For a caffeinated drink with a caffeine mass percent of 0.51% and a density of 1.00 g/ml, how many ml of the drink would be required to reach an ld50 of 170 mg/kg body mass if the person weighed 171 lb?

2,585.47 mL of the drink would be required to reach an LD-50 for 171 lb person.

Explanation:

LD-50 of caffeine = 170 mg/kg body mass

Mass of the person = 171 lb = 77.56 kg

1 lb = 0.453592 kg

Lethal dose for person with mass 171 lb = 77.56 kg × 170 mg/kg =13,185.92 mg

13,185.92 mg = 13.18592 g

(1 mg = 0.001 g)

Percentage of caffeine in drink = 0.51 %

Mass of drink = M

M = 2,585.47 g

Volume of the drink = V

Density of the drink = D = 1.00 g/mL

2,585.47 mL of the drink would be required to reach an LD-50 for 171 lb person.

answer: mole of the gas

answer: mole of the gasexplanation:

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o