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2 years ago

PLEASE HELPPPP ASAPPPPPP!!!!!!!

Chemistry

1 answer:

jek_recluse [69]2 years ago

4 0

Answer:

that the oxygen atom is not equal on both sides

Explanation:

Thus the reaction is not balanced.

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The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol. Part A If an enzyme increases the

emmasim [6.3K]

Answer:

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

Explanation:

From the given information:

The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol.

In this same concentration for the glucose and fructose; the reaction rate can be calculated by the rate factor which can be illustrated from the Arrhenius equation;

Rate factor in the absence of catalyst:

$k_1= A*e^{^{^{ \dfrac {- Ea_1}{RT}}$

Rate factor in the presence of catalyst:

$k_2= A*e^{^{^{ \dfrac {- Ea_2}{RT}}$

Assuming the catalyzed reaction and the uncatalyzed reaction are taking place at the same temperature :

Then;

the ratio of the rate factors can be expressed as:

$\dfrac{k_2}{k_1}={ \dfrac {e^{ \dfrac {- Ea_2}{RT} }} { e^{ \dfrac {- Ea_1}{RT} }}$

$\dfrac{k_2}{k_1}={ \dfrac {e^{[ Ea_1 - Ea_2 ] }}{RT} }}$

Thus;

$Ea_1-Ea_2 = RT In \dfrac{k_2}{k_1}$

Let say the assumed temperature = 25° C

= (25+ 273)K

= 298 K

Then ;

$Ea_1-Ea_2 = 8.314 \ J/mol/K * 298 \ K * In (10^6)$

$Ea_1-Ea_2 = 34228.92 \ J/mol$

$\mathbf{Ea_1-Ea_2 = 34.23 \ kJ/mol}$

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

8 0

3 years ago

For each item, choose whether the change is a physical change or chemical change

Aleonysh [2.5K]

Sorry, but where are the ‘items’?

8 0

3 years ago

Mercury(II) oxide (HgO) decomposes to form mercury (Hg) and oxygen (O2). The balanced chemical equation is shown below.

zavuch27 [327]

15.63 mol. You need 15.63 mol HgO to produce 250.0 g O_2.

Step 1. Convert grams of O_2 to moles of O_2

Moles of O_2 = 250.0 g O_2 × (1 mol O_2/32.00 g O_2) = 7.8125 mol O_2

Step 2. Use the molar ratio of HgO:O_2 to convert moles of O_2 to moles of HgO

Moles of HgO = 0.8885 mol O_2 × (2 mol HgO/1 mol O_2) = 15.63 mol HgO

7 0

3 years ago

Read 2 more answers

When metallic aluminum is added to a solution containing iron(II) sulfate a reaction occurs. What species is being oxidized in t

REY [17]

Answer: Aluminium is getting oxidized in the given chemical reaction.

Explanation:

Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

$X\rightarrow X^{n+}+ne^-$

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

$X^{n+}+ne^-\rightarrow X$

For the given chemical reaction:

$2Al(s)+3FeSO_4(aq.)\rightarrow Al_2(SO_4)_3(aq.)+3Fe(s)$

The half cell reactions for the above reaction follows:

Oxidation half reaction: $Al\rightarrow Al^{3+}+3e^-$

Reduction half reaction: $Fe^{2+}+2e^-\rightarrow Fe$

As, aluminium is loosing 3 electrons to form aluminium cation. Thus, it is getting oxidized. Iron is gaining 2 electrons to form iron anion. Thus, it is getting reduced.

Hence, the oxidized species of the given reaction is aluminium.

4 0

2 years ago

In the Lewis structure for a molecule, one of the nitrogen atoms shares six electrons with a carbon atom and has two electrons t

lutik1710 [3]

Answer: option D. triple bond and one lone pair

Explanation:

3 0

2 years ago