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Ugo [173]
2 years ago
11

ANAL0387: Percent Water in a Hydrate

Chemistry
1 answer:
Reika [66]2 years ago
8 0
<h3>Answer:</h3>

Mass of the hydrate heated = 3.462 g

<h3>Solution and Explanation: </h3>
  • A hydrate is a salt or a compound that contains water of crystallization.  
  • For example, hydrated sodium carbonate which contains 10 moles of water of crystallization, Na₂CO₃ .10H₂O.  
  • In our case we are required to determine the mass of the hydrate;
<h3>Mass of the hydrate heated  </h3>

Mass of the hydrate = mass of crucible and hydrate salt– mass of the crucible  

Mass of crucible and hydrates salt = 21.447 g  

Mass of the crucible = 17.985 g  

Therefore;

Mass of the hydrate = 21.447 g – 17.985 g  

                                    = 3.462 g  

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A flash distillation chamber operating at 101.3 kpa is separating an ethanol water mixture the feed mixture contains z weight et
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The answer is [\frac{LX_1}{46} + \frac{L(1-x)}{18}]\times0.454 mol/hr

[\frac{Vy_1}{46}+\frac{V(1-y)}{18}]\times0.454mol/hr

Explanation:

For flash distillation

F = V+L

\frac{V}{F} + \frac{L}{F} = 1

\frac{F}{V} -\frac{L}{V} = 1

Fz = Vy+Lx

Y = \frac{F}{V}\times Z - \frac{L}{V}\times X                  let, \frac{V}{F} = F

y = \frac{Z}{F} -[ \frac{1}{F} -1]\times X

Highlighted reading

F = 299;  \frac{V}{F} = 0.85 ; z = 0.36

y = \frac{0.36}{0.85} - (-0.15)\times X

 = 0.423 + 0.15x ------------(i)

y^{*} = -43.99713x^{6} + 148.27274x^{5} - 195.46x^{4}+127.99x^{3}-43.3x^{2}+ 7.469x^{}+ 0.02011

At equilibrium, y^{*} = y

0.423+0.15x^{} = y^{*}

-43.99713x^{6}+ 148.27274x^{5} - 195.46x^{4}+127.99x^{3}-43.3x^{2}+ 7.319x^{}-0.403

F(x) for Newton's Law

Let x_{0} = 0

     x_{1}     = \frac{0-[{-0.403}]}{7.319}

             = 0.055

     x_{2}      = \frac{{0.055}-{f(0.055)} {{{{{{{}}}}}}}}{f^{'} (0.055)}

             = \frac{{(0.055)}-(-0.11)}{3.59}

             = 0.085

    x^{3}    = \frac{{0.085}-(0.024)}{2.289}

           = 0.095

   x^{4}     = \frac{{0.095}-(-0.0353)}{-1.410}

            = 0.07

From This x and y are found from equation (i) and L and V are obtained from \frac{V}{F}  and F values

[\frac{LX_1}{46} + \frac{L(1-x)}{18}]\times0.454 mol/hr

[\frac{Vy_1}{46}+\frac{V(1-y)}{18}]\times0.454mol/hr

   

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3 years ago
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