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Arte-miy333 [17]
2 years ago
7

sing any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 25.0°C for the following reaction

. C(s)+ 2Cl2(g)→ CCl4(g) Round your answer to 2 significant digits.
Chemistry
1 answer:
prisoha [69]2 years ago
3 0

Answer:

2.76 × 10⁻¹¹  

Explanation:

I don’t have access to the ALEKS Data resource, so I used a different source. The number may be different from yours.

1. Calculate the free energy of formation of CCl₄

                         C(s)+ 2Cl₂(g)→ CCl₄(g)

ΔG°/ mol·L⁻¹:       0         0         -65.3

ΔᵣG° = ΔG°f(products) - ΔG°f(reactants) = -65.3 kJ·mol⁻¹

2. Calculate K

\text{The relationship between $\Delta G^{\circ}$ and K  is}\\\Delta G^{\circ} = -RT \ln K

T = (25.0 + 273.15) K = 298.15 K

\begin{array}{rcl}-65 300 & = & -8.314 \times 298.15 \ln K \\65300& = & 2479 \ln K\\26.34 & = & \ln K\\K& = & e^{26.34}\\&= & \mathbf{2.76 \times 10}^{\mathbf{11}}\\\end{array}

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Explanation:

Vapor pressure is defined as the pressure exerted by vapors or gas on the surface of a liquid.

Vapor pressure is inversely proportional to the number of solute particles. Hence, more will be the solute particles lower will be the vapor pressure and vice-versa.

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It dissociates to give two particles.

(b)  Ca(ClO_{4})_{2} \rightarrow Ca^{2+} + 2ClO^{-}_{4}

Total number of particles it give upon dissociation are 1 + 2 = 3. Hence, it gives 3 particles.

(c)   Al(ClO_{4})_{3} \rightarrow Al^{3+} + 3ClO^{-}_{4}

Total number of particles it give upon dissociation are 1 + 3 = 4. Hence, it gives 4 particles.

(d)  Surcose being a cobvalent compound doe not dissociate into ions. Therefore, there will be only 1 particle is present.

(e)   NaCl \rightarrow Na^{+} + Cl^{-}

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2 years ago
Part a how many grams of xef6 are required to react with 0.579 l of hydrogen gas at 6.46 atm and 45°c in the reaction shown belo
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First, let us find the corresponding amount of moles H₂ assuming ideal gas behavior.

PV = nRT
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The formula to be used for this problem is as follows:

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6 0
1 year ago
Determine the molar solubility of CuCl in a solution containing 0.050 KCl. Ksp of CuCl is 1.0 x 10-6
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Answer:

2.0x10^{-5}\frac{mol}{L}

Explanation:

Hello!

In this case, since the dissolution of copper (I) chloride is:

CuCl(s)\rightarrow Cu^++Cl^-

And its equilibrium expression is:

Ksp=[Cu^+][Cl^-]

We can represent the molar solubility via the reaction extent as x, however, since there is 0.050 M KCl we immediately add such amount to the chloride ion concentration since KCl is readily ionized; therefore we write:

1.0x10^{-6}=(x)(0.050+x)

Thus, solving for x, we obtain:

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By using the quadratic equation, we obtain:

x_1=2.0x10^{-5}M\\\\x_2=-0.05M

Clearly, the solution is x_1=2.0x10^{-5}M because no negative results are

allowed. Therefore, the molar solubility is:

2.0x10^{-5}\frac{mol}{L}

Best regards!

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