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3 years ago

Write the empirical formulas of the following compounds: (a) Al2Br6, (b) Na2S2O4, (c) N2O5, (d) K2Cr2O7, (e)H2C2O4.

Chemistry

1 answer:

babunello [35]3 years ago

4 0

Answer:1. $AlBr_3$

2. $NaSO_2$

3. $N_2O_5$

4. $K_2Cr_2O_7$

5. $HCO_2$

Explanation:

Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.

1. $Al_2Br_6$ has empirical formula of $AlBr_3$

2. $Na_2S_2O_4$ has empirical formula of $NaSO_2$

3. $N_2O_5$ has empirical formula of $N_2O_5$

4. $K_2Cr_2O_7$ has empirical formula of $K_2Cr_2O_7$

5. $H_2C_2O_4$ has empirical formula of $HCO_2$

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3 years ago

Determine the percent ionization of a 0.225 M solution of benzoic acid. Express your answer using two significant figures.

Assoli18 [71]

Answer:

1.68% is ionized

Explanation:

The Ka of benzoic acid, C₇H₆O₂, is 6.46x10⁻⁵, the equilibrium in water of this acid is:

C₇H₆O₂(aq) + H₂O(l) ⇄ C₇H₅O₂⁻(aq) + H₃O⁺(aq)

Ka = 6.46x10⁻⁵ = [C₇H₅O₂⁻] [H₃O⁺] / [C₇H₆O₂]

<em>Where [] are concentrations in equilibrium</em>

In equilibrium, some 0.225M of the acid will react producing both C₇H₅O₂⁻ and H₃O⁺, the equilibrium concentrations are:

[C₇H₆O₂] = 0.225-X

[C₇H₅O₂⁻] = X

[H₃O⁺] = X

Replacing:

6.46x10⁻⁵ = [X] [X] / [0.225-X]

1.4535x10⁻⁵ - 6.46x10⁻⁵X = X²

1.4535x10⁻⁵ - 6.46x10⁻⁵X - X² = 0

Solving for X:

X = -0.0038. False solution, there is no negative concentrations.

X = 0.00378M. Right solution.

That means percent ionization (100 times Amount of benzoic acid ionized over the initial concentration of the acid) is:

0.00378M / 0.225M * 100 =

<h3>1.68% is ionized</h3>

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