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zavuch27 [327]
3 years ago
5

How does mixture relate to life​

Chemistry
1 answer:
masya89 [10]3 years ago
3 0

Answer:

For example, when we bake a cake, it's a result of a mixture of eggs, flour, sugar, and other ingredients.  Any time two or more items are combined, a mixture is formed. Sometimes, the different parts of a mixture can be separated into individual entities. Other times, they're married for as long as they exist.

Explanation:

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What is the relationship between concentration and rate of reaction?
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. Helium is stored at 293 K and 500 kPa in a 1-cm-thick, 2-m-inner-diamater spherical container made of fused silica. The area w
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Answer:  

(a) 45.17×10^-14 kg/s  

(b) since the amount of helium escaped due to diffusion is insignificant, the final pressure drop in the tank remains the same as the initial pressure 500kPa.  

Explanation:  

Helium gas at temperature T=293k  

Helium gas at pressure P= 500kPa  

The inner diameter of spherical tank is D_1 = 2m  

The inner radius of spherical tank is : r_1 = \frac{D_1}{2}  

= \frac{2}{2}  

=1m  

Thickness of the container r = 1cm =0.01m  

Outer radius of the spherical tank is ;  

t = r_2 - r_1  

-r_2 = -t - r-1  

multiplying through with (-) we have ;  

r_2 = t + r_1  

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r_2 = 1.01m  

From table of binary diffusion coefficients of solids, the diffusion coefficients of helium in silica is noted as  

D_A_B =4.0 ×10^-14 \frac{m^2}{s}  

From table molar mass and gas constant, the molecular weight of helium is:

 

M = 4.003kg/kmol  

The solubility of helium in fused silica is determined from Table of Solubility of selected gases and soilids.  

S_He = 0.00045 kmol/m³. bar  

Considering total molar concentration as constant, the molar concentration of helium inside the container is determined as  

C_B_I = S_H_e×P  

= 0.00045kmol/m³. bar × (5)  

C_B_I = 2.25×10^-3 kmol/m³  

From one dimensional mass transfer through spherical layers is expressed as:

N_di_f_f= 4πr_1 r_2 D_A_B \frac{C_B_I - C_B_2}{r_2 - r_1}

substituting all the values in the above relation, we have;

M_di_f_f= 4π(1) (1.01) (4.0×10^-14) \frac{2.25 × 10^-3 -0}{1.01-1}

M_di_f_f=11.42×10^-14kmol/s

(a) The mass flow rate is expressed as

M_di_f_f = MN_diff

M_di_f_f=4.003×11.42×10^-14

M_di_f_f=45.71×10^-14kg/s

(b) The pressure drop in the tank after a week;

For one week the mass flow rate of helium is

N_di_ff =11.42×10^-14kmol/s

N_di_ff= 11.42×10^-14×7×24×3600 kmol/week

N_di_f_f=6.9×10^-8kmol/week

The volume of the spherical tank is V=\frac{4}{3} πr_1^3

V=\frac{4}{3}π×1^3

V = 4.189m³

The initial mass of helium in the sphere is determined from the ideal gas equation:

PV=NRT

where R is the universal gas constant and its value is R = 8.314 KJ/Kmol.k

N= PV/RT

N= 500 × 4.189/ 8.314 × 293

N= 0.86kmol

The number of moles of helium gas remaining in the tank after one week is:

N_di_f_f-final = 0.86 - 6.9 × 10^-8 kmol/week

N_di_f_f-final ≅ 0.86

therefore, since the amount of helium escaped due to diffusion is insignificant, the final pressure drop in the tank remains the same as the initial pressure 500kPa.  

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