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Serhud [2]
3 years ago
14

What is the value of Keq for the reaction expressed in scientific notation? 2.1 x 10-2 2.1 x 102 1.2 x 103 1.2 x 10-3

Chemistry
2 answers:
Lady_Fox [76]3 years ago
7 0

Complete question:

Consider the reaction.

At equilibrium at 600 K, the concentrations are as follows.

2HF -----> H₂ + F₂

[HF] = 5.82 x 10-2 M

[H2] = 8.4 x 10-3 M

[F2] = 8.4 x 10-3 M

What is the value of Keq for the reaction expressed in scientific notation?

2.1 x 10-2

2.1 x 102

1.2 x 103

1.2 x 10-3

Answer:

2.1 × 10^-2

Explanation:

Kequilibrum(Keq) = product/reactant

Equation for the reaction :

2HF -----> H₂ + F₂

Therefore,

Keq = [H2][F2] / [HF]^2

Keq = [8.4 x 10-3][8.4 x 10-3] / [5.82 x 10-2]^2

Keq = [70.56 × 10^(-3 + - 3)]/[33.8724 × 10^(-2×2)]

Keq = [70.56 × 10^-6] / [33.8724 × 10^-4]

Keq = 2.0665 × 10^(-6 - (-4))

Keq = 2.0665 × 10^(-6 + 4)

Keq = 2.1 × 10^-2

kobusy [5.1K]3 years ago
4 0

Answer:

At equilibrium at 600 K, the concentrations are as follows.

2HF -----> H₂ + F₂

[HF] = 5.82 x 10-2 M

[H2] = 8.4 x 10-3 M

[F2] = 8.4 x 10-3 M

What is the value of Keq for the reaction expressed in scientific notation?

2.1 x 10-2

2.1 x 102

1.2 x 103

1.2 x 10-3

Answer:

2.1 × 10^-2

Explanation:

Kequilibrum(Keq) = product/reactant

Equation for the reaction :

2HF -----> H₂ + F₂

Therefore,

Keq = [H2][F2] / [HF]^2

Keq = [8.4 x 10-3][8.4 x 10-3] / [5.82 x 10-2]^2

Keq = [70.56 × 10^(-3 + - 3)]/[33.8724 × 10^(-2×2)]

Keq = [70.56 × 10^-6] / [33.8724 × 10^-4]

Keq = 2.0665 × 10^(-6 - (-4))

Keq = 2.0665 × 10^(-6 + 4)

Keq = 2.1 × 10^-2

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Answer is: volume of carbon dioxide is 1,84·10⁸ l.
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n(C) = m(C) ÷ M(C).
n(C) = 10⁸ g ÷ 12 g/mol.
n(C) = 8,33·10⁶ mol.
From chemical reaction: n(C) . n(CO₂) = 1 : 1.
n(CO₂) = 8,33·10⁶ mol.
m(CO₂) = 8,33·10⁶ mol · 44 g/mol.
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Hydrazine (N2H4), a rocket fuel , reacts with oxygen to form nitrogen gas and water vapor. The reaction is represented with the
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Answer:

D)  8.40 L H₂O(g).

Explanation:

  • The balanced equation for the mentioned reaction is:

<em>2N₂H₄(l) + O₂(g) → N₂(g) + 2H₂O(g), </em>

It is clear that 2.0 moles of N₂H₄ react with 1.0 mole of O₂ to produce 2.0 moles of N₂ and 2.0 moles of H₂O.

  • At STP, 4.20L of O₂ reacts with N₂H₄:

It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.

<u>using cross multiplication: </u>

1.0 mol of O₂ represents → 22.4 L.

??? mol of O₂ represents → 4.2 L.

∴ 4.2 L of O₂ represents = (1.0 mol)(4.2 L)/(22.4 L) = 0.1875 mol.

  • To find the no. of moles of H₂O produced:

Using cross multiplication:

1.0 mol of O₂ produce → 2.0 mol of H₂O, from stichiometry.

0.1875 mol of O₂ produce → ??? mol of H₂O.

∴ The no. of moles of H₂O = (2.0 mol)(0.1875 mol)/(1.0 mol) = 3.75 mol.

  • Again, using cross multiplication:

1.0 mol of H₂O represents → 22.4 L, at STP.

3.75 mol of H₂O represents → ??? L.

<em>∴ The no. of liters of water vapor will be produced </em>= (3.75 mol)(22.4 L)/(1.0 mol) = <em>8.4 L.</em>

<em></em>

<em>So, the right choice is: D)  8.40 L H₂O(g).</em>

<em></em>

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