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Aleksandr-060686 [28]

4 years ago

8

Help please ❤️ When O2 is added to the melted candy mixture once it has cooled. Why would it not be beneficial to add the CO2 af

ter it is first melted?

1 answer:

amm18124 years ago

8 0

it wouldnt have an effect?

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Andrew [12]

Answer:

0.1

Explanation:

$According \ to \ Henderson \ Hasselbalch \ equation:$

$pH = pka + log \dfrac{[A^-]}{[HA]}$

From the given question;

$log \dfrac{[\varepsilon- NH_2]}{[\varepsilon - NH_3^+]}=pH - pKa$

$\implies log \dfrac{[\varepsilon- NH_2]}{[\varepsilon - NH_3^+]}=9 -10$

$\implies log \dfrac{[\varepsilon- NH_2]}{[\varepsilon - NH_3^+]}=-1$

$\implies \dfrac{[\varepsilon- NH_2]}{[\varepsilon - NH_3^+]}=10^{-1}$

$\implies \dfrac{[\varepsilon- NH_2]}{[\varepsilon - NH_3^+]}=0.1$

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The decomposition of ozone in the upper atmosphere to dioxygen occurs by a two-step mechanism.The first step is a fast reversibl

alex41 [277]

<u>Answer:</u> Step 2 in the given mechanism is the rate determining step

<u>Explanation:</u>

In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.

The intermediate reaction of the mechanism follows:

<u>Step 1:</u> $O_3(g)\rightleftharpoons O_2(g)+O(g);\text{ (fast)}$

<u>Step 2:</u> $O_3(g)+O(g)\rightarrow 2O_2(g);\text{(slow)}$

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Rate law for the reaction follows:

$\text{Rate}=k[O_3][O]$

Hence, step 2 in the given mechanism is the rate determining step

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