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pychu [463]
3 years ago
15

The reaction for the formation of gaseous hydrogen fluoride (HF) from molecular hydrogen (H2) and fluorine (F2) has an equilibri

um constant of 1.15 x 102 at a certain temperature. H2 (g) + F2 (g) ⇋ 2 HF (g) In a particular experiment, 3.00 mol H2 and 6.00 mol F2 are mixed in a 3.00 L flask at this temperature and allowed to come to equilibrium. Given this information, calculate the concentrations of H2, F2, and HF present in the flask at equilibrium.
Chemistry
1 answer:
mestny [16]3 years ago
5 0

Answer:

[H2]eq = 0.0129 M

[F2]eq = 1.0129 M

[HF]eq = 0.9871 M

Explanation:

  • H2(g) + F2(g) ↔ 2HF(g)

∴ Ke = [HF]² / [H2]*[F2] = 1.15 E2

experiment:

∴ n H2 = 3.00 mol

∴ n F2 = 6.00 mol

∴ V sln = 3.00 L

⇒ [H2]i = 3.00 mol / 3.00 L = 1 M

⇒ [F2]i = 6.00 mol / 3.00 L = 2 M

        [ ]i    change      [ ]eq

H2     1         1 - x         1 - x

F2     2        2 - x         2 - x

HF     -            x              x

⇒ K = (x)² / (1 - x)*(2 - x) = 1.15 E2

⇒ x² / (2 - 3x + x²) = 1.15 E2 = 115

⇒ x² = (2 - 3x + x²)(115)

⇒ x² = 230 - 345x + 115x²

⇒ 0 = 230 - 345x + 114x²

⇒ x = 0.9871

equilibrium:

⇒ [H2] = 1 - x = 1 - 0.9871 = 0.0129 M

⇒ [F2] = 2 - x = 2 - 0.9871 = 1.0129 M

⇒ [HF] = x = 0.9871 M

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C3H8 +502 + 3CO2 + 4H2O
astra-53 [7]

Answer:

Explanation:

first, you calculate the amount of O2 in moles:

98.0 ÷ 32 = 3.0625

second, the ratio if O2/C3H8 is 5 so you need to calculate O2 in moles with that:

3.0625 ÷ 5 = 0.6125

third, the amount of CO2 in moles also can be calculate by the ratio of C3H8/CO2 which is 3

0.6125 × 3 = 1.8375

then multiply CO2 in moles by its molar mass which is 44 g/mol

1.8375 × 44 = 80.85g

5 0
3 years ago
What mass of carbon dioxide is produced from the complete combustion of 5.30x10-3 g of methane?
pshichka [43]

Answer:

1.45 x 10⁻² g CO₂

Explanation:

To find the mass of carbon dioxide, you need to (1) convert grams CH₄ to moles CH₄ (via molar mass), then (2) convert moles CH₄ to moles CO₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles CO₂ to grams CO₂ (via molar mass). The final answer should have 3 sig figs to reflect the given value (5.30 x 10⁻³ g).

Molar Mass (CH₄): 12.011 g/mol + 4(1.008 g/mol)

Molar Mass (CH₄): 16.043 g/mol

Combustion of Methane:

1 CH₄ + 2 O₂ ---> 2 H₂O + 1 CO₂

Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)

Molar Mass (CO₂): 44.007 g/mol

5.30 x 10⁻³ g CH₄          1 mole              1 mole CO₂          44.007 g
---------------------------  x  ----------------  x  ---------------------  x  -----------------  =
                                      16.043 g            1 mole CH₄            1 mole

=  0.0145 g CO₂

=  1.45 x 10⁻² g CO₂

8 0
2 years ago
Which of the following is most likely to cause the particles in a chemical change to slow down
Nady [450]
C. Lowering the temperature

At the lower temperature, particles have less kinetic energy, so they move slower.



7 0
3 years ago
If the temperature of a gas increased and the pressure remains the same what happens to the volume of gas?​
podryga [215]

Answer:

the answer is B.

Explanation:

The pressure increases... plus i had this same question

pls give me brainliest

3 0
3 years ago
The reaction, 2C4H10 (g) + 13O2 (g) à 8CO2 (g) + 5H2O (g), is the combustion of butane. What occurs as the reaction proceeds?
Solnce55 [7]
The reaction, 2 C4H10 (g) + 13 O2 (g) = 8 CO2 (g) + 5 H2O (g), is the combustion of butane.   A combustion reaction involves the reaction of a hydrocarbon with oxygen producing carbon dioxide and water. This reaction is exothermic which means it releases energy in the form of heat. Therefore, as the reaction proceeds,a heat energy is being given off by the reaction. This happens because the total kinetic energy of the reactants is greater than the total kinetic energy of the products. So, the excess energy should be given off somewhere which in this case is released as heat.
7 0
3 years ago
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