What will the volume of 2.5 moles of gas be at 281K and 102 kPa?

V = 57.23

Explanation:

Using the ideal gas law (PV = nRT), we plug in the variables for moles, Kelvin, and pressure: (102 * V = 2.5 * R * 281). We plug in the correct R constant value for the type of pressure we are using (atm, Kpa, or mmHg) with values of (0.0821, 8.31, or 62.4) respectively. Since we are using kPa in this example, we use 8.31 for the R constant so we get: (102 * V = 2.5 * 8.31 * 281). We solve the right side to get: (102 * V = 5837.775). We divide 102 sides from both sides to get that (V = 57.233)

water   molecules stay close to each other through cohesion, the collective action of hydrogen bonds between water molecules. cohesion plants draw water upwards from the root through the stem's transport tubes (xylem) to the leaf. this transport occurs even in large trees and against gravity. transpiration, evaporation of water from plant leaves, creates a tension on water molecules being pulled up from the stem and roots.

water also has high (the attraction of one substance to another) properties because of its polar nature. adhesion of water molecules to plant cell walls counteract gravity while water is transported from roots to leaves.

water exhibits a very high surface tension when compared to other liquids. surface tension is a property of a liquid that holds the surface together and allows it to resist an external force. by capillary action, the water forms concave menisci inside the pores. the high surface tension of water pulls the concavity outwards, generating enough force to lift water from the roots to the leaves of plants.

thus, cohesion, along with adhesion and surface tension creates a capillary action that keeps water molecules interacting and moving through the plants out to the leaf cells.

therefore, option d) is the correct answer.

explanation:

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.