Can someone help me with these calculations? I am so lost. The Barometer reading
ID: 515025 • Letter: C
Question
Can someone help me with these calculations? I am so lost.
The Barometer reading is 756.158 torr for both trials
Trial 1 Data Mass of flask A, contents, stopper, and glass tube before reaction Mass of flask A, contents, stopper, and glass tube after reaction Temperature of the gas in flask A Temperature of the gas in flask D mL Volume of oxygen collected Barometer reading (temperature of barometer oC) Aqueous vapor pressure at temperature of gas (average the two temperatures) a torn Trial 2 torrExplanation / Answer
Data
Calculations (We will use trial 1 for sample calculations)
Trial 1
Trial 2
Mass of oxygen
(Mass of flask A, contents, stopper and glass tube before reaction) – (Mass of flask A, contents, stopper and glass tube after reaction) = (153.98 g) – (153.59 g) = 0.39 g
0.39 g
0.45 g
Temperature, absolute
(Temperature in C) + (273 K)
We are asked to average the two temperatures, but which two? Please upload the experiment.
I will use the temperature of flask A, since that is where our oxygen gas is
T = (29 + 273) K = 302 K
302 K
304 K
Corrected barometric pressure
No information is provided about the barometric correction; we are only given the barometric pressure as 756.158 torr
756.158 torr
756.158 torr
Pressure of oxygen alone in flask A
(Barometric pressure) – (aqueous vapor pressure at temperature of gas) = (756.158 torr) – (29.15 torr) = 727.008 torr
727.008 torr
730.158 torr
Volume of oxygen at standard conditions
Use the relation
P1*V1/T1 = P2*V2/T2
where P1 = normal barometric pressure = 760 torr; V1 = volume of gas at standard conditions; T1 = standard temperature = 298 K;
P2 = 727.008 torr; V2 = 27 mL and T2 = 302 K.
Thus,
V1 = P2*V2*T1/(P1*T2)
Plug in values
V1 = (727.008 torr)*(27 mL)*(298 K)/(760 torr).(302 K) = 25.48 mL
25.48 mL
24.11 mL
Moles of dioxygen, O2
Use the relation
n = P*V/RT
where P = 760 torr = 1 atm; V = 25.48 mL = (25.48 mL)*(1 L/1000 mL) = 0.02548 L; T = 298 K and R = 0.082 L-atm/mol.K
===> n = (1 atm)*(0.02548 L)/(0.082 L-atm/mol.K).(298 K) = 0.00104 mole
0.00104 mole
0.000987 mole
Molar volume of O2 at standard conditions
(Volume of oxygen gas under standard conditions in L)/(moles of dioxygen) = (0.02548 L)/(0.00104 mole) = 24.5 L/mol
24.50 L/mol at STP
24.43 L/mol at STP
Data
Calculations (We will use trial 1 for sample calculations)
Trial 1
Trial 2
Mass of oxygen
(Mass of flask A, contents, stopper and glass tube before reaction) – (Mass of flask A, contents, stopper and glass tube after reaction) = (153.98 g) – (153.59 g) = 0.39 g
0.39 g
0.45 g
Temperature, absolute
(Temperature in C) + (273 K)
We are asked to average the two temperatures, but which two? Please upload the experiment.
I will use the temperature of flask A, since that is where our oxygen gas is
T = (29 + 273) K = 302 K
302 K
304 K
Corrected barometric pressure
No information is provided about the barometric correction; we are only given the barometric pressure as 756.158 torr
756.158 torr
756.158 torr
Pressure of oxygen alone in flask A
(Barometric pressure) – (aqueous vapor pressure at temperature of gas) = (756.158 torr) – (29.15 torr) = 727.008 torr
727.008 torr
730.158 torr
Volume of oxygen at standard conditions
Use the relation
P1*V1/T1 = P2*V2/T2
where P1 = normal barometric pressure = 760 torr; V1 = volume of gas at standard conditions; T1 = standard temperature = 298 K;
P2 = 727.008 torr; V2 = 27 mL and T2 = 302 K.
Thus,
V1 = P2*V2*T1/(P1*T2)
Plug in values
V1 = (727.008 torr)*(27 mL)*(298 K)/(760 torr).(302 K) = 25.48 mL
25.48 mL
24.11 mL
Moles of dioxygen, O2
Use the relation
n = P*V/RT
where P = 760 torr = 1 atm; V = 25.48 mL = (25.48 mL)*(1 L/1000 mL) = 0.02548 L; T = 298 K and R = 0.082 L-atm/mol.K
===> n = (1 atm)*(0.02548 L)/(0.082 L-atm/mol.K).(298 K) = 0.00104 mole
0.00104 mole
0.000987 mole
Molar volume of O2 at standard conditions
(Volume of oxygen gas under standard conditions in L)/(moles of dioxygen) = (0.02548 L)/(0.00104 mole) = 24.5 L/mol
24.50 L/mol at STP
24.43 L/mol at STP