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inessss [21]
3 years ago
15

Nitrogen and water react to form nitrogen monoxide and hydrogen, like this: N2(g) + 2H2O(g) → 2NO(g) +2H2(g) Also, a chemist fin

ds that at a certain temperature the equilibrium mixture of nitrogen, water, nitrogen monoxide, and hydrogen has the following composition: compound concentration at equilibrium H20 NO H2 0.25 M 1.3 M 0.33 M 1.2 M Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits
Chemistry
1 answer:
Thepotemich [5.8K]3 years ago
8 0

Answer:

Kc for this reaction is 0.43

Explanation:

This is the equilibrium:

N₂(g) + 2H₂O(g) → 2NO(g) +2H₂(g)

And we have all the concentration at equilibrium:

N₂: 0.25M

H₂ : 1.3M

NO: 0.33M

H₂: 1.2M

They are ok, because they are in MOLARITY. (mol/L)

Let's make the expression for Kc

Kc = ( [NO]² . [H₂]² ) / ([N₂] . [H₂O]²)

Kc = (0.33² . 1.2²) / (0.25 . 1.2²)

Kc = 0.4356

In two significant digits. 0.43

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Read 2 more answers
(2 pts) The solubility of InF3 is 4.0 x 10-2 g/100 mL. a) What is the Ksp? Include the chemical equation and Ksp expression. MW
Aleonysh [2.5K]

Answer:

a) Ksp = 7.9x10⁻¹⁰

b) Solubility is 6.31x10⁻⁶M

Explanation:

a) InF₃ in water produce:

InF₃ ⇄ In⁺³ + 3F⁻

And Ksp is defined as:

Ksp = [In⁺³] [F⁻]³

4.0x10⁻²g / 100mL of InF₃ are:

4.0x10⁻²g / 100mL ₓ (1mol / 172g) ₓ (100mL / 0.1L) = <em>2.3x10⁻³M  InF₃. </em>Thus:

[In⁺³] = 2.3x10⁻³M  InF₃ × (1 mol In⁺³ / mol InF₃) = 2.3x10⁻³M  In⁺³

[F⁻] = 2.3x10⁻³M  InF₃ × (3 mol F⁻ / mol InF₃) = 7.0x10⁻³M F⁻

Replacing these values in Ksp formula:

Ksp = [2.3x10⁻³M  In⁺³] × [7.0x10⁻³M F⁻]³ = <em>7.9x10⁻¹⁰</em>

<em></em>

b) 0.05 moles of F⁻ produce solubility of InF₃ decrease to:

7.9x10⁻¹⁰ = [x] [0.05 + 3x]³

Where x are moles of In⁺³ produced from solid InF₃ and 3x are moles of F⁻ produced from the same source. That means x is solubility in mol / L

Solving from x:

x = -0.018 → False solution, there is no negative concentrations.

x = 6.31x10⁻⁶M → Right answer.

Thus, <em>solubility is 6.31x10⁻⁶M</em>

3 0
3 years ago
Palladium (Pd; Z 46) is diamagnetic. Draw partial orbital diagrams to show which of the following electron configurations is con
nexus9112 [7]

Answer : The electron configurations consistent with this fact is, (b) [Kr] 4d¹⁰  

Explanation :

Electronic configuration : It is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom are determined by the electronic configuration.

Paramagnetic compounds : They have unpaired electrons.

Diamagnetic compounds : They have no unpaired electrons that means all are paired.

The given electron configurations of Palladium are:

(a) [Kr] 5s²4d⁸

In this, there are 2 electrons in 's' orbital and 8 electrons in 'd' orbital. From the partial orbital diagrams we conclude that 's' orbital are paired but 'd' orbital are not paired. So, this configuration shows paramagnetic.

(b) [Kr] 4d¹⁰

In this, there are 10 electrons in 'd' orbital. From the partial orbital diagrams we conclude that electrons in 'd' orbital are paired. So, this configuration shows diamagnetic.

(c) [Kr] 5s¹4d⁹

In this, there are 1 electron in 's' orbital and 9 electrons in 'd' orbital. From the partial orbital diagrams we conclude that 's' orbital and 'd' orbital are not paired. So, this configuration shows paramagnetic.

6 0
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