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MissTica
3 years ago
12

A quantity of 7.480 g of an organic compound is dissolved in water to make 300.0 mL of solution. The solution has an osmotic pre

ssure of 1.43 atm at 27°C. The analysis of this compound shows that it contains 41.8 percent C, 4.7 percent H, 37.3 percent O, and 16.3 percent N. Calculate the molecular formula of the compound
Chemistry
1 answer:
Serga [27]3 years ago
7 0

Answer:

C5H6N2O3

Explanation:

First the empirical formulas

C= 41.8÷12= 2.48

H= 4.7÷1= 4.7

O= 37.3÷ 16= 2.33

N= 16.3÷14 = 1.16

Divide by the smallest

C= 3.48/1.16=3

H= 4.7/1.16=4.1

O= 2.33/1.16=2

N= 1.16/1.16=1

Therefore empirical formula = C3H4NO2

To calculate molecular formula for osmotic pressure,

π= cRT

Or

π=cgRT/M where cg is in g/liter & T is temperature in Kelvin. Thus

π= (7.480*0.0821*300)/ M

M= 184.23/1.43

M= 128.83

To find molecular Formula

Molecular Mass= (empirical mass)n

128.83= (C3H4NO2)n

128.83= 86n

n= 1.5

Therefore the molecular formula

(C3H4NO2)1.5

= C4.5H6N1.5O3

Approximately

C5H6N2O3

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3 0
2 years ago
How many moles of ammonia (nh3) would be produced if 2.5 moles of nitrogen (n2) reacted with excess hydrogen (h2)?
cluponka [151]
Firstly, a balanced equation has to be written for the production of ammonia (NH₃) from hydrogen gas (H₂) and nitrogen gas (N₂):
                    N₂   +   3H₂   →   2NH₃

Now, the mole ratio of N₂ : NH₃ is 1 : 2 based on the coefficients of the balanced equation.

If the moles of N₂ = 2.5 moles

   then the moles of NH₃ produced = 2.5 mol × 2 
                                                          =  5 mol

Thus, the moles of ammonia produced when 2.5 mol of nitrogen gas is combined with excess hydrogen gas is 5 mol.
6 0
3 years ago
How many molecules are in 41.8 g of sulfuric acid
Anton [14]

Answer

× 10²³ molecules are in 41.8 g of sulfuric acid

Explanation

The first step is to convert 41.8 g of sulfuric acid to moles by dividing the mass of sulfuric acid by its molar mass.

Molar mass of sulfuric acid, H₂SO₄ = 98.079 g/mol

Mole=\frac{Mass}{Molar\text{ }mass}=\frac{41.8\text{ }g}{98.079\text{ }g\text{/}mol}=0.426187053\text{ }mol

Finally, convert the moles of sulfuric acid to molecules using Avogadro's number.

Conversion factor: 1 mole of any substance = 6.022 × 10²³ molecules.

Therefore, 0.426187053 moles of sulfuric acid is equal

\frac{0.426187053\text{ }mol}{1\text{ }mol}\times6.022×10²³\text{ }molecules=2.57\times10^{23}\text{ }molecules

Thus, 2.57 × 10²³ molecules are in 41.8 g of sulfuric acid.

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In Part III, the phenolphthalein indicator is used to monitor the equilibrium shifts of the ammonia/ammonium ion system. The phe
AnnZ [28]

The pink color in the solution fades. Some of the colored indicator ion converts to the colorless indicator molecule.

<h3>Explanation</h3>

What's the initial color of the solution?

\text{NH}_4\text{Cl} is a salt soluble in water. \text{NH}_4\text{Cl} dissociates into ions completely when dissolved.

\text{NH}_4\text{Cl} \; (aq)\to {\text{NH}_4}^{+} \; (aq) +{\text{Cl}}^{-} \; (aq).

The first test tube used to contain \text{NH}_4\text{OH}. \text{NH}_4\text{OH} is a weak base that dissociates partially in water.

\text{NH}_4\text{OH} \; (aq) \rightleftharpoons {\text{NH}_4}^{+}  \;(aq)+ {\text{OH}}^{-} \; (aq).

There's also an equilibrium between \text{OH}^{-} and {\text{H}_3\text{O}}^{+} ions.

{\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l).

\text{OH}^{-} ions from \text{NH}_4\text{OH} will shift the equilibrium between \text{OH}^{-} and {\text{H}_3\text{O}}^{+} to the right and reduce the amount of {\text{H}_3\text{O}}^{+} in the solution.

The indicator equilibrium will shift to the right to produce more {\text{H}_3\text{O}}^{+} ions along with the colored indicator ions. The solution will show a pink color.

What's the color of the solution after adding NH₄Cl?

Adding \text{NH}_4\text{Cl} will add to the concentration of {\text{NH}_4}^{+} ions in the solution. Some of the {\text{NH}_4}^{+} ions will combine with \text{OH}^{-} ions to produce \text{NH}_4\text{OH}.

The equilibrium between  \text{OH}^{-} and {\text{H}_3\text{O}}^{+} ions will shift to the left to produce more of both ions.

{\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l)

The indicator equilibrium will shift to the left as the concentration of {\text{H}_3\text{O}}^{+} increases. There will be less colored ions and more colorless molecules in the test tube. The pink color will fade.

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3 years ago
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