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sammy [17]
3 years ago
11

An equimolar mixture of X (g) and Y (g) is placed inside a rigid container at constant temperature. The particle diagram above r

epresents the changes that occur over time. Based on the particle
diagram, which of the following best predicts whether or not the system has reached equilibrium by 300 s?
a. It is not possible to determine that the system has reached equilibrium by 300 s because the stoichiometry of the reaction is not known.

b. it is not possible to determine that the system has reached equilibrium by 300 s because the amounts of X, Y, and XY have continued to change

c. The system has reached equilibrium by 300 s because the rate of formation of XY is constant
d. The system has reached equilibrium by 300 s because the rates of consumption of X and Y are equal

Chemistry
1 answer:
iris [78.8K]3 years ago
6 0

Answer:

B

Explanation:

t= 300s has more XY than t=200s, which means that it continues to change

*i think

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Nitrogen dioxide gas is dark brown in color and remains in equilibrium with dinitrogen tetroxide gas, which is colorless.
sukhopar [10]

Answer:

This reaction is exothermic because the system shifted to the left on heating.

Explanation:

2NO₂ (g) ⇌ N₂O₄(g)

Reactant => NO₂ (dark brown in color)

Product => N₂O₄ (colorless)

From the question given above, we were told that when the reaction at equilibrium was moved from room temperature to a higher temperature, the mixture turned dark brown in color.

This simply means that the reaction does not like heat. Hence the reaction is exothermic reaction.

Also, we can see that when the temperature was increased, the reaction turned dark brown in color indicating that the increase in the temperature favors the backward reaction (i.e the equilibrium shift to the left) as NO₂ which is the reactant is dark brown in color. This again indicates that the reaction is exothermic because an increase in the temperature of an exothermic reaction will shift the equilibrium position to the left.

Therefore, we can conclude that:

The reaction is exothermic because the system shifted to the left on heating.

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What is the speed of a wave with a wavelength of 3 m and a frequency of .1hz?
Ivahew [28]

Answer:

0.671081 mps

Explanation:

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3 years ago
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5 0
3 years ago
A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s
Alla [95]

<u>Answer:</u> The percent yield of the 1-bromobutane is 48.65 %

<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)

  • <u>For NaBr:</u>

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol

The chemical equation for the reaction of 1-butanol and NaBr is:

\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = \frac{1}{1}\times 0.108=0.108 moles of 1-bromobutane

  • Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g

  • To calculate the percentage yield of 1-bromobutane, we use the equation:

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

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%

Hence, the percent yield of the 1-bromobutane is 48.65 %

5 0
2 years ago
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