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jasenka [17]
4 years ago
8

Which of the following is the correct definition of minerals? A a naturally occurring, inorganic solid that has a crystal shape

and a definite chemical composition B a naturally occurring, inorganic solid that forms through crystallization of melted materials C a naturally occurring, organic solid that forms through crystallization of materials dissolving in water D a naturally occurring, organic solid that has a crystal shape and a definite chemical composition
Chemistry
1 answer:
Nataliya [291]4 years ago
6 0
The correct answer for the question that is being presented above is this one: "A a naturally occurring, inorganic solid that has a crystal shape and a definite chemical composition." The correct definition of minerals is that it is a naturally occurring, inorganic solid that has a crystal shape and a definite chemical <span>composition</span>
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: Starting with 0.3500 mol CO(g) and 0.05500 mol COCl2(g) in a 3.050-L flask at 668 K, how many moles of Cl2(g) will be present
Mrac [35]

Answer:

The number of moles of Cl₂ present at equilibrium is 3.94x10⁻⁴ moles.

Explanation:

The reaction is:

CO(g) + Cl₂(g) ⇄ COCl₂(g)  

The equilibrium constant of the above reaction is:

K = 1.2x10³

To find the moles of Cl₂ present at equilibrium, let's evaluate the reverse reaction:

COCl₂(g) ⇄ CO(g) + Cl₂(g)  

The equilibrium constant for the reverse reaction is:

K_{r} = \frac{1}{1.2 \cdot 10^{3}} = 8.3 \cdot 10^{-4}

Now, we need to calculate the concentration of CO and COCl₂:

C_{CO} = \frac{\eta_{CO}}{V} = \frac{0.3500 moles}{3.050 L} = 0.115 M

C_{COCl_{2}} = \frac{\eta_{COCl_{2}}}{V} = \frac{0.05500 moles}{3.050 L} = 0.018 M

Now, from the reaction we have:

COCl₂(g) ⇄ CO(g) + Cl₂(g)  

0.018 - x       0.115+x   x    

The concentration of Cl₂ is:

K_{r} = \frac{[CO][Cl_{2}]}{[COCl_{2}]}

8.3 \cdot 10^{-4} = \frac{(0.115 + x)(x)}{0.018 - x}  

8.3 \cdot 10^{-4}*(0.018 - x) - (0.115 + x)(x) = 0  

By solving the above equation for x we have:

x = 1.29x10⁻⁴ M = [Cl₂]

Finally, the number of moles of Cl₂ present at equilibrium is:

\eta_{Cl_{2}} = C_{Cl_{2}}*V = 1.29 \cdot 10^{-4} mol/L*3.050 L = 3.94 \cdot 10^{-4} moles

Therefore, the number of moles of Cl₂ present at equilibrium is 3.94x10⁻⁴ moles.

I hope it helps you!

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