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e-lub [12.9K]
3 years ago
5

Enter your answer in the provided box. The partial pressure of CO2 gas above the liquid in a bottle of champagne at 20°C is 4.5

atm. What is the solubility of CO2 in champagne? Assume the Henry's law constant is the same for champagne as it is for water: at 20°C, kH = 3.7 × 10−2 mol/(L·atm).
Chemistry
1 answer:
sineoko [7]3 years ago
6 0

<u>Answer:</u> The molar solubility of carbon dioxide gas is 0.17 M

<u>Explanation:</u>

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{CO_2}=K_H\times p_{liquid}

where,

K_H = Henry's constant = 3.7\times 10^{-2}mol/L.atm

p_{CO_2} = partial pressure of carbonated drink = 4.5 atm

Putting values in above equation, we get:

C_{CO_2}=3.7\times 10^{-2}mol/L.atm\times 4.5atm\\\\C_{CO_2}=0.17mol/L=0.17M

Hence, the molar solubility of carbon dioxide gas is 0.17 M

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Determine the hybrid orbital of this molecule.
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The hybrid orbital of this molecule is sp^3. Hence, option C is correct.

<h3>What is hybridisation?</h3>

Hybridization is defined as the concept of mixing two atomic orbitals to give rise to a new type of hybridized orbitals.

In this compound, sp^3 a hybrid orbital makes I-O bonds. Due to sp^3hybridization iodate should have tetrahedral geometry but because of the presence of lone pair of electrons the shape of IO^{3-} the ion is pyramidal.

The hybrid orbital of this molecule is sp^3. Hence, option C is correct.

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brainly.com/question/23038117

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At 1173 K, Keq = 0.0108 for the following reaction: CaCO3(s) ⇄ CaO(s) + CO2(g) The reaction takes place in a 10.0 L vessel at 11
maxonik [38]

Answer:

The amount of calcium carbonate will increase.

Explanation:

K is the constant of a certain reaction when it is in equilibrium, while Q is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

  • K>Q , reaction will move forward by making more product.
  • K<Q , reaction will move backward by making more reactant.

CaCO_3(s)\rightleftharpoons CaO(s) + CO_2(g)

Equilibrium constant of the reaction = K_{eq}=0.0108

Concentration of CaO=\frac{15.0 g}{56 g/mol\times 10.0L}=0.027 M

Concentration of CO_2=\frac{4.25 g}{44 g/mol\times 10.0L}=0.0096 M

Concentration of CaCO_3=\frac{15.0 g}{100 g/mol\times 10.0L}=0.015 M

Q=\frac{[CaO][CO_2]}{[CaCO_3]}

=\frac{0.027 mol/L\times 0.0096 mol/L}{0.015 mol/L}=0.0174

K_{eq}

This means that equilibrium will move in backward direction by which amount of calcium carbonate will increase.

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