that the Concentration from Question 5= .00096 M
To Prepere of commercial aspirin solution, take the following steps:
- Mix 1 aspirin tablet and 10 mL of 1 M NaOH in a 125 mL Erlenheyemer flask, heat to boil.
- Transfer solution to 100 mL volumetric flask and fill to 100 mL mark with deonized water. Cover and mix solution thoroughly.
- Using pipette transfer a 0.200 mL of solution to a 10 mL volumetric flask and dilute it with the 0.02 M buffered iron (III) chloride solution. Transfer it to test tube.
