Can you help me with the explaining and identifying below? A found a link in which the lab is similar to what I did in class. Here it is:
icn2.
umeche.
maine.
edu
/newnav
/NewNavigator
/Labs
/Deter_1493.htm
thanks for the help!
Indicate whether the following conditions would make your calculated of R high, low, or have no effect. Give reasons for you answer.
a)leak in the apparatus
high R value
with larger final mass of water
b)potassium chlorate not dry
I assume that if potassium chlorate is wet, this will effect the decomposition of it. Therefore, potassium chlorate will experience a smaller decrease in mass than supposed to? Therefore, giving me a large R.
c)water levels not equalized before and after heating
either high or low R (depends)
If the pressure in the bottle is greater than beaker, water will flow from the bottle into the beaker. If less, water will be forced into the bottle by the beaker. With the bottle greater than beaker, I will obtain a larger final mass of water, increase the volume of gas, and increase the R. With less, I get the opposite.
d)apparatus not cooled completely before measuring the displaced water
I'm not sure with high temperature objects will weigh less or more. Let's say that the displaced water weighs less. Then, I will obtain a larger final mass of water, increase the volume of gas, and increase the R. If weighs more, then the opposite effect happens.
I've not done this experiment as such AND you said the posted one was similar (but not exactly the same?). You may want to get a second opinion on my answers. Keep in mind the equation.
R = PV/nT where n = g/32
a)leak in the apparatus
high R value
with larger final mass of water
A leak in the apparatus means oxygen can escape to the atmosphere (meaning a lower volume of water is expelled) OR a leak in the Florence flask means less water is expelled. Therefore, lower V means lower R.
b)potassium chlorate not dry
I assume that if potassium chlorate is wet, this will effect the decomposition of it. Therefore, potassium chlorate will experience a smaller decrease in mass than supposed to? Therefore, giving me a large R.
I'm really unsure of this but I think the following. A wet KClO3/MnO2 mixture at the start means we evaporate the water first (before any O2 is generated) but I assume after the water is gone (causing pressure to be generated and water to to collected in the beaker) that O2 will be generated as usual. Then at the end, if we let the system cool to room T, the extra water pushed out of the test tube in the beginning will be collected in the Florence flask and be in equilibrium with the vapor pressure of the water at that T; therefore, I don't think that part will make any difference. I think the difference will come because the difference in mass of the KClO3/MnO2 mixture before and after the experiment (didn't you weigh it both times and take the difference in mass as the mass of oxygen generated?) will be too large (the difference will be too large) and the mass will appear to be too large which will cause the experimental R value to be too low.
c)water levels not equalized before and after heating
either high or low R (depends)
If the pressure in the bottle is greater than beaker, water will flow from the bottle into the beaker. If less, water will be forced into the bottle by the beaker. With the bottle greater than beaker, I will obtain a larger final mass of water, increase the volume of gas, and increase the R. With less, I get the opposite.
This looks ok to me if I understand it correctly.
d)apparatus not cooled completely before measuring the displaced water
I'm not sure with high temperature objects will weigh less or more. Let's say that the displaced water weighs less. Then, I will obtain a larger final mass of water, increase the volume of gas, and increase the R. If weighs more, then the opposite effect happens.
I think the effect here is the effect of pressure. Too high T means pressure in the flask is too high and that means too much water expelled into the beaker.
I note that GK has been answering some of these questions lately; perhaps you can get his opinion, also. Apparently he has seen this done in practice.
Can you help me with the problem above, GK? THanks!
Comment:
Lab questions are difficult to judge because we do not know what the instructions were, how close they were followed, and what assumptions can be made in answering the post-lab questions. Here is my take on the questions on the determination of the Gas Constant, R.
I assume the apparatus used consists of a test tube O2 generator with the KCLO3 in it, connected to a Florence flask full of water. The oxygen generated by heating the KClO3 pushes water out of the Florence flask into a beaker. After evolution of O2 is complete, the apparatus is allowed to cool to room temperature. Then the flask's and/or the beaker's position is raised or lowered so that the water levels are the same in both containers
Indicate whether the following conditions would make your calculated of R high, low, or have no effect. Give reasons for you answer.
a)Leak in the apparatus...
R = PV/nRT, and n = (mass) / (formula mass)
The value of R will be smaller because the value of V in the numerator will be smaller because of the leak.
b)Potassium chlorate not dry...
R = PV/nRT, and n = (mass) / (formula mass). The mass of 2.5g will yield a value of n that is larger than the actual number of moles of KCLO3. Since n is in the denominator of the fraction in the expression for R, it will cause the value of R to be smaller.
c)Water levels not equalized before and after heating ...
R varies in direct proportion to the volume of water transferred which in turn depends on the pressure (water level) in the flask relative to the pressure in the beaker. High or low value for R, depending on which level is higher.
d)apparatus not cooled completely before measuring the displaced water...
A hot O2 gas coming out of the tube will occupy a larger volume before we have thermal equilibrium. How will this affect the value you get for R? Once more, R = PV/nRT. A larger apparent volume will cause a large value for R.
To GK. Thanks for answering with a second opinion. It's nice to know I've not lost all of my reasoning ability. But I wonder about it at times.
It is great working with you again DrBob. I answered this same question on another network that allows links. The student was kind enough to provide a link to the procedure with a diagram, otherwise I would not have been able to visualize the experiment. My usual approach is to look at the functional dependance of the experimental variables and the errors associated with them. That comes up occasionally in the single post retirement class I still teach. Best regards.
Glad to have you check in on this site. You may want to contact Bob Pursley. He can give you details on acquiring the ability to post links. Students can not post links but the tutors may if it has been enabled for them. If you don't know his email address, you can post a new note under post a new question on this board. I may have your address, assuming you are TchrGK from the old AOL boards.
a) Leak in the apparatus: A leak in the apparatus would result in a high R value. This is because if there is a leak, the gas produced during the reaction would escape, leading to a decrease in the final mass of water. Since R is calculated using the final mass of water, a smaller final mass would result in a higher R value.
b) Potassium chlorate not dry: If the potassium chlorate is not dry, it will affect its decomposition. When heated, the wet potassium chlorate will not decompose completely, resulting in a smaller decrease in mass compared to the expected value. This would lead to a larger final mass of water, and ultimately, a high R value.
c) Water levels not equalized before and after heating: The effect on R value in this case depends on the difference in pressure between the bottle and beaker. If the pressure in the bottle is greater, water will flow from the bottle into the beaker, resulting in a larger final mass of water. This would increase the volume of gas and subsequently result in a high R value. On the other hand, if the pressure in the beaker is greater, water will be forced into the bottle, resulting in a smaller final mass of water. This would decrease the volume of gas and lead to a low R value.
d) Apparatus not cooled completely before measuring the displaced water: If the apparatus is not cooled completely, the displaced water may be affected by thermal expansion or contraction. If the displaced water weighs less due to expansion, it would result in a larger final mass of water and a high R value. Conversely, if the displaced water weighs more due to contraction, it would result in a smaller final mass of water and a low R value. Therefore, the effect on R value depends on whether the displaced water experiences expansion or contraction due to incomplete cooling.