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VLD [36.1K]
3 years ago
14

who else has parents that don't want them and mentally abuses them? It's ok I got them to they act all innocent in public but wh

en we are home they make me work all day and they make me clean the mess up I have tried to k I ll myself but then they got rid of me and put me in a mental hospital and said I am crazy and they want me if you are going threw it hang in there true I don't know if it will get better but I hope it does I have also tried to run away and I think I am going to do it again so I can go to jail to get away from them I got battery and runaway assault charges on me.
Chemistry
2 answers:
Ulleksa [173]3 years ago
8 0

Answer: oh my gosh, yes I relate with this so much. Don’t worry, you are not alone and you will get through this.

Explanation:

ivanzaharov [21]3 years ago
8 0

Answer:

YES I RELATE!! And then they just make up rules on the spot just to get you in trouble!?

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Choose specific acid-base conjugate pairs to make the following buffers: (a) pH ≈ 3.5;
leva [86]

Our \ goal \ here\  is \ to \ find \ an \ acid-base \ conjugate \ pair \ that \ have \a \\\$pK_a-p H$. This is because we assume that the acid-base conjugate pair have an equal concentration to be considered an effective buffer.

$$p H-p K_a+\log \frac{\left[A^{-}\right]}{[H A]}$$

$$Where $\left[A^{-}\right]-[H A]$, so $\log \frac{\left[A^{-}\right]}{[H A]}-\log 1-0_{\text {so ... }}$$$

pH-pKa

$$Solve for $K_a$.$$

$$\begin{aligned}p H-p K_a &--\log K_a \\K_a &-10^{-p H} \\&-10^{-3.5} \\K_a &-3.2 \times 10^{-4}\end{aligned}$$

$$The nearest pairs are $\mathrm{HCOCOOH} / \mathrm{HCOCOO}^{-}$with a $\mathrm{K}_{\mathrm{a}}-3.5 \times 10^{-4}$,

$\mathrm{HOCH}_2 \mathrm{CH}(\mathrm{OH}) \mathrm{COOH} / \mathrm{HOCH}_2 \mathrm{CH}(\mathrm{OH}) \mathrm{COO}^{-}$with a $\mathrm{K}_{\mathrm{a}}-2.9 \times 10^{-4}$, and

$\mathrm{CH}_3 \mathrm{COOC}_6 \mathrm{H}_4 \mathrm{COOH} / \mathrm{CH}_3 \mathrm{COOC}_6 \mathrm{H}_4 \mathrm{COO}^{-}$with a $\mathrm{K}_a-3.6 \times 10^{-4}$. We can also check for the $\mathrm{K}_b$. First, solve for $p K_b$ then the $K_b$ -

$$\begin{aligned}p K_w &-p K_b+p K_a \\p K_b &-p K_w-p K_a \\&-14-3.5 \\p K_b &-10.5 \\p K_b &--\log K_b \\K_b &-10^{-p K_b} \\&-10^{-10.5} \\K_b &-3.2 \times 10^{-11}\end{aligned}$$

There \ are\  no \ pairs \ near \  this \  $K_b$ value.

<h3>What is a conjugate acid and base pair?</h3>

An acid-base pair that differs by one proton is referred to as a conjugate pair. A conjugate acid-base pair is a pair of substances that can both absorb and donate hydrogen ions to one another. A proton is added to the compound to create the conjugate acid, and a proton is taken out to create the conjugate base. When a proton is supplied to a base, a conjugate acid is created, and vice versa when a proton is taken away from an acid, a conjugate base is created.

To learn more about conjugate acid and base pair, visit;

brainly.com/question/10468518

#SPJ4

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