In 2009, a team from northwestern university and western washington university reported the preparation of a new "spongy" material composed of nickel, molybdenum, and sulfur that excels at removing mercury from water. the density of this new material is 0.20 g/cm^3, and its surface area is 1242 m^2 per gram of material.

a) calculate the volume of a 40.0-mg sample of this material. (express your answer to two significant figures and include the appropriate units.)

b) calculate the surface area for a 40.0-mg sample of this material.

express your answer to three significant figures and include the appropriate units.

c) a 10.0-ml sample of contaminated water had 7.748 mg of mercury in it. after treatment with 10.0 mg of the new spongy material, 0.001 mg of mercury remained in the contaminated water. what percentage of the mercury was removed from the water?

express your answer to four significant figures.

d) what is the final mass of the spongy material after the exposure to mercury?

express your answer to three significant figures and include the appropriate units

A) 0.20 cm³

B) 49.7 m²

C) 99.99%

D) 17.7 mg

Explanation:

A) The density of a material represents the mass that it occupies in a "piece" of volume. Thus, the density (d) is the mass (m) divided by the volume (v):

d =m/v

If the mass is 40.0 mg = 0.04 g, and the density is 0.20 g/cm³, the volume is:

0.20 = 0.04/v

v = 0.04/0.20

v = 0.20 cm³

B) The surface area (S) is the are that is presented in each gram of the material, so, it's the area (a) divided by the mass (m):

S = a/m

If the mass is 40.0 mg = 0.04 g, and the surface area is 1242 m²/g, so:

1242 = a/0.04

a = 49.7 m²

C) The percent of mercury removed is the mass removed divided by the initial mass, this multiplied by 100%. The mass removed is the initial mass (m0) less the final mass (m), so:

%removed = [(7.748 - 0.001)/7.748] *00%

%removed = 99.99%

D) The final mass of the spongy material is it mass (10 mg) plus the mass removed of the mercury (7.748 - 0.001 = 7.747 mg), so:

m = 10 + 7.747

m = 17.747 mg

m = 17.7 mg

calculate how many grams of ammonia can be produced from 4.21 grams of

n2 + 3h2 = 2nh3

my answer:

25.0g n2 x (1 mol/ 28.0gn2) x (2 mol nh3/ 1 mol n2) x (17.0g nh3/ 1 molnh3) = 30.36g n2

25.0g h2 x (1 mol h2/ 2.00g h2) x (2 mol nh3/ 3 mol h2) x (17.0g nh3/ 1 mol nh3) = 141.67g h2

lr = n2

answer by yourhope:

how do models compare to real-world objects?

d. most models represent a real-world object in a simpler way.

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