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Lelechka [254]
3 years ago
12

What is the compound of KCN called

Chemistry
2 answers:
igomit [66]3 years ago
6 0
Potassium cyanide. Almost like salt, looks like sugar.
tino4ka555 [31]3 years ago
5 0
Hey there!

Here is your answer:

<u><em>The proper answer to this question is "</em></u><span><u><em>Potassium cyanide".</em></u>

Reason:

<u><em>KCN=</em></u></span><span><u><em>Potassium cyanide </em></u>

Also....

</span><span> <u><em>Potassium cyanide is a compound with the formula KCN (remember that Potassium is colorless)</em></u>

If you need anymore help feel free to ask me!

Hope this helps!

~Nonportrit</span>

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Read 2 more answers
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
andreyandreev [35.5K]

Answer:

The coefficient of thermal expansion α is  

      \alpha  =  \frac{1}{T}

The coefficient of compressibility

      \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}

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
3 years ago
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