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vfiekz [6]
3 years ago
15

The extraction of pure copper is expensive. Give one reason why

Chemistry
2 answers:
stiv31 [10]3 years ago
8 0
The ore contains some metals which need alot of energy to get out and energy is expensive
goblinko [34]3 years ago
3 0
Because it requires lots of electricity to turn it into an aqueous solution so that electrolysis can be performed on it.
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The answer to the question?
Artemon [7]

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The nucleus (center) of the atom contains the protons (positively charged) and the neutrons (no charge). The outermost regions of the atom are called electron shells and contain the electrons (negatively charged).

5 0
3 years ago
Consider the following reaction and situations 1 through 10. In the spaces provided, clearly indicate the best response to each
olchik [2.2K]

Answer:

1. C. no change

2. A. increase

3. E. shift to the right

4. A. increase

5. E. shift to the right

6. A. increase

7. F. cannot be determined

8. B increase

9. D. shift to the left

10 F. cannot be determined

Explanation:

<em>According to Le Chaterlier principle, when a reaction is in equilibrium and one of the constraints that affect reactions is applied, the equilibrium will shift so as annul the effects of the constraints.</em>

From the equation: C(s) + H2O(g) ⇌ CO(g) + H2(g),

H is greater than 0, meaning that the system is endothermic, that is energy is absorbed.

1. If the pressure of the system is increased, there would be no change to the system because there are equal number of moles of products and reactants.

2. If H2 concentration is decreased, the equilibrium will shift to the right and more products will be formed. Hence, the concentration of CO will increase.

3. If H2 concentration is decreased, the equilibrium will shift to the right to annul the effects of the decrease in the concentration of a product.

4. If the concentration of H2 is increased, the equilibrium will shift to the left to annul the effects of increased concentration of a product. Hence, more H2O would be formed.

5. If H2 (a product) is removed, and C (a reactant) is added, more of the products will be formed in order to annul the effects of the actions. Hence, equilibrium will shift to the right.

6. If the amount of C (a reactant) is increased, the equilibrium will shift to the right. Hence, more H2 will be formed.

7. The reaction is endothermic, hence an increase in temperature will ordinarily shift the equilibrium to the right. However, the addition of H2 (a product) is supposed to shift the equilibrium to the left. Hence, the effects of simultaneous addition of the two actions become indeterminate.

8. Since the reaction is endothermic, increase in the temperature of the system will shift the equilibrium to the right. Hence, more CO will be formed.

9. If the concentration of H2O (a reactant) is decreased and that of CO (a product) is increased, both actions lead to the equilibrium being shifted to the left.

10. Addition of catalyst to the system will only speed up the rate at which the system reach the equilibrium.

5 0
3 years ago
Determine the value of the equilibrium constant, Kgoal, for the reaction C(s)+12O2(g)+H2(g)⇌12CH3OH(g)+12CO(g), Kgoal=? by makin
Licemer1 [7]

Answer:

1.71x10²⁷

Explanation:

If we sum 1/2 of (3) + 1/2 of (1):

1/2 (3.) C(s) + 1/2O₂(g) ⇌ CO(g), K₃ = √2.10×10⁴⁷  = 4.58x10²³

1/2 (1)   1/2CO₂(g) + 3/2H₂(g) ⇌ 1/2CH₃OH(g) + 1/2H₂O(g), K₁ = √1.40×10² = 11.8

C(s) + 1/2O₂(g) +<u> 1/2CO₂(g) </u>+<u> 3/2H₂(g</u>) ⇌ 1/2CH₃OH(g) + <u>1/2H₂O(g)</u> + <u>CO(g)</u>

K' = 4.58x10²³ * 11.8 = 5.42x10²⁴

+1/2 (2):

<u>1/2 CO(g)</u> +<u> 1/2H₂O(g)</u> ⇌<u> 1/2CO₂(g)</u> + <u>1/2H₂</u> (g), K = √1.00×10⁵ = 316.2

C(s) + 1/2O₂(g) + H₂(g) ⇌ 1/2 CHO₃H(g) + 1/2CO(g)

K'' = 5.42x10²⁴* 316.2 =

<h3>1.71x10²⁷</h3>

5 0
4 years ago
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