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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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High tempature and pressure over time will cause what in rock layers​
Firlakuza [10]

Answer:

Metamorphic rocks form from heat and pressure changing the original or parent rock into a completely new rock. The parent rock can be either sedimentary, igneous, or even another metamorphic rock. The word "metamorphic" comes from Greek and means "To Change Form"

Explanation:

4 0
3 years ago
Read 2 more answers
How much work (in J) is involved in a chemical reaction if the volume decreases from 5.35 to 1.55 L against a constant pressure
Mrac [35]

Answer:- work = 319 J

Solution:- The volume of the gas decreases from 5.35 L to 1.55 L.

\Delta V=1.55L-5.35L  = -3.80 L

external pressure is given as 0.829 atm.

w=-P_e_x_t\Delta V

where w is representing work.

Let's plug in the values and calculate pressure-volume work:

w=-0.829atm(-3.80L)

w = 3.15 atm.L

We need to convert atm.L to J.

1 atm.L = 101.325 J

So, 3.15atm.L(\frac{101.325J}{1atm.L})

= 319 J

So, 319 J of work is involved in a chemical reaction. Positive work means the work is done on the system.

5 0
3 years ago
0.100 mol of CaCO3 and 0.100 mol CaO are placed in an 10.0 L evacuated container and heated to 385 K. When equilibrium is reache
HACTEHA [7]

Answer:

12.531 grams will be the final mass of calcium carbonate.

Explanation:

CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)

Partial pressure of carbon dioxide gas at equilibrium =P=0.220 atm

Volume of carbon dioxide gas = V = 10.0 L

Moles of carbon dioxide gas formed = n

Temperature of the gas = T = 385 K

PV=nRT( ideal gas equation)

n=\frac{PV}{RT}=\frac{0.220 atm\times 10.0 L}{0.0821 atm L/mol K\times 385 K}=0.06960 mol

According to reaction , 1 mole of carbon dioxide gas is formed from 1 mole of calcium carbonate,0.06960 mole of carbon dioxide gas will be obtained from :

\frac{1}{1}\times 0.06960 mol=0.06960 mol calcium carbonate

Moles of calcium carbonate at equilibrium = 0.100 mol - 0.06960 mol = 0.03040 mol

After addition of 0.300 atm of carbon dioxide gas, more amount of calcium carbonate will be be formed.

Here, at the same temperature, the equilibrium pressure of the carbon dioxide gas is 0.220 atm so, entire 0.300 atm of carbon dioxide will get convert to calcium carbonate.

So amount moles of carbon dioxide gas added = moles of calcium carbonate formed after re-establishment of an equilibrium :

Partial pressure of carbon dioxide gas added at equilibrium =P=0.300 atm

Volume of carbon dioxide gas = V = 10.0 L

Moles of carbon dioxide gas formed = n

Temperature of the gas = T = 385 K

PV=nRT( ideal gas equation)

n=\frac{PV}{RT}=\frac{0.300 atm\times 10.0 L}{0.0821 atm L/mol K\times 385 K}=0.09491 mol

Moles of calcium carbonate = 0.09491 mol

Total moles of calcium carbonate in the container :

=0.03040 mol + 0.09491 mol = 0.12531 mol

Mass of 0.12531 moles of calcium carbonate :

0.12531 mol × 100 g/mol = 12.531 g

12.531 grams will be the final mass of calcium carbonate.

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If there is 30 grams of O2 how many grams of CO2 can it make
igomit [66]

Explanation:

hope it helps you with the question

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Draw n-ethyl-3-methylpentanamide. draw your structural formula with all necessary hydrogen atoms.
Elena-2011 [213]
Structure of <span>N-ethyl-3-methylpentanamide is shown below,</span>

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3 years ago
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