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2 years ago

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Based on the diagram, which statement explains how energy is conserved during this chemical reaction? D B Potential energy of a

system - - 1 1 I с Reaction progress A. The potential energy changes indicated by C and B involve energy lost by the surroundings. B. The potential energy lost by the reaction system (B) is gained by the surroundings. C. The potential energy gained by the reaction system (A) is also gained by the surroundings. O D. The potential energy lost by the reaction system (C) is also lost by the surroundings.

1 answer:

Salsk061 [2.6K]2 years ago

6 0

D hope this helps you

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Given the two reactions PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq), K3 = 1.84×10−10, and AgCl(aq)⇌Ag+(aq)+Cl−(aq), K4 = 1.14×10−4, what is the

Nimfa-mama [501]

Answer: The value of equilibrium constant for reaction is, $1.42\times 10^{-2}$

Explanation:

The given chemical equations are:

(1) $PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$ ; $K_3=1.84\times 10^{-10}$

(2) $AgCl(aq)\rightleftharpoons Ag^{+}(aq)+Cl^-(aq)$ ; $K_4=1.14\times 10^{-4}$

Now we have to calculate the equilibrium constant for chemical equation as:

$PbCl_2(aq)+2Ag^{+}(aq)\rightleftharpoons 2AgCl(aq)+Pb^{2+}(aq)$ ; $K=?$

We are reversing reaction 2 and multiplying reaction 2 by 2 and then adding both reaction, we get the final reaction.

The equilibrium constant for the reverse reaction will be the reciprocal of that reaction.

If the equation is multiplied by a factor of '2', the equilibrium constant of that reaction will be the square of the equilibrium constant.

If we are adding equations then the equilibrium constants will be multiplied.

The value of equilibrium constant for reaction is:

$K=(\frac{1}{K_4})^2\times K_3$

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

$K=(\frac{1}{1.14\times 10^{-4}})^2\times (1.84\times 10^{-10})$

$K=1.42\times 10^{-2}$

Hence, the value of equilibrium constant for reaction is, $1.42\times 10^{-2}$

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3 years ago

What is Avagrados Number?

dsp73

6.02x10^23; This represents the number of molecules in 1 mole of a substance.

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A 62.6-gram piece of heated limestone is placed into 75.0 grams of water at 23.1°C. The limestone and the water come to a final

alekssr [168]

Answer:

208.7°C was the initial temperature of the limestone.

Explanation:

Heat lost by limestone will be equal to heat gained by the water

$-Q_1=Q_2$

Mass of limestone = $m_1=62.6 g$

Specific heat capacity of limestone = $c_1=0.921 J/g^oC$

Initial temperature of the limestone = $T_1=?$

Final temperature = $T_2$ =T = 51.9°C

$Q_1=m_1c_1\times (T-T_1)$

Mass of water= $m_2=75.0 g$

Specific heat capacity of water= $c_2=4.186 J/g^oC$

Initial temperature of the water = $T_3=23.1^oC$

Final temperature of water = $T_2$ =T = 51.9°C

$Q_2=m_2c_2\times (T-T_3)$

$-Q_1=Q_2$

$-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)$

On substituting all values:

$-(62.6 g\times 0.921 J/g^oC\times (51.9^oC-T_1))=75.0 g\times 4.186 J/g^oC\times (51.9^oC-23.1^oC)$

$T_1=208.7^oC$

208.7°C was the initial temperature of the limestone.

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Examine the models shown above.where is potential energy the greatest

Lisa [10]

Answer:

A

Explanation:

Because it is the greatest

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