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Westkost [7]
3 years ago
10

Break down the following equations into two half equations(One for oxidation, one or reduction )

Chemistry
1 answer:
enot [183]3 years ago
7 0

Answer:

see explanation

Explanation:

Break down the following equations into two half equations(One for oxidation, one or reduction )

1. Cu + 2Ag+ ➡️Cu2+ +2Ag

2. Cl2 +2I- ➡️I2 + 2Cl​-

in reaction "1", the copper is being oxidized from 0 to +2 by losing 2 electrons

Cu---->Cu2+   + 2e-in reaction "2

in reaction "1" the silver is being reduced by gaining those 2 e-.

2Ag1+  + 2e----------->2Ag

In reaction "2", the iodine is being oxidized by losing 2 e-

2I- -------------> I2 + 2e

In reaction "2" the chlorine is being reduced by gaining those 2 e

Cl2 + 2e------------> 2Cl-

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The coefficient of thermal expansion α = (1/V)(∂V/∂T)p. Using the equation of state, compute the value of α for an ideal gas. Th
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      \alpha  =  \frac{1}{T}

The coefficient of compressibility

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Now  considering (\frac{ \delta P }{\delta  T} )V

From equation (1) we have that

       \frac{ \delta P}{\delta  T}  =  \frac{n R }{V}

From  ideal equation

         nR  =  \frac{PV}{T}

So

     \frac{\delta P}{\delta  T}  =  \frac{PV}{TV}

=>  \frac{\delta  P}{\delta  T}  =  \frac{P}{T}

=>   \frac{\delta  P}{\delta  T}  =  \frac{\alpha }{\beta}

Explanation:

From the question we are told that

   The  coefficient of thermal expansion is \alpha  =  \frac{1}{V} *  (\frac{\delta V}{ \delta  P})  P

    The coefficient of compressibility is \beta  =  - (\frac{1}{V} ) *  (\frac{\delta V}{ \delta P} ) T

Generally the ideal gas is  mathematically represented as

        PV  =  nRT

=>      V  =  \frac{nRT}{P}  --- (1)

differentiating both side with respect to T at constant P

       \frac{\delta V}{\delta T }  =  \frac{ n R }{P}

substituting the equation above into \alpha

       \alpha  =  \frac{1}{V} *  ( \frac{ n R }{P})  P

        \alpha  = \frac{nR}{PV}

Recall from ideal gas equation  T =  \frac{PV}{nR}

So

          \alpha  =  \frac{1}{T}

Now differentiate equation (1) above with respect to  P  at constant T

          \frac{\delta  V}{ \delta P}  =  -\frac{nRT}{P^2}

substituting the above  equation into equation of \beta

        \beta  =  - (\frac{1}{V} ) *  (-\frac{nRT}{P^2} ) T

        \beta =\frac{ (\frac{n RT}{PV} )}{P}

Recall from ideal gas equation that

       \frac{PV}{nRT}  =  1

So

       \beta   =  \frac{1}{P}

Now  considering (\frac{ \delta P }{\delta  T} )V

From equation (1) we have that

       \frac{ \delta P}{\delta  T}  =  \frac{n R }{V}

From  ideal equation

         nR  =  \frac{PV}{T}

So

     \frac{\delta P}{\delta  T}  =  \frac{PV}{TV}

=>  \frac{\delta  P}{\delta  T}  =  \frac{P}{T}

=>   \frac{\delta  P}{\delta  T}  =  \frac{\alpha }{\beta}

5 0
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