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GenaCL600 [577]
3 years ago
8

At the candy store, gourmet jelly beans sell for $1.15 per ounce. There are 28 students in Alfred's class. If each student eats

0.035 kg of jelly beans, how much would it cost to feed the class a gourmet jelly bean treat?
Chemistry
1 answer:
saw5 [17]3 years ago
8 0

Answer:

Total cost = $40.25 (Approx)

Explanation:

Given:

Per ounce = $1.15

Number of student = 28

Each student eat = 0.035 kg

Find:

Total cost

Computation:

Total weight of candy = 28 × 0.035 kg

Total weight of candy = 0.98 kg

1 ounce = 0.028 kg (approx).

Total weight of candy = 0.98 kg / 0.028

Total weight of candy = 35 ounce (Approx)

Total cost = 35 × $1.15

Total cost = $40.25 (Approx)

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Answer:

Cross-pollination

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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
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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 :

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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} }}

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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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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}

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