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STatiana [176]
3 years ago
11

PLEASE ANSWER ASAP I WILL GIVE BRAINLIEST!!

Chemistry
2 answers:
anzhelika [568]3 years ago
5 0
It’s lower, hope this helps with your issue and helps solve your problem, no problem
kirill115 [55]3 years ago
4 0

Answer: It’s Lower

What can you say about the pH of a carbonic acid solution compared to that of a sulfuric acid soloution with the same contention: Strong acids dissociate fully in water to produce the maximum number of H^{+}ions. The pH levels of strong acids are smaller than those of low acids, which is why the reactions rates for strong acids are greater than the reactions of weak acids with products like metals, metal carbonates, etc. Meaning The pH Level is lower.

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Brilliant_brown [7]

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26. Which waves have wavelengths longer than those of visible light? Give an example of how each kind
Kipish [7]

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Determine the equilibrium constant for the acid-base reaction between ethanol and hydrobromic acid? Acid pKa Hydrobromic Acid −5
siniylev [52]

Answer:

10^{-3.4

Explanation:

Let us first take a look at the image below;

In the acid - base  reaction; we can see the transfer of electrons that takes place;

We can also see that the reaction goes in the direction which converts the stronger acid and the stronger base to the weaker acid and the weaker base.

The stronger acid is shown with the one with more negative P_{Ka} Value.

∴ The equilibrium constant for the acid-base reaction is expressed as:

K_{eq}= \frac{K_a of reactant acid}{K_a of product acid}

      = \frac{10^{-pK} of reactant acid}{10^{-pk} of product acid}

From  P_{Ka} Value (shown in the image below), it is clear and vivid that hydrobromic acid is a stronger acid than the ethyloxonium ion, therefore the equilibrium lies to the right.

From the chemical equation (shown in the attached image); the equilibrium constant for the acid-base reaction can be expressed as:

K_{eq}=\frac{10^{-pK} of hydrobromic acid}{10^{-pk} of Ethyloxonium acid}

K_{eq}=\frac{10^{-5.8} of hydrobromic acid}{10^{-2.4} of Ethyloxonium acid}

       = 10^{-3.4}

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