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Radda [10]
3 years ago
6

2. Heat is:

Chemistry
2 answers:
amid [387]3 years ago
7 0
The answers will have to be A
nlexa [21]3 years ago
5 0
A. Thermal energy good job
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We have air (21% O2 and 79% N2) at 23 bar and 30 C. 4. What is the ideal molar volume (m^3/kmol)? a. b. What is the Z factor? Wh
k0ka [10]

Answer:

The  ideal molar volume is  \frac{V}{n}  =V_z=  0.001095 \ m^3/mol  

The  Z factor is  Z = 0.09997

The  real molar volume is \frac{V_r}{n} = V_k=   0.0001095\ \frac{m^3}{mol}

Explanation:

From the question we are told that

    The pressure is  P  = 23 \ bar =  23 *10^5 Pa

      The temperature is  T  =  30 ^ oC  = 303 \ K

According to the ideal gas equation we have that

          PV  =  nRT

=>      \frac{V}{n}=V_z= \frac{RT}{P}

Where  \frac{V}{n } is the molar volume  and  R is the gas constant with value

            R  =  8.314 \ m^3 \cdot Pa \cdot K^{-1}\cdot mol^{-1}

substituting values

            \frac{V}{n}  =V_z=  \frac{ 8.314 *  303}{23 *10^{5}}

             \frac{V}{n}  =V_z=  0.001095 \ m^3/mol            

The  compressibility factor of the gas is mathematically represented  as

            Z = \frac{P *  V_z}{RT}

substituting values        

          Z = \frac{23 *10^{5} *   0.001095}{8.314 * 303}

          Z = 0.09997

Now the real molar volume is evaluated as

         \frac{V_r}{n} = V_k=  \frac{Z *  RT }{P}

substituting values

             \frac{V_r}{n} = V_k=   \frac{0.09997 *  8.314 *  303}{23 *10^{5}}

             \frac{V_r}{n} = V_k=   0.0001095\ \frac{m^3}{mol}

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Explanation:

Selenium has an atomic number of 34

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