In a laboratory, you produce a quantity of the radioactive isotope unobtainium-253. Over the course of several weeks, the unstable isotope decays, and you measure the amount of unobtainium-253 remaining in the sample. You obtain the following data: The half-life of unobtainium-253 is _Incorrect. days.

After 2 half-lives, there will be _Incorrect. grams of unobtainium-253 remaining in this sample.

After _Incorrect. days, there will be 1.5 grams of unobtainium-253 remaining in this sample.

it has to be understood that the sea water contains a huge amount of sodium and chlorine ions and if they enter the cell of the atlantic cod fish, the cell will shrivel and the fish will die. to avoid such a condition, the fish uses active transport for the purpose of removing the ions against the gradient.

explanation:

the atlantic cod is not a maritime osmoconformer. its native application of salts is less than that of the seawater it lives in. it is continually falling water from the gills to the ocean via osmosis.

countering that, the fish is going to be swallowing seawater. but since its internal collection of salts is lower than seawater, the fish will want to go free of excess salt, and since that's going upon a gradient, then active transport, requiring energy, must be used to pump out the salt. much marine fish use the organs to filter out the salt and these fish discharge very straight urine.

silty loam

hope this

to this one it would be receptive field