How many grams of KNO3 are required to make 100.0 mL of a 0.350 M solution? The molecular weight of KNO3 is 101.1 g/mole

look in attachments

step-by-step explanation:

a = (-2, 0)

b= (-4, 0)

c = (-5,2)

d = (-1,2)

ending amount = beginning amount / (2^n)

where "n" = number of half-lives (or elapsed time / half-life)

ending amount = 50 mg / (2 ^10 / 5.3)

ending amount = 50 mg / (2 ^ 1.8867924528)

ending amount = 50 m / 3.6981210597

ending amount = 13.5203794557 mg

or about 13.5 mg remain

source:  

well, because the $0.25 is an added cost depending on how many minutes, the answer would be y = 30 + 0.25x.