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pochemuha
3 years ago
8

How many formula units make up 25.8g of magnesium chloride (MgCl2)? Express the number of formula units numerically.

Chemistry
1 answer:
Verdich [7]3 years ago
8 0

Answer:

Explanation:

Kandndkjwnww

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What is the ph of 0.20 m nitrous acid at equilibrium (ka= 4.6×10-4)?
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Nitrous acid is HNO2. Its dissociation is HNO2 = H(+) + NO2(-). The equilibrium constant, Ka = [H+][NO2-]/[HNO2] => x^2 / (0.20 - x). Given that it is a weak acid (Ka = 4.6 * 10^ -4),you can use a very good approxiamation: 0.20 >> x => 0.20 - x = 0.20, and this permits you to solve the value of x more easily. => 4.6 * 10^-4 = (x^2) / 0.20 => x^2 = 0.20 * 4.6 * 10^ -4 = 0.92 * 10^-4 => x = 9.59 * 10^ -3. Now,calculate pH = log { 1 / [H+] } = log { 1 / (9.59 * 10 ^-3) } = 2.02. <span>Answer pH = 2.02</span>
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How many grams hydrogen in 4.5 mol H2SO4
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H2SO4 ---> 2H^+ + SO4^2-

Hence n H+ = 9 mols

Mass of H = nM = (9*1) = 9g


Alternately

mass of H2SO4= nM= 4.5*98= 441

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Which property describes the shininess of an element?
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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
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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