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avanturin [10]
3 years ago
14

Classify each amino acid by the chemical properties of its side chain (R group) at pH 7·Select the amino acid that fits best in

each category. Each amino acid will be used only once. 1. This amino acid has a positively charged R group: Select answer 2. This amino acid has a negatively charged R group! Select answer 3. This amino acid has a neutral polar R group: Sect answer 4 This amino acid has a nonpolar aliphatic R tryptophan aspartate valine arginine Select answer 5. This amino acid has an aromatic R group:
Chemistry
1 answer:
Leno4ka [110]3 years ago
7 0

Answer:

. 1. This amino acid has a positively charged R group: ARGININE

2. This amino acid has a negatively charged R group: ASPARTATE

3. This amino acid has a neutral polar R group: NONE

4. This amino acid has a nonpolar aliphatic R: VALINE

5. This amino acid has an aromatic R group: TRYTOPHAN

Explanation:

1) Arginine contains an extra amino group bearing a positive charge, in its chain which imparts basic properties to it

2) Aspartate contains an extra carboxyl group with a dissociable protron. Once the Protron is dissociated, it carries an extra negative charge in its side chain (R)

3) NONE of the amino acids given belong to this group because amino acids with neutral polar R groups contain functional groups that form hydrogen bonds with water. But, this is not the case with tryptophan aspartate valine or arginine

4) Valine has a R group that is hydrocarbon in nature and thus hydrophobic.

5) Trytophan has a benzene ring in its side chain

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Calculate the freezing point of a solution containing 5. 0 grams of kcl and 550. 0 grams of water. the molal-freezing-point-depr
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The freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is  - 0.45°C

Using the equation,

ΔT_{f} = iK_{f}m

where:

ΔT_{f} = change in freezing point (unknown)

i = Van't Hoff factor

K_{f} = freezing point depression constant

m = molal concentration of the solution

Molality is expressed as the number of moles of the solute per kilogram of the solvent.

Molal concentration is as follows;

MM KCl = 74.55 g/mol

molal concentration = \frac{5.0g*\frac{1mol}{74.55g} }{550.0g*\frac{1kg}{1000g} }

molal concentration = 0.1219m

Now, putting in the values to the equtaion ΔT_{f} = iK_{f}m we get,

ΔT_{f} = 2 × 1.86 × 0.1219

ΔT_{f} = 0.4536°C

So, ΔT_{f} of solution is,

ΔT_{f_{solution} } = 0.00°C - 0.45°C

ΔT_{f_{solution} } =  - 0.45°C

Therefore,freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is  - 0.45°C

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