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2 years ago

What is the empirical formula for CaCo3?*

1 answer:

5 0

This problem is requiring the empirical formula for CaCO₃, which is its molecular formula, and turns out to be equal, this is A. CaCO3 according to the following:

<h3>Empirical formulas:</h3><h3 />

In chemistry, molecular formulas show both the actual type and number of atoms in a chemical compound, based on the elements across the periodic table and the subscripts standing for the number of atoms in the compound.

However, the empirical formula is a reduced expression of the molecular one, which shows the minimum number of atoms in a compound after simplifying to the smallest whole numbers.

In such a way, since the given compound is CaCO₃ and both Ca and C have a one as their subscript, it is not possible to simplify any further and therefore the empirical formula equals the molecular one this time, making the answer to be A. CaCO3.

Learn more about empirical formulas: brainly.com/question/1247523

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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

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