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xeze [42]
2 years ago
5

4.80 mol of solid A was placed in a sealed 1.00-L container and allowed to decompose into gaseous B and C. The concentration of

B steadily increased until it reached 1.30 M, where it remained constant.
A(s) <===> B(g) + C(g)
Then, the container volume was doubled and equilibrium was re-established. How many moles of A remain?
Chemistry
1 answer:
ipn [44]2 years ago
5 0

Answer:

A = 3.5 moles

Explanation:

In order to do this, we need first to write the innitial conditions. The reaction is as follow:

A(s) <-------> B(g) + C(g)

From here, we can write an expression for the equilibrium constant which is:

Kc = [B]*[C]

We don't take A in the expression, because is solid, and solid and liquids do not contribute in the equilibrium.

We know the concentration of B, which is the same for C, because the moles of C are coming from A when it's decomposed, so if B is 1.3 M we can assume C is the same.

Kc for this innitial condition is:

Kc = (1.3)*(1.3) = 1.69

Now, in the second part of reaction the volume is doubled, which means that concentrations are halved:

A: 4.8/2 = 2.4 M

B and C: 1.3/2 = 0.65 M

So conditions for the second part will be like this:

   A(s) <----> B(g) + C(g)   Kc = 1.69

I:   2.4           0.65   0.65

C:  -x              +x       +x

E: 2.4-x         0.65+x   0.65+x

Replacing in Kc we have:

1.69 = (0.65+x)²   solving for x here

√1.69 = 0.65+x

1.3 = 0.65+x

x = 1.3 - 0.65

x = 0.65 M

Therefore, the remaining moles of A would be:

A = 2.4 - 0.65

[A] = 1.75 M

moles A = 1.75 * 2 = 3.5 moles

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Using the mass/volume percentage method for percentages of the solution, you simply divide the grams of solute by the volume of the solution and multiply by 100 to get your percentage.
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3 years ago
Predict the sign of the entropy change,Delta S, for each of the following reactions:The signs are either going to be pos or nega
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Answer: a) Pb^{2+}(aq)+2Cl^-(aq)\rightarrow PbCl_2(s):  negative

b)  CaCO_3(s)\rightarrow CaO(s)+CO_2(g) : positive

c) 2NH_3(g)\rightarrow N_2(g)+3H_2(g): positive.

d) P_4(g)+5O_2(g)\rightarrow P_4O_{10}(s) : negative

e) C_4H_8(g)+6O_2(g)\rightarrow 4CO_2(g)+4H_2O(g):  positive.

f) I_2(s)\rightarrow I_2(g) : positive.

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from  an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

\Delta S is positive when randomness increases and \Delta S is negative when randomness decreases.

a) Pb^{2+}(aq)+2Cl^-(aq)\rightarrow PbCl_2(s)

As ions are moving to solid form , randomness decreases and thus sign of \Delta S is negative.

b) CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

As solid is changing to gas, randomness increases and thus sign of \Delta S is positive.

c) 2NH_3(g)\rightarrow N_2(g)+3H_2(g)

As 2 moles of reactants are converted to 4 moles of products , randomness increases and thus sign of \Delta S is positive.

d) P_4(g)+5O_2(g)\rightarrow P_4O_{10}(s)

As gas is changing to solid, randomness decreases and thus sign of \Delta S is negative.

e) C_4H_8(g)+6O_2(g)\rightarrow 4CO_2(g)+4H_2O(g)

As 7 moles of reactants are converted to 8 moles of products , randomness increases and thus sign of \Delta S is positive.

f) I_2(s)\rightarrow I_2(g)

As solid is changing to gas, randomness increases and thus sign of \Delta S is positive.

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3 years ago
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Answer:

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6 0
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Eading and
4vir4ik [10]

Answer: 25.8 g of Cl_2 will be produced from the decomposition of 73.4 g of AuCl_3

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} AuCl_3=\frac{73.4g}{303g/mol}=0.242moles

The balanced chemical reaction is:

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According to stoichiometry :

2 moles of AuCl_3 produce =  3 moles of Cl_2

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