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Andru [333]
4 years ago
13

utamate is a polarGlutamate is a blank acidic amino acid, whereas proline is blank. The blank of glutamate may form blank with a

nother amino acid R group. Glutamate may also move to the blank of the protein where it would form blank interactions with blank. However, proline would move to the blank of the protein. acidic amino acid, whereas proline is Glutamate is a blank acidic amino acid, whereas proline is blank. The blank of glutamate may form blank with another amino acid R group. Glutamate may also move to the blank of the protein where it would form blank interactions with blank. However, proline would move to the blank of the protein.. The Glutamate is a blank acidic amino acid, whereas proline is blank. The blank of glutamate may form blank with another amino acid R group. Glutamate may also move to the blank of the protein where it would form blank interactions with blank. However, proline would move to the blank of the protein. of glutamate may form Glutamate is a blank acidic amino acid, whereas proline is blank. The blank of glutamate may form blank with another amino acid R group. Glutamate may also move to the blank of the protein where it would form blank interactions with blank. However, proline would move to the blank of the protein. with another amino acid R group. Glutamate may also move to the Glutamate is a blank acidic amino acid, whereas proline is blank. The blank of glutamate may form blank with another amino acid R group. Glutamate may also move to the blank of the protein where it would form blank interactions with blank. However, proline would move to the blank of the protein. of the protein where it would form Glutamate is a blank acidic amino acid, whereas proline is blank. The blank of glutamate may form blank with another amino acid R group. Glutamate may also move to the blank of the protein where it would form blank interactions with blank. However, proline would move to the blank of the protein. interactions with Glutamate is a blank acidic amino acid, whereas proline is blank. The blank of glutamate may form blank with another amino acid R group. Glutamate may also move to the blank of the protein where it would form blank interactions with blank. However, proline would move to the blank of the protein.. However, proline would move to the Glutamate is a blank acidic amino acid, whereas proline is blank. The blank of glutamate may form blank with another amino acid R group. Glutamate may also move to the blank of the protein where it would form blank interactions with blank. However, proline would move to the blank of the protein. of the protein.
Chemistry
1 answer:
Vanyuwa [196]4 years ago
6 0

Answer:

Polar, nonpolar,polar acidic group R,hydrogen,outer surface,hydrophilic,water and inside.

Explanation:

Glutamate is a <u>polar</u> acidic amino acid, whereas proline is <u>non polar</u>. The <u>polar acidic group R</u> of glutamate may form <u>hydrogen</u> with another amino acid R group. Glutamate may also move to the <u>outer surface</u> of the protein where it would form <u>hydrophilic</u> interactions with <u>water</u>. However, proline would move to the <u>inside</u> of the protein.

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3 years ago
Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH
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Answer:

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Explanation:

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C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole..[1]

H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole..[2]

CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2  then adding all the equations, Using Hess's law:

We get :

C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole..[1]

2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol..[2]

CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole [3]

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3

\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)

\Delta H=-238.7kJ/mole

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

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3 years ago
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