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2 years ago

What is the equilibrium constant for I2+Cl2=2ICl if I2=0.50m,Cl2=o.60m and ICl=5.0

Chemistry

1 answer:

Murljashka [212]2 years ago

8 0

Answer:

0.5

Explanation:

because equilibrium constant is products over reactants

substituting values will give you 0.5

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There are a lot of empty _____ between the particles

Furkat [3]

There are a lot of empty space between the particles

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2 years ago

The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

0.1044 = $\frac{x}{0.1 - x}$

x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

7 0

3 years ago

How many milliliters of a 1.5 m h2so4 are needed to neutralize 35ml sample of a 1.5 m solution?

DochEvi [55]

Answer:

1) 17.5 mL

Explanation:

Hello,

In this case, the reaction between sulfuric acid and potassium hydroxide is:

$H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O$

In such a way, we notice a 1:2 molar ratio between the acid and the base, therefore, at the equivalence point we have:

$2*n_{acid}=n_{base}$

And in terms of concentrations and volumes:

$2*M_{acid}V_{acid}=M_{base}V_{base}$

Thus, we solve for the volume of acid:

$V_{acid}=\frac{M_{base}V_{base}}{2*M_{acid}} =\frac{35mL*1.5M}{2*1.5M} \\\\V_{acid}=17.5mL$

Best regards.

5 0

3 years ago

For the reaction C2H4(g) + H2O(g) --> CH3CH2OH(g)

Dominik [7]

Answer : The value of equilibrium constant for this reaction at 262.0 K is $3.35\times 10^{2}$

Explanation :

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

$\Delta H^o$ = standard enthalpy = -45.6 kJ = -45600 J

$\Delta S^o$ = standard entropy = -125.7 J/K

T = temperature of reaction = 262.0 K

Now put all the given values in the above formula, we get:

$\Delta G^o=(-45600J)-(262.0K\times -125.7J/K)$

$\Delta G^o=-12666.6J=-12.7kJ$

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln k$

where,

$\Delta G^o$ = standard Gibbs free energy = -12666.6 J

R = gas constant = 8.314 J/K.mol

T = temperature = 262.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

$-12666.6J=-(8.314J/K.mol)\times (262.0K)\times \ln k$

$k=3.35\times 10^{2}$

Therefore, the value of equilibrium constant for this reaction at 262.0 K is $3.35\times 10^{2}$

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3 years ago

An element is a pure substance. It can not easily be broken down into parts by physical means.

liberstina [14]

Answer: An element is a pure substance. It can not easily be broken down into parts by physical means.

A)True

B)False

<u>The answer is true. </u>

7 0

3 years ago