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3 years ago

Explain how you know the following reaction is balanced, 2Hgo 2Hg + 02

Chemistry

1 answer:

Lelu [443]3 years ago

7 0

I'm not gonna write 200-500 words, but, it's balanced because both side have the same number of elements. this relates to law of conservation of matter bc both sides are equal.

hope this helped!!

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3 years ago

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3 years ago

Which of the following equations can be used to calculate pOH?

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POH = 14 - pH
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3 years ago

When cyclohexene is treated with KMnO4, H2O, the syn-1,2-diol is produced. What reaction occurs when benzene is similarly treate

Ulleksa [173]

Answer:

No reaction is observed

Explanation:

The benzene ring is aromatic. Being an aromatic ring, the benzene ring is remarkably stable to all reactions that destroy the aromatic ring.

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7 0

3 years ago

An unknown compound, X is thought to have a carboxyl group with a pKa of 2.0 and another ionizable group with a pKa between 5 an

Westkost [7]

Answer:

7.3

Explanation:

By Henderson Hasselbalch equation we can calculate the pH or the pOH of a solution by its pKa. Remember that $pH = -log[H^{+}]$ , and pKa = -logKa. Ka is the equilibrium constant of the acid.

Henderson Hasselbalch equation :

$pH = pKa - log \frac{[HA]}{[A^{-}]}$

Where [HA] is the concentration of the acid, and $[A^{-}]$ is the concentration of the anion which forms the acid.

So, acid X, has two ionic forms, the carboxyl group and the other one. First, we have 0.1 mol/L of the acid, in 100 mL, so the number of moles of X

n1 = (0.1 mol/L)x(0.1 L) = 0.01 mol

When it dissociates, it forms 0.005 mol of the carboxyl group and 0.005 mol of the other group. Assuming same stoichiometry.

Adding NaOH, with 0.1 mol/L and 75 mL, the number of moles of $OH^-$ will be

n2 = (0.1 mol/L)x(0.075 L) = 0.0075 mol

So, the 0.0075 mol of $OH^-$ reacts with 0.005 mol of carboxyl, remaining 0.0025 mol of $OH^-$ , which will react with the 0.005 mol of the other group. So, it will remain 0.0025 mol of the other group.

The final volume of the solution will be 175 mL, but both concentrations (the acid form and ionic form) have the same volume, so we can use the number of mol in the equation.

Note that, the number of moles of the acid form is still 0.01 mol because it doesn't react!

So,

$6.72 = pKa - log \frac{0.01}{0.0025}$

6.72 = pKa - log 4

pKa - log4 = 6.72

pKa = 6.72 + log4

pKa = 6.72 + 0.6

pKa = 7.3

8 0

4 years ago