pro at chem broseph

[velkira]

Molar Mass of a Gas -- Volume of Flask

The following data were obtained while doing the experiment using three Erlenmeyer flasks, each equipped with a septum stopper, and labeled #1, #2 and #3.
Flask Weight of
flask
+ stopper
+ air (g) Weight of
flask
+ stopper
+ compound (g) Mass of
water to
fill flask
(no stopper)
#1

103.0640

103.1397

149.20
#2

101.8738

101.9620

151.20
#3

105.7170

105.8315

152.00
est. unc.

±0.0002

±0.0002

±0.10

Given the following additional data,

pressure = 746.11 ± 0.10 mmHg

temperature = 17.50 ± 0.10 °C

density of water = 0.9986883 g/mL (uncertainty negligible)

volume displaced by septum stopper (Vs)= 5.90 ± 0.10 mL


NOTE: (In the actual experiment you will be expected to determine all of the above values and estimate their uncertainties. In particular, you should think carefully about the steps involved in determining the mass of water required to fill the flask.)

Now calculate:

The volume of water to fill flask #2.

1pts
Tries 0/5

The estimated uncertainty in the volume of water to fill flask #2.
±
1pts
Tries 0/5

The gas volume in flask #2, when stoppered.

1pts
Tries 0/5

The estimated uncertainty of the gas volume in flask #2, when stoppered.
±
1pts
Tries 0/5











Mass of a Volume of Air

Flask Weight of
flask
+ stopper
+ air (g) Weight of
flask
+ stopper
+ compound (g) Mass of
water to
fill flask (g)
(no stopper)
#1

103.0640

103.1397

149.20
#2

101.8738

101.9620

151.20
#3

105.7170

105.8315

152.00
est. unc.

±0.0002

±0.0002

±0.10

pressure = 746.11 ± 0.10 mm of Hg

temperature = 17.50 ± 0.10 °C

density of water = 0.9986883 g/mL (uncertainty negligible)

volume displaced by septum stopper (Vs)= 5.90 ± 0.10 mL

Calculate:

The density of dry air.

dair(g/mL) = [0.001293/(1 + 0.00367×T(°C))]×[P(mmHg)/760]
± 6×10-7 g/mL (using the numbers given here) An Example
1pts
Tries 0/5

The mass of dry air in flask #2, when stoppered.

1pts
Tries 0/5

The estimated uncertainty in the mass of dry air in flask #2, when stoppered.
±
1pts
Tries 0/5







Mass of a Volume of Unknown Gas

Flask Weight of
flask
+ stopper
+ air (g) Weight of
flask
+ stopper
+ compound (g) Mass of
water to
fill flask (g)
(no stopper)
#1

103.0640

103.1397

149.20
#2

101.8738

101.9620

151.20
#3

105.7170

105.8315

152.00
est. unc.

±0.0002

±0.0002

±0.10

pressure = 746.11 ± 0.10 mm of Hg

temperature = 17.50 ± 0.10 °C

density of water = 0.9986883 g/mL (uncertainty negligible)

volume displaced by septum stopper (Vs)= 5.90 ± 0.10 mL

Calculate:

The mass of empty (hypothetically evacuated) flask #2, when stoppered (mass of stopper included).

1pts
Tries 0/5

The estimated uncertainty in the mass of empty (hypothetically evacuated) flask #2, when stoppered (mass of stopper included).
±
1pts
Tries 0/5

The mass of compound in (stoppered) flask #2.

1pts
Tries 0/5

The estimated uncertainty in the mass of compound in (stoppered) flask #2.
±
1pts
Tries 0/5






Molar Mass of a Gas -- Calculation

Flask Weight of
flask
+ stopper
+ air (g) Weight of
flask
+ stopper
+ compound (g) Mass of
water to
fill flask (g)
(no stopper)
#1

103.0640

103.1397

149.20
#2

101.8738

101.9620

151.20
#3

105.7170

105.8315

152.00
est. unc.

