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sukhopar [10]
3 years ago
7

Pesticide concentrations in the Rhine River between Germany and France between 1969 and 1975 averaged 0.55 mg/L of hexachloroben

zene (C6Cl6), 0.060 mg/L of Dieldrin (C12H8Cl6O), and 1.02 mg/L of hexachlorocyclohexane (C6H6Cl6). Express these concentrations in millimoles per liter.
Chemistry
1 answer:
sertanlavr [38]3 years ago
6 0

Answer:

1.93×10⁻³ mmoles/L of C₆Cl₆; 1.58×10⁻⁴ mmoles/L of C₁₂H₈Cl₆O; 3.51×10⁻³ mmoles/L of C₆H₆Cl₆

Explanation:

We have to find out the molar mass of each pesticide to calculate the moles, and then the milimoles

C₆Cl₆ → 12. 6 + 35.45 .6 = 284.7 g/m

C₁₂H₈Cl₆O →  12 . 12 + 8.1 + 35.45 .6 + 16 = 380.7 g/m

C₆H₆Cl₆ → 12.6 + 6.1 + 35.45 .6 = 290,7 g/m

Let's convert mg to g (/1000)

0.55 mg / 1000 = 5.5×10⁻⁴ g

0.060 mg / 1000 = 6×10⁻⁵ g

1.02 mg / 1000 = 1.02×10⁻³ g

Now we can know the moles (mass / molar mass)

5.5×10⁻⁴ g / 284.7 g/m = 1.93×10⁻⁶ moles of C₆Cl₆

6×10⁻⁵ g / 380.7 g/m = 1.58×10⁻⁷ moles of C₁₂H₈Cl₆O

1.02×10⁻³ g / 290,7 g/m = 3.51×10⁻⁶ moles of C₆H₆Cl₆

Milimoles = Mol . 1000

1.93×10⁻⁶ . 1000 = 1.93×10⁻³ mmoles of C₆Cl₆

1.58×10⁻⁷ . 1000 = 1.58×10⁻⁴ mmoles of C₁₂H₈Cl₆O

3.51×10⁻⁶ . 1000 = 3.51×10⁻³ mmoles of C₆H₆Cl₆

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<h2>calculation</h2><h3>find the empirical formula first as in step 1 and 2</h3>

Step 1: f<em>ind the moles of C and H</em>

  • moles =  % composition/molar mass
  •       from periodic table molar mass of C= 12 g/mol while that of H= 1 g/mol
  • moles is  C is therefore = 85.6/12=  7. 13 moles
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  Step 2:  <em>calculate the mole  fraction  by dividing each mole by smallest number of mole(7.13)</em>

  • that is C= 7.13/7.13 = 1

         H=  14.4/7.13 =2

the empirical formula is therefore = CH2

<h2>Then calculate the molecular formula from empirical formula</h2>

step 3: divide the  grams molar mass  by empirical formula mass

               empirical formula mass =  12+(1 x2) = 14 g/mol

    = 84.2/ 14 = 6

step 4: multiply  each of the subscript  within the empirical  formula with the value gotten in step 3

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<u>Answer:</u> The above reaction is non-spontaneous.

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Here, nickel is getting reduced because it is gaining electrons and iron is getting oxidized because it is loosing electrons.

We know that:

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Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=-0.23-0.77=-1.0V

Relationship between standard Gibbs free energy and standard electrode potential follows:

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As, the standard electrode potential of the cell is coming out to be negative for the above cell. Thus, the standard Gibbs free energy change of the reaction will become positive making the reaction non-spontaneous.

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