Calculate the number of pounds of CO2 released into the atmosphere when a 12.0-gallon tank of gasoline is burned in an automobile engine. Assume that gasoline is primarily octane, C8H18 and that the density of gasoline is 0.692 g. mL-1 (this assumption ignores additives). Also assume complete combustion.

Useful conversion factors:
1 gallon = 3.785L
1 Kg = 2.204 lb

lb

106.95 lb of CO2

Explanation:

Equation of reaction:

2C8H18 + 25O2 = 8CO2 + 18H2O

Mass of C8H18 = density × volume = 0.692×12×3.785×1000 = 31430.64 g

From the equation of reaction above,

2 moles of C8H18 (228 g) produced 8 moles of CO2 (352 g)

31430.64 g of C8H18 will produce 31430.6×352/228 = 48524.50 g of CO2 = 48524.50/1000 = 48.52450 kg of CO2 = 48.52450×2.204 = 106.95 lb of CO2

213.89 lb of CO2.

Explanation:

Equation for the reaction:

C8H18 + 25/2O2(g) --> 8CO2(g) + 9H2O(g)

Given:

Volume of gasoline = 12 gallon

Converting gallon to ml,

12 gallon * 3.785 l/1 gallon * 1000 ml/1 l

= 45420 ml

Density of the gasoline = 0.692 g/ml

Mass = density * volume

= 45420 * 0.692

= 31430 g

Molar mass of octane = (8*12) + (18*1)

= 114 g/mol.

Number of moles = mass/molar mass

= 31430/114

= 275.702 mol.

From the above equation, 1 mole of octane was completed burnt to give off 8 moles of CO2.

By stoichiometry,

Number of moles of CO2 = 275.702 * 8

= 2205.614 mol of CO2.

Molar mass of CO2 = 12 + (2*16)

= 44 g/mol

Mass of CO2 = number of moles * molar mass

= 2206.614 * 44

= 97047.02 g

Converting g to pound,

= 97047.02 g *1 kg/1000 g * 2.204 lb/1kg

= 213.89 lb of CO2.

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.

c. 75.7

i think that is the right