Pf3i2 is a nonpolar molecule. based on this information, determine the f−p−f bond angle, the i−p−i bond angle, and the f−p−i bond angle

The nonpolar molecule is with an electrostatic potential that is spherically symmetric if averaged across time, since electrons fluctuate, temporary dipoles exist anywhere at a given instant in a nonpolar molecule.

bond angle is

bond angle is

bond angle is

Its combination is made up of a mixture bound together by chemical bonds & arranged in such a way that the arrangement of charge is symmetrical. The Nonpolar substances lack polarity. In a solution, they will not transform into ions.When certain bond angles occur in molecules, its net dipole moment cancels out and the molecule becomes non-polar.

Therefore, the angle is " , , and ".

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PF3I2 is a trigonal bipyramidal molecule whose image is shown in the answer below.

The molecule PF3I2 is trigonal bipyramidal. A trigonal pyramidal molecule has two axial bonds and three equatorial bonds. The two iodine atoms are positioned at axial positions while the fluorine atoms are positioned at equatorial positions.

The molecule is nonpolar because the individual dipoles due to fluorine and iodine atoms cancel out. Looking at the structure of the molecule;

The F - P - F bond angle is 120°The I - P - I bond angle is 180°The I - P - F bond angle is 90°

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We can’t you if you only tell us that you have to show us what the answers are

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.