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# The Molar Volume of Gas (Coursework Sample)

Instructions:

INSTRUCTIONS: FILL IN THE WORKSHEET ABOUT THE MOLAR VOLUME OF GAS. SHOW RELEVANT COMPUTATIONS.
THE COURSEWORK ON CHEMISTRY WAS MORE ON COMPUTATIONS THAN ESSAY, IT TACKLED COMPARISONS OF ACCURACY AND PRECISION BASED ON THE RESULTS ACQUIRED. DISCUSSION ON RESULTS WHERE FACTORS SUCH AS TIME TO COLLECT DATA WAS RELEVANT.

source..
Content:

Name: _________________________ Partner: ________________________ Lab Period: ______
Data and Calculations: The Molar Volume of Gas

Trial 1

Trial 2

1. Weight of Mg ribbon

0.0342 g

0.0350 g

2. Moles of Mg

1.41×103 mol

1.44×103 mol

3. Temperature of H2O

21.2°C

21.2°C

4. Barometric Pressure

729.308 mmHg

729.308 mmHg

5. Volume of Hydrogen

36.3 mL

37.2 mL

6. Vapor Pressure of H2O

18.892 mmHg

18.892 mmHg

7. Pressure of dry H2

mmHg

mmHg

8. Volume of dry H2 at STP

0.032 L

0.033 L

9. Molar Volume of H2 (V0)

22.83 L/mol

22.86 L/mol

10. Average of Molar volume of H2

22.85 L/mol

11. Percent Deviation

1.92 %

2.05 %

In the space below, show your work, including units, for finding the molar volume of dry H2 at STP.
31318201079502HCl(aq) + Mg(s) H2(g) + MgCl2(aq)
Trial 1: 0.0342 g Mg1 mol Mg24.305 g Mg=0.00140171179 mol Mg1 mol H21 mol Mg=1.41×103mol H2
Trial 2: 0.0350 g Mg1 mol Mg24.305 g Mg=0.0014400329 mol Mg1 mol H21 mol Mg=1.44×103mol H2
33985201270TRIAL 2Volume of dry H2 at STP:P1V1T1=P2V2T2V2=P1V1T1(T2)P2V2=729.308 mmHg×0.0372 L296 K(273 K)760 mmHgV2=0.032923899 LMolar Volume of H2 at STP0.032923899 L0.0014400329 mol H2=22.86 L/molPercent Deviation(Accepted molar volume at 21.2°C= 22.4 L)%Deviation= 22.4 L-22.86 L 22.4 L100%=2.05%00TRIAL 2Volume of dry H2 at STP:P1V1T1=P2V2T2V2=P1V1T1(T2)P2V2=729.308 mmHg×0.0372 L296 K(273 K)760 mmHgV2=0.032923899 LMolar Volume of H2 at STP0.032923899 L0.0014400329 mol H2=22.86 L/molPercent Deviation(Accepted molar volume at 21.2°C= 22.4 L)%Deviation= 22.4 L-22.86 L 22.4 L100%=2.05%-152401270TRIAL 1Volume of dry H2 at STP:P1V1T1=P2V2T2V2=P1V1T1(T2)P2V2=729.308 mmHg×0.0363 L296 K(273 K)760 mmHgV2=0.032127 LMolar Volume of H2 at STP0.032127 L 0.0014071179 mol H2=22.83 L/molPercent Deviation(Accepted molar volume at 21.2°C= 22.4 L)%Deviation= 22.4 L-22.83 L 22.4 L100%=1.92%020000TRIAL 1Volume of dry H2 at STP:P1V1T1=P2V2T2V2=P1V1T1(T2)P2V2=729.308 mmHg×0.0363 L296 K(273 K)760 mmHgV2=0.032127 LMolar Volume of H2 at STP0.032127 L 0.0014071179 mol H2=22.83 L/molPercent Deviation(Accepted molar volume at 21.2°C= 22.4 L)%Deviation= 22.4 L-22.83 L 22.4 L100%=1.92%
1 After the reaction was complete you equalized the pressures between the room and the tube.
1 Before you equalized the pressures, on your first trial, assume that the level of the gas in the tube was above the level of the water in the smaller jar (before transferring to the large tank to equalize the pressure).
2 Which was at a higher pressure, the lab room, or the gases in the collection tube?
The lab room has higher pressure.
2 After the reaction was completed, you waited 5 minutes to measure the gas volume.
3 Why was it important to wait rather than collect data immediately after the reaction ended? Give specific reason(s).
It was important to wait to let the reaction complete. Without doing so, letting the reaction remain incomplete will lead to erroneous results as there i

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