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love history [14]
3 years ago
14

Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it

is used to make the refrigerant carbon tetrafluoride, CF4 via the reaction 2COF2(g)⇌CO2(g)+CF4(g), Kc=6.30 If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2 remains at equilibrium?
Chemistry
1 answer:
lions [1.4K]3 years ago
4 0

<u>Answer:</u> The equilibrium concentration of COF_2 is 0.332 M

<u>Explanation:</u>

We are given:

Initial concentration of COF_2 = 2.00 M

The given chemical equation follows:

                2COF_2(g)\rightleftharpoons CO_2(g)+CF_4(g)

<u>Initial:</u>          2.00

<u>At eqllm:</u>     2.00-2x          x      x

The expression of K_c for above equation follows:

K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}

We are given:

K_c=6.30

Putting values in above expression, we get:

6.30=\frac{x\times x}{(2.00-2x)^2}\\\\x=0.834,1.25

Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible

So, equilibrium concentration of COF_2=(2.00-2x)=[2.00-(2\times 0.834)]=0.332M

Hence, the equilibrium concentration of COF_2 is 0.332 M

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attashe74 [19]

Answer:

2.29 g of N2

Explanation:

We have to start with the <u>chemical reaction</u>:

NaN_3~->~Na~+~N_2

The next step is to <u>balance the reaction</u>:

2NaN_3~->~2Na~+~3N_2

We can continue with the <u>mol calculation</u> using the molar mass of

NaN_3 (65 g/mol), so:

3.55~g~NaN_3\frac{1~mol~NaN_3}{65~g~NaN_3}=0.054~mol~NaN_3

Now, with the<u> molar ratio</u> between NaN_3  and N_2  we can <u>calculate the moles</u> of N_2  (2:3), so:

0.054~mol~NaN_3\frac{3~mol~N_2}{2~mol~NaN_3}=0.0819~mol~N_2

With the molar mass of N_2 we can <u>calculate the grams</u>:

0.0819~mol~N_2=\frac{1~mol~N_2}{28~g~N_2}=2.29~g~N_2

I hope it helps!

5 0
3 years ago
The blue color of the sky results from the scattering of sunlight by air molecules. The blue light has a frequency of about 7.5
vazorg [7]

A) c = 3 x 10^8 m/s 
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B) E = h  f 
H = Planck's constant = 6.63 x 10^-34 J/s 
E = 6.63 x 10^-34 x 7.15 x 10^14 = 4.74 x 10^-19 J

Read more on Brainly.com - brainly.com/question/5760368#readmore
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3 years ago
CH3CO2H(aq) + H2O(l) ⇄ CH3CO2-(aq) + H3O+(l)
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Correct Answer: Option C:<span> The equilibrium position will shift to the right toward the products. 

Reason:
1) This problem is based on </span>Le Chatelier's principle. It is stated as '<em>any</em><span><em> changes in the temperature, volume, or concentration of a system will result in predictable and opposing changes in the system in order minimize this change and achieve a new equilibrium state.</em>' 

2) In present case, the reaction involved is: 
</span><span>                 CH3CO2H(aq) + H2O(l) ⇄ CH3CO2-(aq) + H3O+(l) 
</span>Hence, when the concentration of acetic acid (reactant) is increased, the equilibrium will shift to right to minimize the effect of change in concentration of reactant. 
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2 years ago
(IMAGE ATTACHED)<br>(pls help this is due tonight)<br>​
miv72 [106K]

Answer:

You can see that the line is going up and is curved in a positive direction.

Explanation:

When an object is speeding up, the acceleration is in the same direction as the velocity. Thus, this object has a positive acceleration.

5 0
2 years ago
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For the reaction shown, find the limiting reactant for each of the initial quantities of reactants.
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Answer:

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