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Pepsi [2]
3 years ago
11

The pressure at the center of the earth is probably greater than 3x106 atm, and the temperature there is about 4000 °C. What is

the change in the Gibbs function of reaction on going from crust to core for a reaction in which Vm = 1 cm3 mol-1 and Sm = 2.1 J K-1 mol-1
Chemistry
1 answer:
andrezito [222]3 years ago
7 0

Answer:

The change in the Gibbs function of reaction on going from crust to core, dG=291442.4 J/mol

Explanation:

dG=vdp-sdT

Where T is the temperature .

P is the pressure.

S is entropy

V= 1cm3 =10-6 m3

P= 3*10∧6 atm = 3*10∧11 pa

The temperature at the center of the earth, Tc = 4*103 °C. : The temperature at the sufrace (crust) of the earth, Ts =298K.

Subistuting the values

dG = (10-6 *(3*1011-105) - (2.1(4373-298))

dG=291442.4 J/mol.

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6 0
3 years ago
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Explanation:

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3 years ago
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8 0
3 years ago
When 40.5 g of Al and 212.7 g of Cl2 combine in the reaction: 2 Al (s) + 3 Cl2 (g) --> 2 AlCl3 (s) c) How many moles of the e
sineoko [7]

Answer:

The answer to your question is 1.49 mol of Cl₂  

Explanation:

Data

mass of Al = 40.5 g

mass of Cl₂ = 212.7 g

moles of excess reactant = ?

- Balanced chemical reaction

                2 Al(s)  +  3Cl₂(g)  ⇒   2AlCl₃

Process

a) Calculate the molar mass of the reactants  

molar mass of Al = 2 x 26.98 = 53.96 g

molar mass of Cl₂ = 6 x 35.45 = 212.7 g

b) Calculate the theoretical proportion  Al/Cl₂ = 53.96/212.7 = 0.254

   Calculate the experimental proportion Al/Cl₂ = 40.5/212.7 = 0.19

As the experimental proportion is lower than the theoretical proportion we conclude that the limiting reactant is Aluminum.

c) Calculate the grams of excess reactant

                    53.96 g of Al ------------------ 212.7 g of Cl₂

                     40.5 g of Al -------------------  x

                      x = (40.5 x 212.7) / 53.96

                      x = 8614.35 / 53.96

                      x = 159.64 g of Cl₂

Excess Cl₂ = 212.7 - 159.64

                  = 53.057 g

d) Calculate the moles of Cl

                       35.45 g of Cl ----------------- 1 mol

                       53.057 g of Cl ---------------  x

                        x = (53.057 x 1)/35.45

                       x = 1.49 mol of Cl₂                    

6 0
3 years ago
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