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9966 [12]
2 years ago
13

A sample of a compound is analyzed and found to contain 0.420 g nitrogen, 0.480g oxygen, 0.540 g carbon and 0.135 g hydrogen. Wh

at is the empirical formula of the compound? a. C2H5NO b. CH3NO c. C3H9N2O2 d. C4HN3O4 e. C4H13N3O3
Chemistry
1 answer:
Crazy boy [7]2 years ago
7 0

Answer:

c. C3H9N2O2

Explanation:

The empirical formula of a compound is defined as the simplest whole number ratio of atoms present in a molecule. To solve this question we need to convert the mass of each atom to moles. With the moles we can find the ratio as follows:

<em>Moles N -Molar mass: 14.01g/mol-</em>

0.420g N * (1mol/14.01g) = 0.0300 moles N

<em>Moles O -Molar mass: 16g/mol-</em>

0.480g O * (1mol/16g) = 0.0300 moles O

<em>Moles C -Molar mass: 12.01g/mol-</em>

0.540g C * (1mol/12.01g) = 0.0450 moles C

<em>Moles H -Molar mass: 1.0g/mol-</em>

0.135g H * (1mol/1g) = 0.135moles H

Dividing in the moles of N (Lower number of moles) the ratio of atoms is:

N = 0.0300 moles N / 0.0300 moles N = 1

O = 0.0300 moles O / 0.0300 moles N = 1

C = 0.0450 moles C / 0.0300 moles N = 1.5

H = 0.135 moles H / 0.0300 moles N = 4.5

As the empirical formula requires whole numbers, multiplying each ratio twice:

N = 2, O = 2, C = 3 and H = 9

And the empirical formula is:

<h3>c. C3H9N2O2</h3>

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Answer:

Requirements for a correctly written chemical equation are reactants and products, their formula and valency

Explanation:

Formula of the given compound are -

1 - Potassium Hydroxide - KOH

2 - Calcium Nitrate - Ca(NO_3)_2

The requirements for a correctly written chemical equation are -

  • Identifying reactants and products
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Example of word equation, formula equation, and chemical equation is as follows -

Aluminium + iron9(III)oxide ⇒ aluminium oxide + iron (word equation)

Al_(_s_)  + Fe_2O_3_(_s_)    ⇒   Al_2O_3_(_s_)  + Fe_(s) (formula equation)              

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3 years ago
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3 years ago
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Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant o
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Answer:

4.26 %

Explanation:

There is some info missing. I think this is the original question.

<em>Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is  4.50  ×  10 ⁻⁴.</em>

<em />

Step 1: Given data

Initial concentration of the acid (Ca): 0.249 M

Acid dissociation constant (Ka): 4.50  ×  10 ⁻⁴

Step 2: Write the ionization reaction for nitrous acid

HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)

Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])

We will use the following expression.

[A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4}  } = 0.0106 M

Step 4: Calculate the percent ionization of nitrous acid

We will use the following expression.

\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%

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