Asample of methane gas having a volume of 2.8 l at 25°c and 1.65 atm was mixed with a sample of oxygen gas having a volume of 35l at 31°c and 1.25 atm. the mixture was ignited to form carbon dioxide and water. calculate the volume of carbon dioxide formed at a pressure of 2.5 atm and a temperature of 125°c

When the balanced reaction equation for the reaction of methane gas and O2 is:

CH4(g) + 2O2(g)  → CO2(g) + 2H2O(g)

we will use the ideal gas equation to get the moles of each gas:

1) moles of methane:

PV= n RT 

when p is the pressure =1.65 atm

V is the volume = 2.8 L

R is the ideal gas constant = 0.0821

T is the temperature in Kelvin = 25+ 273=  298 K

∴n of CH4 = 1.65 * 2.8 / 0.0821 * 298 K = 0.189 mole

2) moles of O2:

PV = nRT

when P = 1.25 atm

and V= 35L

and R = 0.0821 L atm / mol K

T = 31 + 273 = 304 K

so, by substitution:

n of O2= 1.25 atm * 35 L / 0.0821 * 304 = 1.75 mol

from the balanced reaction equation, we can see that the molar ratio between O2 and CH4 is 2: 1 

∴ moles of O2 = 2 * .0189 moles of methane = 0.378 mole

so the O2 exists in excess and CH4 is the limiting reactant

when 1 mole of CH4 → 1 mole ofCO2

 1 mol of CO2 → 22.4 L CO2

 volume of CO2 at STP =( 0.189 molesCH4 /1)*(1mol CO2)/(1 mole CH4)*(22.4L CO2)/(1molCO2) = 4.23 L

then we will use this formula to get the correct value of V of CO2 at laboratory conditions:

P1V1 /T1 = P2V2/T2

when at STP conditions:

P1= 1 atm 

V1 = 4.23 L 

T1 = 273

and the laboratories conditions are:

P2 = 2.5 atm 

T2 = 125 + 273 = 398 K

∴ V2 = 1 atm * 4.23 L *398 / 273 * 2.5 atm

∴ V2 = 15.4 L

∴ the volume of CO2 formed at pressure of 2.5 atm and T = 125°C is 15.4 L

Answer: abeexplanation:

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.

What are the options? i can you if you give them.