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pychu [463]
3 years ago
14

Mount Everest is approximately 8,000 meters high. How many miles high is Mount Everest?

Chemistry
2 answers:
anygoal [31]3 years ago
4 0
Although its precise height is disputed, Everest rises to some 29,108 feet (8872 meters), or about 5.5 miles high, the tallest in the greatest chain of mountains on earth, the Himalayas." "Exact Mt. Everest Measurement Made." Hope it helps
strojnjashka [21]3 years ago
3 0

Answer:

Mount Everest is  4.97 miles high

Explanation:

we know,

1 meter = 0.0006213712 mile

so,

8000 meters  

=8000 x  0.0006213712 miles

=4.97097 miles

       

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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
Ivahew [28]

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  

r_2 = 1 + 0.01m  

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.  

3 0
3 years ago
What is the mass of 5489 moles of selenium
blondinia [14]
To find the mass in grams of Selenium, you use the moles you have and multiply it by the molar mass.

5489 Moles Se * (79g Se / 1mole Se) =433631g of Se
6 0
3 years ago
Read 2 more answers
Identify the compound that has polar bonds, but a dipole moment of 0. identify the compound that has polar bonds, but a dipole m
matrenka [14]
CF4 is the compound that has polar bonds, but dipole moment of O. C-f bond is a polar bond which is covalent. It has dipole of 4 polar C.F bonds and results in the overall monopolar molecule.
CF4 it has no net dipole moment.
4 0
3 years ago
Read 2 more answers
The object represented by this graph is moving? A.away from the origin at a constant velocity. B.away from the origin at a decre
kumpel [21]

C. toward the origin at a constant velocity

8 0
3 years ago
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Diluting concentrated 1.74M acetic acid to prepare 200 ml of 8M stick solution. What volume of concentrated acid must be used
TiliK225 [7]
<h3>Answer:</h3>

43.5 mL

<h3>Explanation:</h3>

We are given;

  • Molarity of dilute acetic acid is 1.74 M
  • Volume of the solution that needs to be prepared is 200 mL
  • Molarity of the stock solution 8 M

We are required to determine the volume of acid that must be used.

  • We need to know the dilution formula;
  • According to the dilution equation;

M1V1 = M2V2

Rearranging the formula;

V1 = M2V2 ÷ M1

   = (1.74 M × 200 mL) ÷ 8 M

   = 43.5 mL

Therefore, the volume of the concentrated acid that must be added is 43.5 mL

6 0
3 years ago
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