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3 years ago

What is the role of the activated complex in a chemical reaction?

Chemistry

1 answer:

brilliants [131]3 years ago

8 0

The activated complex is the intermediate between reactants and products in a chemical reaction

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3 years ago

Consider the following chemical reaction: 2KCl + 3O2 --> 2KClO3. If you are given 100.0 moles of KCl and 100.0 moles of O2...

g100num [7]

Answer:

O₂; KCl; 33.3

Explanation:

We are given the moles of two reactants, so this is a limiting reactant problem.

We know that we will need moles, so, lets assemble all the data in one place.

2KCl + 3O₂ ⟶ 2KClO₃

n/mol: 100.0 100.0

1. Identify the limiting reactant

(a) Calculate the moles of KClO₃ that can be formed from each reactant

(i)From KCl

$\text{Moles of KClO}_{3} = \text{100.0 mol KCl} \times \dfrac{\text{2 mol KClO}_{3}}{\text{2 mol KCl}} = \text{100.0 mol KClO}_{3}$

(ii) From O₂

$\text{Moles of KClO}_{3} = \text{100.0 mol O}_{2} \times \dfrac{\text{2 mol KClO}_{3}}{\text{3 mol O}_{2}} = \text{66.67 mol KClO}_{3}$

O₂ is the limiting reactant, because it forms fewer moles of the KClO₃.

KClO₃ is the excess reactant.

2. Moles of KCl left over

(a) Moles of KCl used

$\text{Moles used} = \text{100.0 mol O}_{2} \times \dfrac{\text{2 mol KCl}}{\text{3 mol O}_{2}} = \text{66.67 mol KCl}$

(b) Moles of KCl left over

n = 100.0 mol - 66.67 mol = 33.3 mol

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3 years ago

Reactions cannot occur without a certain minimum amount of energy. What is this minimum amount of energy called? A.the activatio

madam [21]

A. the activation energy

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3 years ago

Read 2 more answers

When aqueous solutions of manganese(II) iodide and sodium phosphate are combined, solid manganese(II) phosphate and a solution o

Feliz [49]

Answer:

The net ionic equation for the given reaction :

$3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)$

Explanation:

$3MnI_2(aq)+2Na_3PO_4_2(aq)\rightarrow Mn_3(PO_4)_2(s)+6NaI(aq)$ ...[1]

$MnI_2(aq)\rightarrow Mn^{2+}(aq)+2I^-(aq)$ ..[2]

$Na_3PO_4(aq)\rightarrow 3Na^{+}(aq)+PO_4^{3-}(aq)$ ...[3]

$NaI(aq)\rightarrow Na^+(aq)+I^-(aq)$

Replacing $MnI_2(aq)$ , NaI and $Na_3PO_4(aq)$ in [1] by usig [2] [3] and [4]

$3Mn^{2+}(aq)+6I^-(aq)+6Na^{+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)+6Na^+(aq)+6I^-(aq)$

Removing the common ions present ion both the sides, we get the net ionic equation for the given reaction [1]:

$3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)$

8 0

4 years ago