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ArbitrLikvidat [17]
2 years ago
6

A compound is found to contain 42.88 % carbon and 57.12 % oxygen by weight. To answer the questions, enter the elements in the o

rder presented above. 1. What is the empirical formula for this compound
Chemistry
1 answer:
Ipatiy [6.2K]2 years ago
3 0

Answer:

The empirical formule is CO

Explanation:

Step 1: Data given

Suppose the mass of a compound is 100 grams

Suppose the compound contains:

42.88 % C = 42.88 grams C

57.12 % O = 57.12 grams O

Molar mass C = 12.01 g/mol

Molar mass O = 16.0 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles C = mass C / molar mass C

Moles C = 42.88 grams / 12.01 g/mol

Moles C = 3.57 moles

Moles O = 57.12 grams / 16.0 g/mol

Moles O = 3.57 moles

Step 3: Calculate the mol ratio

We divide by the smallest amount of moles

C: 3.57 moles / 3.57 moles = 1

O: 3.57 moles / 3.57 moles = 1

The empirical formule is CO

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2 years ago
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

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Then ;

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

Ea_1-Ea_2 = 34228.92 \ J/mol

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

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What is the smallest division on this balance?
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0.001 would be the smallest.

Good Luck! :)
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