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Alekssandra [29.7K]
4 years ago
8

Consider: CO(g) + Cl2 (g) ⇌ COCl2 (g) Kc = 1.2×103 at 395 °C. If the equilibrium concentrations of Cl2 and COCl2 are the same at

395 °C, find the equilibrium concentration of CO in the reaction.
Chemistry
1 answer:
ad-work [718]4 years ago
4 0

Answer : The equilibrium concentration of CO in the reaction is, 8.3\times 10^{-4}M

Explanation :

The given chemical reaction is:

CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)

The expression for equilibrium constant is:

K_c=\frac{[COCl_2]}{[CO][Cl_2]}

As we are given:

Concentration of Cl_2 at equilibrium = Concentration of COCl_2

So,

K_c=\frac{[Cl_2]}{[CO][Cl_2]}

K_c=\frac{1}{[CO]}

1.2\times 10^3=\frac{1}{[CO]}

[CO]=8.3\times 10^{-4}M

Therefore, the equilibrium concentration of CO in the reaction is, 8.3\times 10^{-4}M

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How many molecules are there in 5H20?
Svetach [21]

Answer:

The number before any molecular formula applies to the entire formula. So here you have five molecules of water with two hydrogen atoms and one oxygen atom per molecule. Thus you have ten hydrogen atoms and five oxygen atoms in total.

3 0
3 years ago
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2-- What is the [H3O+] in a solution with [OH-] = 1 x 10-12 M?<br>​
Arte-miy333 [17]

Answer:

The answer is 0.01 M

Explanation:

The problem is solved by applying the expression for ionic product of water as follows:

Kw = [H₃O⁺] [OH⁻]

Where Kw is the ionic product of water and it is 1.10⁻¹⁴ M at⁴ 25ºC. As [OH⁻]= 1.10⁻¹² M, [H₃O⁺] will be:

[H₃O⁺]= Kw/ [OH⁻]= 1.10⁻¹⁴M/1.10⁻¹²M= 0.01 M

5 0
3 years ago
Based on the following equation, how many moles of hydrochloric acid (HCl)are needed to react with 0.44 moles of potassium
inysia [295]

Answer:

1.76 moles of HCl.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2KMnO₄ + 8HCl —> 3Cl₂ + 2MnO₂ + 4H₂O + 2KCl

From the balanced equation above,

2 moles of KMnO₄ reacted with 8 moles of HCl.

Finally, we shall determine the number of mole of HCl needed to react with 0.44 moles of KMnO₄. This can be obtained as follow:

From the balanced equation above,

2 moles of KMnO₄ reacted with 8 moles of HCl.

Therefore, 0.44 moles of KMnO₄ will react with = (0.44 × 8)/2 = 1.76 moles of HCl.

Thus, 1.76 moles of HCl is needed for the reaction.

5 0
3 years ago
Based on the molar masses. how can you tell that an equation is balanced
harkovskaia [24]
You can tell if each side of the equation has the same molar mass.
3 0
3 years ago
Please help I’m timed !!!
Rashid [163]

Answer:

b. \% diss =4.4x10^{-2}\%

Explanation:

Hello there!

In this case, given the ionization reaction of HClO as weak acid:

HClO\rightleftharpoons H^++ClO^-

We can write the equilibrium expression as shown below:

Ka=3.5x10^{-8}=\frac{[H^+][ClO^-]}{[HClO]}

In such a way, via the definition of x as the reaction extent, we can write:

3.5x10^{-8}=\frac{x^2}{[HClO]}

As long as Ka<<<<1 so that the x on the bottom can be neglected. Thus, we solve for x as shown below:

x=\sqrt{3.5x10^{-8}*0.18} =\\\\x=7.94x10^{-5}M

And finally the percent dissociation:

\% diss=\frac{x}{[HClO]} *100\%\\\\\% diss=\frac{7.94x10^{-5}M}{0.18}*100\% \\\\\% diss =0.044\%=4.4x10^{-2}\%

Which is choice b.

Best regards!

6 0
3 years ago
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