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forsale [732]
3 years ago
13

Methane, a highly flammable gas, is placed in a piston that has a volume of 2.1

Chemistry
1 answer:
kupik [55]3 years ago
4 0
Use the formula:
P1V1/T1N1 = P2V2/T2N2
You can cross out T1,N1,T2, and N2 because you are working with pressure and volume in this equation.
Now, you are left with Boyle’s Law:
P1V1 = P2V2
Substitute the values in
STP = 1 atm for pressure
(1 atm)(2.1L) = P2 (0.125 L) [I converted 125 ml to liters so they would be the same unit]
Now, divide both sides by 0.125 to find P2
P2 = (1 atm)(2.1 L)/(0.125 L)
Liters cancel out
P2 = 16.8 atm

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

The concentration of COF₂ at equilibrium is 0.296 M.

Explanation:

To solve this equilibrium problem we use an ICE Table. In this table, we recognize 3 stages: Initial(I), Change(C) and Equilibrium(E). In each row we record the <em>concentrations</em> or <em>changes in concentration</em> in that stage. For this reaction:

   2 COF₂(g) ⇌ CO₂(g) + CF₄(g)

I      2.00              0              0

C      -2x              +x            +x

E   2.00 - 2x         x              x

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Kc=8.30=\frac{[CO_{2}] \times [CF_{4}] }{[COF_{2}]^{2} } =\frac{x^{2} }{(2.00-2x)^{2} } \\8.30=(\frac{x}{2.00-2x} )^{2} \\\sqrt{8.30} =\frac{x}{2.00-2x}\\5.76-5.76x=x\\x=0.852

The concentration of COF₂ at equilibrium is 2.00 -2x = 2.00 - 2 × 0.852 = 0.296 M

6 0
3 years ago
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