Question

Question 9 (1 point) The maximum acceptable level for a certain pollutant in soil is 15.0 parts per million (ppm; that is 15.0 grams pollutant in 1 million grams of soil). What is the maximum number of milligrams of pollutant that could be in 10.0 pounds of soil? (11b=454g) 04 68.1 0.15 0.0681 Question 10 (1 point)

please answer all parts!!

Answer 1

9) 15 gm of pollutant can be in 10^-6 g or 1 million grams of soil

So, in 4540 * 10^-6 million gram  of soil , amount of pollutant present can be = 15* 4540 *10^-6 g

= 15* 4540* 10^-3 mg

68.1 mg of pollutant option 2

10) for 1 kg of body weight , amount of dose is 2 mg

therfore for 145 pounds or 145*454/1000 kg = 65.3 kg , amount of dose required is 65.2*2 = 132 mg

4) Molarity = no. of moles/ volume of solution in litre

.400 M = x*1000/200

moles of HCl = .08 mole

2) 65 bags weighs 3.4 pounds

therfore, 1 bag weighs 3.4/ 65 pounds = .052 pounds

1 bas has 30 candy , i. e weight of 30 candy = .052 pound

weight of 1 candy = .052/30 = .001744

1) 3.4 pounds of weight in 65 bags

therefore 1 pound of weight will be of 65/3.4= 19.1 bag or 19 bags

1 bag has 30 candy

therfore, 19bags will have = 19*30= 574 candy