±0.0002

±0.0002

±0.10

pressure = 746.11 ± 0.10 mm of Hg

temperature = 17.50 ± 0.10 °C

density of water = 0.9986883 g/mL (uncertainty negligible)

volume displaced by septum stopper (Vs)= 5.90 ± 0.10 mL

Calculate:

The molar mass of the compound, determined using flask #2.
(enter units of g/mol)

1pts
Tries 0/5

The estimated uncertainty in the molar mass of the compound, determined using flask #2.
±
1pts
Tries 0/5









Molar Mass of a Gas -- Half-Range, Mean and Standard Deviation

Flask Weight of
flask
+ stopper
+ air (g) Weight of
flask
+ stopper
+ compound (g) Mass of
water to
fill flask (g)
(no stopper)
#1

103.0640

103.1397

149.20
#2

101.8738

101.9620

151.20
#3

105.7170

105.8315

152.00
est. unc.

±0.0002

±0.0002

±0.10

pressure = 746.11 ± 0.10 mm of Hg

temperature = 17.50 ± 0.10 °C

density of water = 0.9986883 g/mL (uncertainty negligible)

volume displaced by septum stopper (Vs)= 5.90 ± 0.10 mL

The above calculations could be repeated using the data for the other two flasks to give:

The molar mass of the compound, determined using flask #3 = 47.99 g/mol.

The molar mass of the compound, determined using flask #1 = 41.80 g/mol).

Now calculate:

The half-range of the determined molar masses (using the molar mass values determined from all three flasks).
±
1pts
Tries 0/5

The mean molar mass, rounded to the appropriate number of significant figures (based on the magnitude of the standard deviation -- see below).

1pts
Tries 0/5

The standard deviation rounded to either two or one "significant figures", depending upon whether or not the first non-zero figure is less than 2.
±
1pts
Tries 0/5

[Broseph]

Your prelabs suck lol.

[lancer]

Originally Posted by Broseph

Your prelabs suck lol.

[lancer]

and nigga, we ain't no yahoo answers here. do your own homework or go to tutoring sessions.

[Clancer]

fuck chemistry. I passed it with like a 70 even

[lancer]

Originally Posted by Clancer

fuck chemistry. I passed it with like a 70 even

real science>computer science

[velkira]

so clancer is into computer science. HMMmmmmmmmmmmmmmmmmmm

also yahoo answer fails. unless for lulz like

"what's a good comeback when a guy tells you to go make a sandwich?"

standard female responses to make the OP feel good

pro dude
"bitch better comeback with the damn sandwich."

[patt]

shove that flask up ur ass

[velkira]

ty everyone. but mostly broseph.

[Clancer]

Originally Posted by lancer

real science>computer science

FEEL MY ALGORITHMS ALL OVER YOU

[velkira]

my first c++ lab. went terrible

1. state the problem
2. analyze the problem
3. make an algorithm

You are designing a highway bridge. In particular you are trying to determine how to put the road surface on the bridge. You live in an area where the maximum summertime temperature is 40 degrees centigrade and the minimum winter temperature is -30 degrees centigrade. The coefficient of linear expansion, α, measures the fractional change in length per unit change in temperature (fractional change = change in length per unit length). A coefficient of linear expansion will be no larger than 0.0001. Consider a section of road surface with length L1 (or L2) at temperature T. If the temperature T increases by ∆T the length of the section of road surface with L1=3 will increase by 3∆L to L1+ 3∆L. If the temperature decreases by ∆T a section of road surface with length L2=5 will decrease in length by 5∆L to L2-5∆L. The amount the length changes (per unit length of road surface L=1) by is ∆L = ∆Tα . You are to write a program to determine two things, the maximum size of the space between road sections and the amount of space to leave between the sections of road surface on the day they are installed. At the maximum temperature there should be no space between the sections of road surface. You must leave enough room between sections of road surface for the road surface to expand to the length it would have at the maximum temperature of 40 degrees. The user will provide the temperature on the day the installation takes place, the length of each section of road surface to be installed (at the temperature when the installation occurs) and the coefficient of linear expansion of the material being used in the road surface.

[Clancer]

Is that for a first year programming class? In my C++ labs we never had anything that complex.

also, subtle brag post: this was a project I had to do for my concepts of programming languages class last semester. wrote it in C++, got a 96 on it

[velkira]

it's a first year course that is mandatory for engineers. i took it for lulz.

was not very lulz at all. terrible.