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DochEvi [55]
2 years ago
15

Consider the equation: 2NO2(g) N2O4(g). Using ONLY the information given by the equation which of the following changes would in

crease the molar concentration at equilibrium of the product N2O4(g)?
increase the pressure
increase the temperature
decrease the concentration of NO2(g)
decrease the temperature
decrease the pressure
Chemistry
2 answers:
ratelena [41]2 years ago
8 0

The answers A. increase the pressure

Explanation:

Reika [66]2 years ago
7 0
I believe the correct answer is the first option. To increase the molar concentration of the product N2O4, you should increase the pressure of the system. You cannot determine the effect of changing the temperature since we cannot tell whether it is an endothermic or an exothermic reaction. Also, decreasing the number of NO2 would not increase the product rather it would shift the equilibrium to the left forming more reactants. The only parameter we can change would be the pressure. And, since NO2 takes up more space than the product increasing the pressure would allow the reactant to collide more forming the product.
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Name the region of the atom where protons and neutrons are located. This region is indicated by label B. Name the region of the
11Alexandr11 [23.1K]

Protons and neutrons are located at the nucleus.

Electrons are located on the outer shells, or orbitals.

Subatomic Particles are located inside of the atoms. Protons, neutrons, and electrons.

7 0
2 years ago
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A graduated cylinder initially has 32.5 mL of water in it. After a 75.0 g piece of lead (Pb) is added to the graduated cylinder,
Mice21 [21]

Answer:

39.1-32.5 and you will find your answer it always like that, you subtract your starting point from your ending point

Explanation:

6 0
1 year ago
Given that the rate constant is 4.0×10−4 m−1 s−1 at 25.0 ∘c and that the rate constant is 2.6×10−3 m−1 s−1 at 42.4 ∘c, what is t
stepladder [879]
So here's how you find the answer:

Given: (rate constants)

K₁ = 4.0 x 10⁻⁴ M⁻¹s⁻¹.
T₁ = 25.0 C = 293 K.
k₂ = 2.6 x10⁻³ M⁻¹s⁻¹.
T₂ = 42.4 C = 315.4 K.
R = 8,314 J/K·mol.

Use the equation:

ln(k₁/k₂) = Ea/R (1/T₂ - 1/T₁).

Transpose:

Ea = R·T₁·T₂ / (T₁ - T₂) · ln(k₁/k₂)

Substitute within the given transposed equation:

<span>Ea = 8,314 J/K·mol · 293 K · 315.4 K ÷ (293 K - 315.4 K) · ln (4.0 x 10⁻⁴ M⁻¹s⁻¹/ 2.6 x 10⁻³ M⁻¹s⁻¹).
</span>
Continuing the solution we get:

<span>Ea = 768315 J·K/mol ÷ (-22,4 K) * (-1.87)
</span>
The value of EA is:

<span>Ea = 64140.58 J/mol ÷ 1000 J/kJ = 64.140 kJ/mol.</span>


4 0
2 years ago
Rank the boiling points of the following molecules from highest to lowest. butanone diethyl ether butane and butanol.
Gekata [30.6K]

Answer:

From highest to lowest:

butanol: 117.7 degree Celsius

butanone: 79.64 degree Celsius

diethyl ether: 34.6 degree Celsius

n-butane: -0.4 degree Celsius

7 0
2 years ago
g You observed the formation of several precipitates in the Reactions in Solution lab exercise. Identify the precipitate in each
RUDIKE [14]

<u>Answer:</u>

<u>For a:</u> Lead iodide is a yellow precipitate.

<u>For b:</u> Barium sulfate is a white precipitate.

<u>For c:</u> Ferric hydroxide is a brown precipitate.

<u>For d:</u> Copper (II) hydroxide is a blue precipitate.

<u>Explanation:</u>

Precipitation reaction is defined as the reaction where a solid precipitate (solid substance) is formed at the end of the reaction. It is insoluble in water.

For the given options:

  • <u>For (a):</u>

The chemical reaction between KI and lead (II) nitrate follows:

2KI(aq)+Pb(NO_3)_2(aq)\rightarrow PbI_2(s)+2KNO_3(aq)

The iodide of lead is generally insoluble in water. Thus, lead iodide is a yellow precipitate.

  • <u>For b:</u>

The chemical reaction between barium chloride and sulfuric acid follows:

BaCl_2(aq)+H_2SO_4(aq)\rightarrow BaSO_4(s)+2HCl(aq)

The sulfate of barium is insoluble in water. Thus, barium sulfate is a white precipitate.

  • <u>For c:</u>

The chemical reaction between NaOH and ferric chloride follows:

3NaOH(aq)+FeCl_3(aq)\rightarrow Fe(OH)_3(s)+3NaCl(aq)

The hydroxide of iron is insoluble in water. Thus, ferric hydroxide is a brown precipitate.

  • <u>For d:</u>

The chemical reaction between NaOH and copper sulfate follows:

CuSO_4+2NaOH\rightarrow Cu(OH)_2+Na_2SO_4

The hydroxide of copper is insoluble in water. Thus, copper (II) hydroxide is a blue precipitate.

6 0
1 year ago
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