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love history [14]
2 years ago
4

NaCl is purified by adding HCl to a saturated solution of NaCl (317 g/L). Will pure NaCl precipitate when 30 mL of 4.5 M HCl is

added to 0.15 L of saturated solution? Group of answer choices NaCl will precipitate from solution No NaCl will precipitate
Chemistry
1 answer:
allsm [11]2 years ago
6 0

Answer:

If you add concentrated HCl solution to a saturated NaCl solution, you will drive the equilibrium of equation 1) back to the left by increasing the concentration of Cl-(aq). As a result, NaCl will re-crystallize in the solution. This is a demonstration of Le Châtelier’s principle that states: if a system in equilibrium is disturbed, it will shift the equilibrium to counteract the disturbance.

1) NaCl(s) ↔ Na+(aq) + Cl-(aq)

2) HCl(aq) ↔ H+(aq) + Cl-(aq)

K= Ksp = [Na+(aq)][Cl-(aq)]

Ka =

[Cl-(aq)] = Ka[HCl(aq)]/[H+(aq)]

Ksp / [Na+(aq)] = Ka[HCl(aq)]/[H+(aq)]

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ethylene glycol used in automobile antifreeze and in the production of polyester. The name glycol stems from the sweet taste of
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<u>Answer:</u> The empirical and molecular formula for the given organic compound is CH_3O and C_4H_{12}O_4

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=9.06g

Mass of H_2O=5.58g

We know that:

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Molar mass of water = 18 g/mol

  • <u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 9.06 g of carbon dioxide, \frac{12}{44}\times 9.06=2.47g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 5.58 g of water, \frac{2}{18}\times 5.58=0.62g of hydrogen will be contained.

  • Mass of oxygen in the compound = (6.38) - (2.47 + 0.62) = 3.29 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{2.47g}{12g/mole}=0.206moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.62g}{1g/mole}=0.62moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{3.29g}{16g/mole}=0.206moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.206 moles.

For Carbon = \frac{0.206}{0.206}=1

For Hydrogen  = \frac{0.62}{0.206}=3

For Oxygen  = \frac{0.206}{0.206}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 3 : 1

Hence, the empirical formula for the given compound is C_1H_{3}O_1=CH_3O

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

n=\frac{\text{molecular mass}}{\text{empirical mass}}

We are given:

Mass of molecular formula = 124 amu = 124 g/mol

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Putting values in above equation, we get:

n=\frac{124g/mol}{31g/mol}=4

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(1\times 4)}H_{(3\times 4)}O_{(1\times 4)}=C_4H_{12}O_4

Thus, the empirical and molecular formula for the given organic compound is CH_3O and C_4H_{12}O_4

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Now we convert those masses into moles:

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Those amounts of moles represent the amount of each component in 1 mol of cacodyl, thus, the molecular formula of cacodyl is C₄H₁₂As₂.

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