4/325 = 2/unknown temperature
unknown temperature= 2/(4/325)=162.5k
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>
Answer:
0.6257 M is the molarity of solution that is 5.50 percentage by mass oxalic acid.
Explanation:
Mass percentage of oxalic acid = 5.50%
This means that in 100 grams of solution there are 5.50 grams of oxalic acid.
Mass of solution , m = 100
Volume of the solution = V
Density of the solution = d = 1.024 g/mL

V = 97.66 mL = 0.09766 L
(1 mL = 0.001 L)
Moles of oxalic acid = 

The molarity of the solution :

0.6257 M is the molarity of solution that is 5.50 percentage by mass oxalic acid.