Use the perfodic table to fill in the table below.

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

lions [1.4K]

Answer: The equilibrium concentration of $COF_2$ is 0.332 M

Explanation:

We are given:

Initial concentration of $COF_2$ = 2.00 M

The given chemical equation follows:

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

Initial: 2.00

At eqllm: 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

4 0

3 years ago

Which of the following is used to show sound waves?

Ipatiy [6.2K]

I think it’s c bc it makes more sense

3 0

2 years ago

What is the balanced chemical equation for the reaction of potassium with sulfur?

Aneli [31]

K + S = K2S

Potassium reacts with sulfur to produce potassium sulfide

8 0

3 years ago

Nitrogen and hydrogen combine to form ammonia in the Haber process. Calculate (in kJ) the standard enthalpy change ∆H° for the r

Kazeer [188]

Answer:

∆H° rxn = - 93 kJ

Explanation:

Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds broken minus bonds formed (H according to Hess Law.

We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.

N₂ (g) + 3H₂ (g) ⇒ 2NH₃ (g)

1 N≡N = 1(945 kJ/mol) 3 H-H = 3 (432 kJ/mol) 6 N-H = 6 ( 389 kJ/mol)

∆H° rxn = ∑ H bonds broken - ∑ H bonds formed

∆H° rxn = [ 1(945 kJ) + 3 (432 kJ) ] - [ 6 (389 k J]

∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ

be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond, N≡N, we have to use this one .

8 0

3 years ago

The density of water at 30.0 °C is 0.9956 g/mL. If the specific gravity of acetic acid is 1.040 at 30.0 °C, what is the density

mash [69]

Answer:

The density of acetic acid at 30°C = 1.0354_g/mL

Explanation:

specific gravity of acetic acid = (Density of acetic acid at 30°C) ÷ (Density of water at 30°C)

Therefore, the density of acetic acid at 30°C = (Density of water at 30°C) × (Specific gravity of acetic acid at 30°C)

= 0.9956 g/mL × 1.040

= 1.0354_g/mL

Specific gravity, which is also known as relative density, is the ratio of the density of a substance to the density of a specified standard substance.

Generally the standard substance of to which other solid and liquid substances are compared is water which has a density of 1.0 kg per litre or 62.4 pounds/cubic foot at 4 °C (39.2 °F) while gases are normally compared with dry air, with a density of 1.29 grams/litre or 1.29 ounces/cubic foot under standard conditions of a temperature of 0 °C and one standard atmospheric pressure

7 0

3 years ago