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jekas [21]
3 years ago
14

Which part of the energy profile of a chemical reaction that goes to completion is typically the highest in energy? A. The activ

ation energy B. The products C. The reactants D. The transition state
Chemistry
1 answer:
Helen [10]3 years ago
4 0
The answer is D transition state. In the energy profile, the transition state is the highest point. For a reaction, the activation energy is the minimal energy needed to trigger a reaction. The reactants are the start of the reaction and the products are the end of the reaction.
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Consider the following reaction:
adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

8 0
3 years ago
Which of the following best describes what causes the phases of the moon? The alignment of the sun, Earth, and moon The speed of
BlackZzzverrR [31]
The alignments of the planets would be the correct answer.<span />
6 0
3 years ago
What does a student need to know about double bonds and triple bonds when predicting molecular geometry of molecules?
zhuklara [117]

This problem is asking for an explanation of what we need to know about double and triple bonds to successfully predict molecular geometries in molecules. At the end, one comes to the conclusion that double and triple bonds contribute to the degree in which an atom is bonded and they also determine the lone pairs, which, at the same time,  define the molecular geometry.

<h3>Molecular geometry:</h3>

In chemistry, molecules are not necessarily flat arrangements of atoms, yet they have specific bond angles, orientations and shapes, which define the molecular geometry. In such a way, we can use the VSEPR theory in order to know the molecular geometry of a molecule; however, we first need its Lewis structure or at least the number and type of bonds to do so.

Consider water and carbon dioxide; the former has two hydrogen to oxygen bonds (O-H) and 2 lone pairs because O has six valence electrons but just 2 are bonded to complete the octet, so 4 unpaired electrons lead to two lone pairs. On the other hand, the latter has two double bonds (C=O) and 0 lone pairs because carbon has four valence electrons and they are all bonded to complete the octet.

In such a way, one can see how the double bond affected the bonding in CO2 in contrast to the H2O; situation that also applies to triple bonds, because CO2 has a linear molecular geometry whereas water has a bent one (see attached picture)

Hence, one comes to the conclusion that double and triple bonds contribute to the degree in which an atom is bonded and they also determine the lone pairs, which, at the same time, define the molecular geometry.

Learn more about molecular geometry: brainly.com/question/7558603

Learn more about the VSEPR theory:  brainly.com/question/14225705

5 0
2 years ago
Write a balanced equation for the complete oxidation reaction that occurs when ethane burns in air
damaskus [11]
<h3>Balanced equation : 2C₂H₆ (g) + 7O₂ (g) ⟶ 4CO₂ (g) + 6H₂O (ℓ)</h3><h3>Further explanation</h3>

Alkanes are saturated hydrocarbons that have single bonds in chains

General formula for alkanes :

\tt \large{\bold{C_nH_{2n+2}}

Hydrocarbon combustion reactions (specifically alkanes)

\large {\box {\bold{C_nH _ (_2_n _ + _ 2_) + \dfrac {3n + 1} {2} O_2 \Rightarrow nCO_2 + (n + 1) H_2O}}}

So that the burning of ethane with air (oxygen):

\tt C_2H_6+\dfrac{7}{2}O_2\rightarrow 2CO_2+3H_2O

2C₂H₆ (g) + 7O₂ (g) ⟶ 4CO₂ (g) + 6H₂O (ℓ)

or we can use mathematical equations to solve equilibrium chemical equations by giving the coefficients for each compound involved in the reaction

C₂H₆ (g) + aO₂ (g) ⟶ bCO₂ (g) + cH₂O (ℓ)

C : left 2, right b ⇒ b=2

H: left 6, right 2c⇒ 2c=6⇒ c= 3

O : left 2a, right 2b+c⇒ 2a=2b+c⇒2a=2.2+3⇒2a=7⇒a=7/2

6 0
3 years ago
The circles in the diagrams below represent energy levels in an atom, and the arrows show electron (blue dot) transitions from o
Hoochie [10]
Ranking of the atom from highest to lowest is as follows:
Highest
Arrow = from outer edge to center
2nd Highest
Arrow = second closest ring to the outer edge to center
3rd Highest
Arrow = middle circle to center
Lowest
Arrow = outer edge to middle circle

6 0
3 years ago
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