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aliya0001 [1]
3 years ago
9

Phosphorous pentachloride decomposes according to the reaction

Chemistry
1 answer:
guajiro [1.7K]3 years ago
4 0

Answer: The equilibrium constant, K_c, for the reaction is 0.061.

Explanation:

Initial concentration of PCl_5  = \frac{\text {given mass}}{\text {Molar mass}\times Volume in L}}=\frac{12.3g}{208.2g/mol\times 1.50L}=0.039M  

Equilibrium concentration of PCl_5 = \frac{31.8}{100}\times 0.039=0.012M  

The given balanced equilibrium reaction is,

                            PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)

Initial conc.              0.039 M                0 M        0 M

At eqm. conc.     (0.039-x) M              (x) M   (x) M

Given : (0.039-x) = 0.012

x = 0.027

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[Cl_2]\times [PCl_3]}{[PCl_5]}

Now put all the given values in this expression, we get :

K_c=\frac{0.027\times 0.027}{0.012}=0.061

The equilibrium constant, K_c, for the reaction is 0.061.

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During a combustion reaction, 9.00 grams of oxygen reacted with 3.00 grams of CH4.
Monica [59]

Answer:

0.74 grams of methane

Explanation:

The balanced equation of the combustion reaction of methane with oxygen is:

  • CH₄ + 2 O₂ → CO₂ + 2 H₂O

it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.

firstly, we need to calculate the number of moles of both

for CH₄:

number of moles = mass / molar mass = (3.00 g) /  (16.00 g/mol) = 0.1875 mol.

for O₂:

number of moles = mass / molar mass = (9.00 g) /  (32.00 g/mol) = 0.2812 mol.

  • it is clear that O₂ is the limiting reactant and methane will leftover.

using cross multiplication

1 mol of  CH₄ needs → 2 mol of O₂

???  mol of  CH₄  needs → 0.2812 mol of O₂

∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol

so 0.14 mol will react and the remaining CH₄

mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol

now we convert moles into grams

mass of CH₄ left over = no. of mol of CH₄ left over *  molar mass

                                    = 0.0469 mol * 16 g/mol = 0.7504 g

So, the right choice is 0.74 grams of methane

3 0
3 years ago
Which group contains elements that have the following characteristics:
alekssr [168]

Answer:

1)

Explanation:

the answer to you question Is 1)

6 0
3 years ago
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7 0
2 years ago
Nitrogen dioxide is a red-brown gas responsible for the brown color of smog. A container of nitrogen dioxide that is at low pres
____ [38]

Answer:

Initially 1.51\times 10^{-2} moles of nitrogen dioxide were in the container .

Explanation:

Volume of the container at low pressure and at room temperature =V_1=3.4 L

Number of moles in the container = n_1

After more addition of nitrogen gas at the same pressure and temperature.

Volume of the container after addition = V_2=5.11 L

Number of moles in the container after addition=n_2=2.28\times 10^{-2} mol

Applying Avogadro's law:

\frac{Volume}{Moles}=constant (at constant pressure and temperature)

\frac{V_1}{n_1}=\frac{V_2}{n_2}

n_1=\frac{V_1\times n_2}{V_2}=\frac{3.4 L\times 2.28\times 10^{-2} mol}{5.11 L}

n_1=1.51\times 10^{-2} mol

Initially 1.51\times 10^{-2} moles of nitrogen dioxide were in the container .

8 0
3 years ago
A sample of oxygen occupies 47.2 liters under a pressure of 1240 torr at 298K. What volume would it occupy at 303K if the pressu
kupik [55]

Answer:

81.5 L

Explanation:

We can use the combined gas law equation that gives the relationship among pressure, temperature and volume of gases for a fixed amount of gas.

P1V1 / T1 = P2V2 / T2

where P1 - pressure, V1 - volume and T1 - temperature at the first instance

P2 - pressure, V2 - volume and T2 - temperature at the second instance

substituting the values in the equation

1240 Torr x 47.2 L / 298 K = 730 Torr x V2 / 303 K

V2 = 81.5 L

the new volume the gas would occupy when the conditions have changed is 81.5 L

3 0
3 years ago
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