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Nookie1986 [14]
3 years ago
13

Imagine that you are observing an enzyme-catalyzed reaction in lab. every time you add more enzyme, the reaction rate increases

proportionally until the reaction rate suddenly levels off. no other chemicals were added, and no modifications were made in the experimental setting. why do you think the reaction rate stopped increasing?
Chemistry
1 answer:
tester [92]3 years ago
6 0

Enzymes catalyze the chemical reactions, they act upon the reaction substrates and speed up the reaction. Enzymes have active sites, the places where the reaction substrates interact with the enzyme bringing about the conversion of substrates to products. So, as the enzyme concentration increases the rate of reaction increases till a point where the rate is leveled off. The rate does not further increase, as the substrate might have become limiting at that point. All the available amount of substrate would have been associated with the active sites of the enzymes. So, at that point although there is enough catalyst, lack of substrate would limit the rate of reaction.

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Answer:

Elements having same valence electrons are placed in <u>same group.</u>

Explanation:

First, let's start with some basic concepts of modern periodic table:

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So, elements showing same valency are placed in same group.

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Read 2 more answers
A sample of octane undergoes combustion according to the equation 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O ΔH°rxn = -11018 kJ. What mas
Anna71 [15]

Answer:

\large \boxed{\text{528.7 g} }

Explanation:

It often helps to write the heat as if it were a reactant or a product in the thermochemical equation.

Then you can consider it to be 11018 "moles" of "kJ"  

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

M_r:                      32.00

              2C₈H₁₈ + 25O₂ ⟶ 16CO₂ + 8H₂O + 11 018 kJ

n/mol:                                                                  7280

1. Moles of O₂

The molar ratio is 25 mol O₂:11 018 kJ

\text{Moles of O}_{2} = \text{7280 kJ} \times \dfrac{\text{25 mol O}_{2}}{\text{11 018 kJ}} = \text{16.52 mol O}_{2}

2. Mass of O₂

\text{Mass of C$_{8}$H}_{18} = \text{16.52 mol O}_{2} \times \dfrac{\text{32.00 g O}_{2}}{\text{1 mol O}_{2}} = \textbf{528.6 g O}_{2}\\\text{The reaction requires $\large \boxed{\textbf{528.67 g O}_{2}}$}

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3 years ago
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