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k0ka [10]
4 years ago
8

When limestone (calcium carbonate) is strongly heated, it breaks down into calcium oxide and carbon dioxide. If 88 grams of carb

on dioxide is released when 200 grams of limestone is heated, how much calcium oxide is produced in this reaction? A. 56 grams B. 88 grams C. 112 grams D. 200 grams
Chemistry
2 answers:
professor190 [17]4 years ago
8 0

<u>Answer:</u> The correct answer is Option C.

<u>Explanation:</u>

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

The chemical equation for the decomposition of limestone follows:

CaCO_3\rightarrow CaO+CO_2

Let the mass of calcium oxide be 'x' grams

We are given:

Mass of carbon dioxide = 88 grams

Mass of calcium carbonate = 200 grams

Total mass on reactant side = 200 g

Total mass on product side = 88 + x

So, by applying law of conservation of mass, we get:

200=88+x\\\\x=112g

The mass of calcium oxide that would be produced is 112 grams.

Hence, the correct answer is Option C.

xxTIMURxx [149]4 years ago
4 0
Answer: C. 112 grams

Using the law of conservation of mass, you can say that the total mass of the reactant would be equal to total mass of the product. The reactant of this reaction would be 200g CaCO3(calcium carbonate). The product would be x g <span>calcium oxide(CaO) and 88g carbon dioxide(CO2). The calculation would be:

reactant = product
200g= 88g+ x
x= 200g-88g= 112g

</span>
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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  

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}

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M_di_f_f = MN_diff

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

For one week the mass flow rate of helium is

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

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V = 4.189m³

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

PV=NRT

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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
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