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KonstantinChe [14]
3 years ago
5

What’s the answer to this science problem?

Chemistry
1 answer:
Law Incorporation [45]3 years ago
5 0

Answer:

The image is not accurate because

the heavier atoms are moving faster than the lighter ones

but the lighter ones must move faster than the heavier ones.

Explanation:

The image depicts the atoms of two different elements; atoms of lighter and heavier elements.

And the rate of diffusion (movement of atoms) of an element is inversely proportional to the square root of its molecular.

⇒ As the weight increases, the rate of diffusion (movement of atoms) decreases.

And as the weight decreases, the rate of diffusion (movement of atoms) increases.

so, heavier atoms move slowly compared to lighter atoms.

The image is not accurate because

the heavier atoms are moving faster than the lighter ones

but the lighter ones must move faster than the heavier ones.

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What is the answer to this question
Stels [109]

Answer:

38 kg/m³

0.038 g/mL

Explanation:

Volume of a cube is the side length cubed.

V = s³

Given s = 0.65 m:

V = (0.65 m)³

V ≈ 0.275 m³

The mass is 10.5 kg.  The density is the mass divided by volume:

ρ = (10.5 kg) / (0.275 m³)

ρ ≈ 38 kg/m³

Or:

ρ ≈ 0.038 g/mL

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interstitium

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What is the formula of calculating density?
Salsk061 [2.6K]

density = mass/volume

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A sample has a mass of 15 g and a volume of 3 mL calculate the density
zlopas [31]
Since density is mass/volume, the density is 5g/ml.
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3 years ago
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
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