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trasher [3.6K]
4 years ago
10

Most of Uranus's atmosphere is composed of

Chemistry
1 answer:
AlekseyPX4 years ago
5 0
Uranus's atmosphere is mostly composed of hydrogen and helium
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Consider the following reaction at equilibrium: 2NH3 (g) N2 (g) + 3H2 (g) Le Châtelier's principle predicts that the moles of H2
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Answer:

A decrease in the total volume of the reaction vessel (T constant)

Explanation:

  • Le Châtelier's principle predicts that the moles of H2 in the reaction container will increase with a decrease in the total volume of the reaction vessel.
  • <em><u>According to the Le Chatelier's principle, when a chnage is a applied to a system at equilibrium, then the equilibrium will shift in a way that counteracts the effect causing it.</u></em>
  • In this case, a decrease in volume means there is an increase in pressure, therefore the equilibrium will shift towards the side with the fewer number of moles of gas.
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4.33 g of 3-hexanol were obtained from 5.84 g of hex-3-ene. Determine the percentage yield of 3-hexanol. a Determine the moles o
Vilka [71]

<u>Answer:</u> The amount of hex-3-ene used is 0.0711 moles and the percent yield of 3-hexanol is 59.56 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of hex-3-ene = 5.84 g

Molar mass of hex-3-ene = 82.14 g/mol

Putting values in equation 1, we get:

\text{Moles of hex-3-ene}=\frac{5.84g}{82.14/mol}=0.0711mol

The chemical equation for the conversion of hex-3-ene to 3-hexanol follows:

\text{hex-3-ene}+H_2O\xrightarrow []{10\% H_2SO_4} \text{3-hexanol}

By Stoichiometry of the reaction:

1 mole of hex-3-ene produces 1 mole of 3-hexanol

So, 0.0711 moles of hex-3-ene will produce = \frac{1}{1}\times 0.0711=0.0711mol of 3-hexanol

Now, calculating the mass of 3-hexanol from equation 1, we get:

Molar mass of 3-hexanol = 102.2 g/mol

Moles of 3-hexanol = 0.0711 moles

Putting values in equation 1, we get:

0.0711mol=\frac{\text{Mass of 3-hexanol}}{102.2g/mol}\\\\\text{Mass of 3-hexanol}=(0.0711mol\times 102.2g/mol)=7.27g

To calculate the percentage yield of 3-hexanol, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of 3-hexanol = 4.33 g

Theoretical yield of 3-hexanol = 7.27 g

Putting values in above equation, we get:

\%\text{ yield of 3-hexanol}=\frac{4.33g}{7.27g}\times 100\\\\\% \text{yield of 3-hexanol}=59.56\%

Hence, the amount of hex-3-ene used is 0.0711 moles and the percent yield of 3-hexanol is 59.56 %

4 0
3 years ago
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