What mass of Calcium Bromide (below) is needed to make 0.500 L of
0.554 M solution? (hint: find moles first, then convert to grams)

55.36 is the mass of Calcium Bromide is needed to make 0.500 L of  0.554 M solution.

Explanation:

Data given:

mass of Ca = ?

molarity of the calcium bromide solution = 0.554 M

Volume of the calcium bromide solution= 0.5 L

First the moles of calcium bromide is calculated as:

molarity =

number of moles = 0.554 x 0.5

number of moles = 0.277 moles of calcium bromide

atomic mass of calcium bromide = 199.89 grams/mole

mass = atomic mass x number of moles

mass of calcium bromide = 199.89 x 0.277

                                           = 55.36 grams

mass of calcium bromide required is 55.36 grams to make solution of 0.554 M in a solution of 0.5 litres.

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.

question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation:   if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

question answered by

(jacemorris04)