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3 years ago

14

The molar concentrations for the reactants and products at equilibrium are found to be [CCl4]=1.0 M, [O2]=0.3 M, [COCl2]=4.0 M,

and [Cl2]=2.0 M. What is the value of the equilibrium constant for this reaction? 2CCl4(g)+O2(g)⇌2COCl2(g)+2Cl2(g) Express your answer numerically using two significant figures.

1 answer:

kirill115 [55]3 years ago

6 0

Answer:

$2.1\times 10^2$ is the value of the equilibrium constant for this reaction.

Explanation:

$2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)$

Molar concentrations of $CCl_4$ at equilibrium = $[CCl_4]=1.0M$

Molar concentrations of $O_2$ at equilibrium = $[O_2]=0.3M$

Molar concentrations of $COCl_2$ at equilibrium = $[COCl_2]=4.0M$

Molar concentrations of $Cl_2$ at equilibrium = $[Cl_2]=2.0M$

The equilibrium constant is given as:

$K_c=\frac{[COCl_2]^2[Cl_2]^2}{[CCl_4]^2[O_2]}=\frac{(4.0 M)^2\times (2.0M)^2}{(1.0M)^2\times (0.3 M)}$

$K_c=213.33\approx 2.1\times 10^2$

$2.1\times 10^2$ is the value of the equilibrium constant for this reaction.

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What expression approximates the volume of O2 consumed, measure at STP, when 55 g of Al reacts completely with excess O2?2 Al(s)

Genrish500 [490]

Answer:

34.28 L ( 1.5*22.4 L)

Explanation:

Calculation of the moles of aluminum as:-

Mass = 55 g

Molar mass of aluminum = 26.981539 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{55\ g}{26.981539\ g/mol}$

$Moles= 2.0384\ mol$

According to the reaction:-

$4Al+3O_2\rightarrow 2Al_2O_3$

4 moles of aluminum react with 3 moles of oxygen gas

1 mole of aluminum react with $\frac{3}{4}$ moles of oxygen gas

2.0384 moles of aluminum react with $\frac{3}{4}\times 2.0384$ moles of oxygen gas

Moles of oxygen gas = 1.5288 moles

At STP,

Pressure = 1 atm

Temperature = 273.15 K

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1 atm × V = 1.5288 mol × 0.0821 L.atm/K.mol × 273.15 K

⇒V = 34.28 L ( 1.5*22.4 L)

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3 years ago

Why is electrolysis used not carbon to extract metal

ss7ja [257]

Because metals that are more reactive than carbon wont react with it.

8 0

3 years ago

Is work done whenever you hold a heavy object for a long time?

AnnZ [28]

No, work is not done whenever you hold a heavy object for a long time

<h3>What is work done ?</h3>

The result of a force's displacement and its component of force exerted by the object in the direction of displacement is what is known as the force's work. When we push a block with some force, the body moves quickly and work is completed.

No work, as that term is used here, is done until the object is moved in some way and a component of the force travels along the path that the object is moved. Because there is no displacement when holding a heavy object still, energy is not transferred to it.

Learn more about Work done here:

brainly.com/question/25573309

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7 0

1 year ago

A 7.00-mL portion of 8.00 M stock solution is to be diluted to 0.800 M. What will be the final volume after dilution? Enter your

amm1812

Explanation:

The number of moles of solute present in liter of solution is defined as molarity.

Mathematically, Molarity = $\frac{\text{no. of moles}}{\text{Volume in liter}}$

Also, when number of moles are equal in a solution then the formula will be as follows.

$M_{1} \times V_{1} = M_{2} \times V_{2}$

It is given that $M_{1}$ is 8.00 M, $V_{1}$ is 7.00 mL, and $M_{2}$ is 0.80 M.

Hence, calculate the value of $V_{2}$ using above formula as follows.

$M_{1} \times V_{1} = M_{2} \times V_{2}$

$8.00 M \times 7.00 mL = 0.80 M \times V_{2}$

$V_{2} = \frac{56 M. mL}{0.80 M}$

= 70 ml

Thus, we can conclude that the volume after dilution is 70 ml.

7 0

3 years ago

ratelena [41]

Answer:

a) galvanic cell

b)electrolytic cell

c) i) K=6.27x10'34

ΔG°=198790 J

ii) K=3.58x10'-34

ΔG°= 191070 J

d) E°=0.278 v

ΔG°= -26827 J

Explanation:

a) There are two kinds of an electrochemical cell, the first is called "galvanic cells", and the second "electrolytic cell".

The fuel cells are capable of produce electric energy through chemical reactions. These reactions are often spontaneous. So, the galvanic cell has a negative value for Gibbs free energy.

b) The electrolytic cell increases the value of Gibbs energy, to positive values, due to the reactions are not spontaneous.

c) i) look image attached

ii) k = look image attached

ΔG° = -nFE° = - 6 X 95500 J/vmole x (-0.33 v)

ΔG° =-191070

d) E°= 0.0592 v/n x lg K

E°= 0.0592V / 1 X log 5.0X10'4

E°= 0.278 v

ΔG° = -nFE° = -1 x 96500 J/ vmole x 0.278v

ΔG° = -26827 J

5 0

3 years ago