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2 years ago

Which of these scenarios involve a reaction that is at equilibrium

Chemistry

2 answers:

Ber [7]2 years ago

5 0

The correct answer is, the amount of products and reactants is constant.

Tasya [4]2 years ago

3 0

Reaction is producing more reactants than products

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Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation N2O4(g) 2N

Amanda [17]

Answer : The correct option is, (a) 0.44

Explanation :

First we have to calculate the concentration of $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

Now we have to calculate the dissociated concentration of $N_2O_4$ .

The balanced equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(aq)$

Initial conc. 1.0 M 0

At eqm. conc. (1.0-x) M (2x) M

As we are given,

The percent of dissociation of $N_2O_4$ = $\alpha$ = 28.0 %

So, the dissociate concentration of $N_2O_4$ = $C\alpha=1.0M\times \frac{28.0}{100}=0.28M$

The value of x = $C\alpha$ = 0.28 M

Now we have to calculate the concentration of $N_2O_4\text{ and }NO_2$ at equilibrium.

Concentration of $N_2O_4$ = 1.0 - x = 1.0 - 0.28 = 0.72 M

Concentration of $NO_2$ = 2x = 2 × 0.28 = 0.56 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.56)^2}{0.72}=0.44$

Therefore, the equilibrium constant $K_c$ for the reaction is, 0.44

8 0

2 years ago

How does the equilibrium change to counter the removal of A in this reaction? A + B ⇌ AB

atroni [7]

To counter the removal of A the equilibrium change by shifting toward the left

explanation

If the reaction is at equilibrium and we alter the condition a new equilibrium state is created

The removal of A led to the shift of equilibrium toward the left since it led to less molecules in reactant side which favor the backward reaction.( equilibrium shift to the left)

6 0

2 years ago

Read 2 more answers

A student is given a sample of a blue copper sulfate hydrate. He weighs the sample in a dry covered porcelain crucible and got a

Nata [24]

Answer:

There are present 5,5668 moles of water per mole of CuSO₄.

Explanation:

The mass of CuSO₄ anhydrous is:

23,403g - 22,652g = 0,751g.

mass of crucible+lid+CuSO₄ - mass of crucible+lid

As molar mass of CuSO₄ is 159,609g/mol. The moles are:

0,751g × $\frac{1mol}{159,609g}$ = 4,7052x10⁻³ moles CuSO₄

Now, the mass of water present in the initial sample is:

23,875g - 0,751g - 22,652g = 0,472g.

mass of crucible+lid+CuSO₄hydrate - CuSO₄ - mass of crucible+lid

As molar mass of H₂O is 18,02g/mol. The moles are:

0,472g × $\frac{1mol}{18,02g}$ = 2,6193x10⁻² moles H₂O

The ratio of moles H₂O:CuSO₄ is:

2,6193x10⁻² moles H₂O / 4,7052x10⁻³ moles CuSO₄ = 5,5668

That means that you have 5,5668 moles of water per mole of CuSO₄.

I hope it helps!

5 0

2 years ago

Mr. Rutherford's chemistry class was collecting data in a neutralization study. Each group had 24 test tubes to check each day f

ycow [4]

I’m pretty sure it is A at least that’s what we did at our school to test this

6 0

2 years ago

A 0.1025-g sample of copper metal is dissolved in 35 ml of concentrated hno3 to form cu2 ions and then water is added to make a

GalinKa [24]

Molarity is defined as number of moles of solute in 1 L of solution.

Here, 0.1025 g of Cu is reacted with 35 mL of HNO_{3} to produced Cu^{2+} ions.

The balanced reaction will be as follows:

Cu+3HNO_{3}\rightarrow Cu(NO_{3})_{2}+NO_{2}+H_{2}O

From the above reaction, 1 mole of Cu produces 1 mole of Cu^{2+}, convert the mass of Cu into number of moles as follows:

n=\frac{m}{M}

molar mass of Cu is 63.55 g/mol thus,

n=\frac{0.1025 g}{63.55 g/mol}=0.0016 mol

Now, total molarity of solution, after addition of water is 200 mL or 0.2 L can be calculated as follows:

M=\frac{n}{V}=\frac{0.0016 mol}{0.2 L}=0.008 mol/L=0.008 M

Thus, molarity of Cu^{2+} is 0.008 M.

7 0

2 years ago