Answer:
Explanation:
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In this case, we are asked to compute the by mass percent for the given 3.2 g of ethylene glycol in 43.5 g of water. In such a way, since the by mass percent is computed as follows:
Whereas the solute is the ethylene glycol and the solvent the water, therefore we obtain:
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Answer : The value of equilibrium constant (K) is, 424.3
Explanation : Given,
Concentration of at equilibrium = 0.067 mol
Concentration of at equilibrium = 0.021 mol
Concentration of at equilibrium = 0.040 mol
The given chemical reaction is:
The expression for equilibrium constant is:
Now put all the given values in this expression, we get:
Thus, the value of equilibrium constant (K) is, 424.3
If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.
The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. We can calculate the concentration of CO₂ using Henry's law.
We can calculate the mass of CO₂ in 1.1 L considering its molar mass is 44.01 g/mol.
Now, we will repeat the same procedure for a partial pressure of 1.28 atm.
The mass of CO₂ released will be equal to the difference in the masses at the different pressures.
If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.
Learn more: brainly.com/question/18987224
<em>The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO₂ gas (in g) will be released from 1.1 L of the carbonated water when the partial pressure of CO2 is lowered to 1.28 atm? At 25 ºC, the Henry’s law constant for CO₂ dissolved in water is 1.65 x 10⁻³ M/atm, and the density of water is 1.0 g/cm³.</em>