Answer:
<h3>A) a chemical change started immediately and finished at 8 minutes!!</h3>
Explanation:
it is stated that the substance <em>slowly </em>starts turning blue, not that it suddenly started changing!! and, this couldn't be a non-chemical change because there are very few chemicals that can turn a completely different color when the two substances were the same color!!! ^_^ please let me know if messed anything up
Answer: Molecule
Step-by-step-explanation
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First, we write out a balanced equation.
HA <--> H(+) + A(-)
Next, we create an ICE table
HA <--> H+ + A-
[]i 0.40M 0M 0M
Δ[] -x +x +x
[]f 0.40-x x x
Next, we write out the Ka expression.
Ka = [H+][A-]/[HA]
Ka = x*x/(0.40-x)
However, because Ka is less than 10^-3, we can assume the amount of dissociation is negligible. Thus,
Assume 0.40-x ≈ 0.40
Therefore, 1.2x10^-6 = x^2/0.40
Then we solve for the [H+] concentration, or x

x=6.93x10^-4
Next, to find pH we do
pH = -log[H+]
pH = -log[6.93x10^-4]
pH = 3.2
We have that for the Question it can be said that the NaOH combines with CH_3COOH to produce CH_3COONa (Salt)
From the question we are told
how should the ph of a 0.1 m solution of <em>nac2h3o2</em> compare with that of a 0.1 m <u>solution </u>of kc2h3o2?
Generally
with the ph of a 0.1 m solution of <em>nac2h3o2</em> compared with that of a 0.1 m <u>solution </u>of kc2h3o2 ,we see that the salt produce is a weak acid and strong akali salt
We see that the salt produced in water gives a base from the derived weak acid
The salt produce is CH_3COONa
Therefore
the NaOH combines with CH_3COOH to produce CH_3COONa (Salt)
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