Green plants use light from the sun to drive photosynthesis. photosynthesis is a chemical reaction in which water (h2o) and carbon dioxide (co2) chemically react to form the simple sugar glucose (c6h12o6) and oxygen gas (o2). what mass of oxygen gas is produced by the reaction of 5.2 g of water? be sure to have the correct # of sig figs.

Answer : The mass of oxygen gas produced will be 4.45 grams

Explanation : Given,

Mass of water = 5.2 g

Molar mass of water = 18 g/mole

Molar mass of = 32 g/mole

The balanced chemical reaction will be,

First we have to calculate the moles of water.

Now we have to calculate the moles of oxygen.

From the balanced chemical reaction, we conclude that

As, 6 moles of water react to give 6 moles of oxygen gas

So, 0.139 moles of water react to give 0.139 moles of oxygen gas

Now we have to calculate the mass of oxygen gas.

Therefore, the mass of oxygen gas produced will be 4.45 grams

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

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.