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
1 answer:
<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:
where,
= Henry's constant =
= partial pressure of carbonated drink = 4.5 atm
Putting values in above equation, we get:
Hence, the molar solubility of carbon dioxide gas is 0.17 M
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Answer:
1-Propene has eight sigma bonds and one pi bond.
Explanation:
The lewis structure of 1-Propene is shown below. All sigma bonds are highlighted by blue color and the pi bond is highlighted by red color.
Carbon having four unpaired electrons in its excited state can form four covalent (sigma bonds) bonds. In given structure the orbitals containing unpaired electron in sp³ hybridized C undergo head to head overlap with three hydrogen atoms having single unpaired electron and orbital of other sp² hybridized carbon in same fashion. So, four sigma bonds are formed by sp³ hybridized carbon atom.
The doubly bonded carbon atoms also form sigma bond but two unpaired electrons of these two carbons undergo the formation of pi bond. This pi bond forms when the orbitals overlap along the plane perpendicular to the existing sigma bond resulting in the formation of single pi bond.
1-Propene has eight sigma bonds and one pi bond.
Explanation:
The lewis structure of 1-Propene is shown below. All sigma bonds are highlighted by blue color and the pi bond is highlighted by red color.
Carbon having four unpaired electrons in its excited state can form four covalent (sigma bonds) bonds. In given structure the orbitals containing unpaired electron in sp³ hybridized C undergo head to head overlap with three hydrogen atoms having single unpaired electron and orbital of other sp² hybridized carbon in same fashion. So, four sigma bonds are formed by sp³ hybridized carbon atom.
The doubly bonded carbon atoms also form sigma bond but two unpaired electrons of these two carbons undergo the formation of pi bond. This pi bond forms when the orbitals overlap along the plane perpendicular to the existing sigma bond resulting in the formation of single pi bond.