5) What is the difference between CO2 and XeCl2? The difference between SCl2, SCl4, and SCl6?

CO₂ has 4 + 2(6) = 16 valence electrons

CO₂ have linear molecular structures with a 180° bond angle. CO₂ is nonpolar  because the individual bond dipoles cancel each other out

XeCl₂ has 8 + 2(7) = 22 valence electrons.

There are five pairs of electrons about the central Xe atom. The structure will be based  on a trigonal bipyramid geometry. The most stable arrangement of the atoms in XeCl₂ is  a linear molecular structure with a 180° bond angle. It is nonpolar because the Xe−Cl bond dipoles  and lone pairs around Xe are arranged in such a manner that they cancel each other out.

(b)

SCl₆ is nonpolar because it has 6 bonding pairs and no lone pairs, giving it an octahedral shape. When you draw the vectors, you can see that the resultant is 0, making it nonpolar.

SCl₄ on the other hand is polar because it has 4 bonding pairs and one lone pair, giving it a trigonal pyramidal shape.

SCl₂ is also polar due to reasons similar to those in the case of SCl₄.

passed in september 1935, these laws are known as nuremberg laws.   they forbid jews from marrying but not directly forced them to divorce aryans. what is most noticeable is the fact that the jews were defranchised, meaning by this that they were not expected to have any civil rights, including the right to vote. the right answer is deprived of the right to vote.

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.