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kramer
3 years ago
12

What is the molar concentration of a sucrose solution prepared by dissolving 350.25 g of sucrose is enough deionized water to yi

eld a final solution volume of 500.00 mL?
Chemistry
1 answer:
vovangra [49]3 years ago
3 0

The  molar concentration of a sucrose solution prepared by dissolving 350.25 g of sucrose is enough deionized water to yield a final solution volume of 500.00 mL is 2.048 M.

We must first obtain the number of moles of sucrose in the solution as follows;

Number of moles = mass/ molar

Mass of sucrose = 350.25 g

Molar mass of sucrose = 342 g/mol

Number of moles of sucrose= 350.25 g / 342 g/mol

= 1.024 moles

Recall that;

Number of moles = concentration × volume

concentration = Number of moles/volume

volume of solution = 500.00 mL or 0.5 L

concentration = 1.024 moles/0.5 L

concentration = 2.048 M

Learn more: brainly.com/question/13385951

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In Section 5.6, we learned that triple bonds are stronger and shorter than single bonds. For example, a C−CC−C single bond has a
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3 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

= 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
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