I need it soon please Table 2: Temperature, Pressure, and Volume Data Temperatur
ID: 304574 • Letter: I
Question
I need it soon please
Table 2: Temperature, Pressure, and Volume Data
Temperature of Tap Water (°C)
Room (or regional) Pressure (atm)
Initial Volume
of Air (mL)
Final Volume of Air
(after reaction) (mL)
Volume of O2Collected
(Final Volume - Initial Volume)
12
1
88mL
95
7
Table 3: Reaction Time Data
Time Reaction Started
Time Reaction Ended
Total Reaction Time
7;00
7:03
3 minutes
What do you think would happen if you added more than 5 mL of yeast solution to the H2O2?
What do you think would happen if you added more than 5 mL of H2O2to the 5 mL of yeast solution?
What was going on in the graduated cylinder as the H2O was pushed out?
How would the number of moles (n) of O2change if the atmospheric pressure was reduced in half but all other variables stayed the same?
Using the Ideal Gas Law (PV = nRT), calculate the grams of O2 produced in the reaction. (Hint: solve for n, and then convert moles to grams. Don’t forget to convert your temperature from Celsius to Kelvin.) Show your work.
Temperature of Tap Water (°C)
Room (or regional) Pressure (atm)
Initial Volume
of Air (mL)
Final Volume of Air
(after reaction) (mL)
Volume of O2Collected
(Final Volume - Initial Volume)
12
1
88mL
95
7
Explanation / Answer
The reaction that is occurring is the decomposition of H2O2 catalysed with yeast, this is represented by:
2 H2O2 = 2 H2O + O2
If H2O2 is added to an excess of yeast, it would have all possible means to fulfill the reaction, so it would be faster.
If you add H2O2 in excess to the yeast, it would be a little slower and less reaction, since it would require more yeast to complete the relationship.
As we already said, a yeast-catalyzed H2O2 decomposition reaction was taking place in the graduated cylinder.
The pressure is directly proportional to the moles, this can be seen in the equation of ideal gases (PV = nRT), therefore, if the pressure is reduced by half, because the moles are also produced in half.
For the ideal gas equation cleared we have, n = PV / RT and replacing:
n = 1 atm * 0.007 L / 0.082 atm * L / mol * K * 285 K = 3x10 ^ -4 moles
Mass O2 = 3x10 ^ -4 moles * (32 g / 1 mol) = 0.0096 gr O2.