1 Normal No Spaci. Heading 1 Heading 2Title Subtitile Subtle Em.... Emphasis Sty

Question

1 Normal No Spaci. Heading 1 Heading 2Title Subtitile Subtle Em.... Emphasis Styles 1 A sample of an industrial waste water is analyzed and found to contain 36.4 ppb Fe" How many grams of iron could be recovered from 828 kg of this waste water? grams Fe 2 A sample of an industrial waste water is analyzed and found to contain 30.4 ppb Cu How many grams Cu 3. In the laboratory, a student dilutes 20.2 mL of a 8.62 M nitric acid solution to a total volume of 300.0 mL What is the concentration of the diluted solution? Concentration - low many milliliters of 10.2 M perchloric acid solution should be used to prepare 3.00 L of 0.100 M? mL

Explanation / Answer

1)

ppb = mass of Fe2+ * 10^9 / mass of waste

36.4 = mass of Fe2+ * 10^9 / (828 Kg)

mass of Fe2+ = 3.01*10^-5 Kg

mass of Fe2+ = 3.01*10^-2 g

Answer: 3.01*10^-2 g

2)

ppb = mass of Cu2+ * 10^9 / mass of waste

30.4 = mass of Cu2+ * 10^9 / (600 Kg)

mass of Cu2+ = 1.82*10^-5 Kg

mass of Cu2+ = 1.82*10^-2 g

Answer: 1.82*10^-2 g

3)

use dilution formula

M1*V1 = M2*V2

Here:

M1 is molarity of solution before dilution

M2 is molarity of solution after dilution

V1 is volume of solution before dilution

V2 is volume of solution after dilution

we have:

M1 = 8.62 M

V1 = 20.2 mL

V2 = 300.0 mL

we have below equation to be used:

M1*V1 = M2*V2

M2 = (M1*V1)/V2

M2 = (8.62*20.2)/300

M2 = 0.580 M

Answer: 0.580 M

Feel free to comment below if you have any doubts or if this answer do not work

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