How many liters of h2o are produces @ stp by the combustion of 30g of ch4 ch4+> co2 +2h2o

CH4+2O2———> CO2 +2H2O

M(CH4)=16.0 g/mol
30g CH4*1mol/16.0 g    - mol of CH4
30g CH4*(1mol/16.0 g)*(2 mol H2O/1 mol CH4) =15  mol H2O
30g CH4*(1mol/16.0 g)*(2 mol H2O/1 mol CH4)*22.4L/1mol=84 L H2O as gas, but at STP water should be liquid,so

15 mol H2O*18g H2O/1 mol=270g of H2O 
270g*1ml/1g= 270 ml =0.27 L of liquid water, in reality it should be like this, 
but I do not know what people who created this problem wanted H2O as liquid or as gas

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.

the catch is that the top numbers are atomic masses the bottom numbers are the numbers of protons and electrons in an uncharged atom. they are the number of that atoms as you would find them on the periodic table. further, the numbers on the left are used to determine the missing numbers on the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer