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Which statements are true? Δ???? for an exothermic reaction is positive. Δ???? for an endothermic reaction is positive. The evap

LuckyWell [14K]

Answer: The true statements are $\Delta H$ for an endothermic reaction is positive, a combustion reaction is exothermic and when energy is transferred as heat from the system to the surroundings, $\Delta H$ is negative.

Explanation:

There are 2 types of chemical reactions categorized into heat change:

Endothermic reactions: These reactions are defined as the reactions in which energy is absorbed by the system from the surroundings. The $\Delta H$ for these reactions is always positive. For Example: Changing of water into water vapor.
Exothermic reactions: These reactions are defined as the reactions in which energy is released by the system to the surroundings. The $\Delta H$ for these reactions is always negative. For Example: Combustion reactions.

Combustion reactions are defined as the reactions in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide and water. Heat is released during these reactions. Thus, it is considered as exothermic reactions.

Hence, the true statements are $\Delta H$ for an endothermic reaction is positive, a combustion reaction is exothermic and when energy is transferred as heat from the system to the surroundings, $\Delta H$ is negative.

6 0

4 years ago

What is the conjugate acid in the following equation hbr + H2O yields h30 positive + BR negative

kirill115 [55]

Answer:

HBr + H2O = H3O+ + Br-

So our conjugate acid is the H3O+ to H2O

Explanation:

A conjugate acid of a base results when the base accepts a proton.

Consider ammonia reacting with water to form an equilibrium with ammonium ions and hydroxide ions:

NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)

Ammonium, NH4+, acts as a conjugate acid to ammonia, NH3.

3 0

3 years ago

The Galapagos Islands formed over a hot spot. The majority of the volcanoes here are ________________ volcanoes. A) shield B) ci

Diano4ka-milaya [45]

I think the answer would be shield volcanoes.

5 0

3 years ago

Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimen

weqwewe [10]

Answer:

Rate = (0.64 M²/min) [A]¹[B]²

Explanation:

1) Determination of the orders of A & B:

The rate law of the reaction = k [A]ᵃ[B]ᵇ

where, k is the rate law constant,

a is the order of the reaction with respect to reactant A,

b is the order of the reaction with respect to reactant B.

This is initial rate method problem:

From trial 1 & 2:

Reactant [A] has the same concentration in both trials, but [B] has different concentrations and the rate of the reaction changes, so the reaction rate depends on [B].

From trial 1, Rate₁ = k [A₁]ᵃ[B₁]ᵇ, [1.2 × 10⁻² M/min] = k [0.30 M]ᵃ[0.25 M]ᵇ

From trial 2, Rate₂ = k [A₂]ᵃ[B₂]ᵇ, [4.8 × 10⁻² M/min] = k [0.30 M]ᵃ[0.50 M]ᵇ

By dividing Rate₁ / Rate₂:

Rate₁ / Rate₂ = k [A₁]ᵃ[B₁]ᵇ / k [A₂]ᵃ[B₂]ᵇ

[1.2 × 10⁻² M/min] / [4.8 × 10⁻² M/min] = k [0.30 M]ᵃ[0.25]ᵇ / k [0.30 M]ᵃ[0.50 M]ᵇ

0.25 = [0.50]ᵇ

Taking log for both sides; log(0.25) = b log(0.5)

b = log(0.25) / log(0.5) = 2.

The reaction is second order with respect to reactant B.

By the same way for reactant A:

From trial 2 & 3:

Reactant [B] has the same concentration in both trials, but [A] has different concentrations and the rate of the reaction changes, so the reaction rate depends on [A].

From trial 2, Rate₂ = k [A₂]ᵃ[B₂]ᵇ, [4.8 × 10⁻² M/min] = k [0.30 M]ᵃ[0.50 M]ᵇ

From trial 2, Rate₃ = k [A₃]ᵃ[B₃]ᵇ, [9.6 × 10⁻² M/min] = k [0.60 M]ᵃ[0.50 M]ᵇ

By dividing Rate₂ / Rate₃:

Rate₂ / Rate₃ = k [A₂]ᵃ[B₂]ᵇ / k [A₃]ᵃ[B₃]ᵇ

[4.8 × 10⁻² M/min] / [9.6 × 10⁻² M/min] = k [0.30 M]ᵃ[0.50 M]ᵇ / k [0.60 M]ᵃ[0.50 M]ᵇ

0.50 = [0.50]ᵃ

Taking log for both sides; log(0.50) = a log(0.5)

a = log(0.50) / log(0.50) = 1.

The reaction is first order with respect to reactant A.

The rate law of the reaction will be: Rate = k [A]¹[B]²

The overall order of the reaction is third order reaction.

2) Determining the rate law constant:

Taking data of trial 1:

Rate = k [A]¹[B]²

[1.2 × 10⁻² M/min] = k [0.30 M]¹[0.25 M]²

k = [1.2 × 10⁻² M/min] / [0.30 M]¹[0.25 M]² = 0.64 M²/min.

So, the rate law will be: Rate = (0.64 M²/min) [A]¹[B]²

3 0

3 years ago

Convert 512 kilograms to milligrams. 0.000512 0.512 512,000 512,000,000

olga55 [171]

The answer is D! 512000000

5 0

4 years ago

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