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Natasha_Volkova [10]
3 years ago
8

Two beakers are placed in a closed container. Beaker A initially contains 0.15 mole of naphthalene (C10H8) in 100 g of benzene (

C6H6), and beaker B initially contains 31 g of an unknown compound dissolved in 100 g of benzene. At equilibrium, beaker A is found to have lost 7.0 g of benzene. Assuming ideal behavior, calculate the molar mass of the unknown compound. State any assumptions made.
Chemistry
1 answer:
GalinKa [24]3 years ago
3 0

Answer:

Molar mass of the unknown compound is 192.2 g/mol

Explanation:

Assumption:

At equilibrium, molality of beaker A = molality of beaker B

Molality of beaker A = moles of naphthalene/kg of benzene

Mass of benzene = 100 - 7 = 93 g = 93/1000 = 0.093 kg

Molality of beaker A = 0.15/0.093 = 1.613 m

Molality of beaker B = moles of unknown compound (y)/kg of benzene

Mass of benzene = 100 g = 100/1000 = 0.1 kg

Molality of beaker B = (y/0.1) m

y/0.1 = 1.613

y = 0.1×1.613 = 0.1613 mol of unknown compound

Molar mass of unknown compound = mass/number of moles = 31 g/ 0.1613 mol = 192.2 g/mol

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The absorbance of an equilibrium mixture containing FeSCN2 was measured at 447 nm and found to be 0.347. What is the equilibrium
Ksenya-84 [330]

Answer:

The concentration is C = 1.11 mol/L

Explanation:

From the question we are told that

     The absorbance is  A = 0.347

       The length is  l =  447 nm  =  447 *10^{-7} \ cm

     

Generally absorbance is mathematically represented as

        A =  \epsilon*  C * l

where \epsilon is the molar absorptivity of  FeSCN2  with a value \epsilon  =  7.0*10^3 L/cm/mol

 and  C is the equilibrium concentration of FeSCN2

So  

       C = \frac{A}{\epsilon *  l  }

substituting values

        C = \frac{0.347}{7.0*10^{3} *  447 *10^{-7}  }

         C = 1.11 mol/L

5 0
3 years ago
What are the 3 forces that make the tectonic plates<br> move
Molodets [167]

Answer:

Viscous Drag.

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Explanation:

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8 0
3 years ago
The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:
aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

                         H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium:  0.50M - x          x            x

The equilibrium constant (Ka₁) is:

K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M

Similarly, from the second dissociation of sulfurous acid we have:

                              HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium:  7.94x10⁻²M - x          x            x

The equilibrium constant (Ka₂) is:  

K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}  

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0
3 years ago
suppose you wanted to compare two different stain removers to learn which one was better at removing food stains from clothing.
Dovator [93]

Answer:

The amount/type of stain

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8 0
1 year ago
Acetylene is hyrodgenated to form ethane. The feed to the reactor contains 1.60 mol H2/mol C2H2. The reaction proceeds to comple
Black_prince [1.1K]

Answer:

Explanation:

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1 mole     2 mole        1 mole

Feed of reactant is 1.6 mole H₂ / mole C₂H₂

or 1.6 mole of H₂ for 1 mole of C₂H₂

required ratio as per chemical reaction written above

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So H₂ is in short supply . Hence it is limiting reagent .

1.6 mole of H₂ will react with half of 1.6 mole or .8 mole of C₂H₂ to form .8 mole of C₂H₆

a )Calculate the stoichiometric reactant ratio =  mole H₂ reacted/mole C₂H₂ reacted

= 1.6 / .8 = 2 .

b )

yield ratio = mole C₂H₆ formed / mole H₂ reacted ) = 0.8 / 1.6 = 1/2 = 0.5 .

4 0
3 years ago
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