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Tems11 [23]
3 years ago
11

What is the volume of .04 mol of a gas at .9 kPa and 20 degrees celsius

Chemistry
1 answer:
siniylev [52]3 years ago
8 0
Pv=nrt
just make sure to convert temperature unit from Celsius to kalvin. r is constat
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Blood, trail mix, and salad are all examples of ______________.
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homogeneous and heterogeneous mixtures

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What is the principle feature of a metallic bond?
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Electrons are free to move throughout metal substance, shared throughout so electricity and heat are conducted well
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A sheet of polyethylene 1.5-mm thick is being used as an oxygen barrier at a temperature of 600K. If the flux is 2.48 x 10-5 kg/
Eddi Din [679]

Complete Question

The complete question is shown on the first uploaded image  

Answer:

The concentration of  high-pressure side is  C_1 =   0.722 \ kg/m^3

Explanation:

From the question we are told that

    The thickness of the polyethylene is  d  =  1.5 \ mm = 0.0015 \ m

     The  temperature is  T  =  600 \   K

      The flux is  JA  =  2.48 *10^{-5} \  kg/m^2\cdot s

      The concentration on the low-pressure side is  C_2 =  0.5 \ kg/m^3

       The initial diffusivity  is  D_o  =  6.2 *10^{-4} \ m^2 /s

       The activation energy for  diffusion is   Q_d  =  41 \ kJ /mol  =  41*10^3 J /mol

Generally the diffusivity  of the oxygen at 600 K can be  mathematically evaluated  as

        D   = D_o * e^{- \frac{Q_d}{R * T  } }  

substituting values  

         D   = 6.2*10^{-4} * e^{- \frac{41 *10^3}{8.314 * 600  } }  

          D   = 1.671 *10^{-7} \ m^2 /s  

Generally the flux is mathematically represented as

          JA  =  D  *  \frac{C_1 -C_2}{d}

Where C_1 is the concentration of oxygen at the higher side

       So  

             C_1 =    d  * \frac{JA}{D}  + C_2

substituting values  

             C_1 =    0.0015   * \frac{2.48*10^{-5}}{1.671*10^{-7}}  + 0.5

              C_1 =   0.722 \ kg/m^3

 

6 0
3 years ago
Suppose that 25g of each substance is initially at 27.0grade centígrados what is the final temperatura of each substance upon ab
sertanlavr [38]

The temperature rise of each substance depends on its specific heat capacity.

<h3>What determine the rise in temperature of equal masses of substances when heated?</h3>

The rise in temperature in a substance depends on the specific heat capacity of the substance.

The formula to determine the temperature rise is give as:

Q = mcΔT

Where:

  • Q = quantity of heat
  • m = mass
  • c = specific heat capacity
  • ΔT = temperature rise

Substances with large heat capacities have the lowest temperature rises while those with small heat capacities have the biggest temperature increase.

In conclusion, the temperature rises depends on the specific heat capacity.

learn more about specific heat capacity at: brainly.com/question/27862577

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Question 3 of 10
miss Akunina [59]

Answer:

the answer is B

Explanation:

B is the answer

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