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mart [117]
3 years ago
14

Classify each amino acid into the following categories. Select the amino acid that fits best in each category. Each amino acid w

ill be used only once. 1. A basic amino acid: 2. An acidic amino acid: 3. A neutral polar amino acid: 4. A nonpolar aliphatic amino acid: 5. An aromatic amino acid:
Chemistry
1 answer:
mel-nik [20]3 years ago
3 0

Answer:

1. Lysine

2. Aspartic acid

3. Serine

4. Alanine

5. Tryptophan

Explanation:

Amino acids are biomolecules that contain two functional groups and one R side chain. The two functional groups are: carboxyl group and amino group.

The α-amino acids are the amino acids in which the two functional groups and the R side chain are attached to the α-carbon of the amino acid. They are total 22 α-amino acids.

1. A basic amino acid: Lysine is a positively charged, polar basic amino acid with a lysyl side chain.

2. An acidic amino acid: Aspartic acid is a negatively charged, polar acidic amino acid with an acidic carboxymethyl group.

3. A neutral polar amino acid: Serine is a polar and neutral amino acid with a hydroxymethyl group.

4. A non-polar aliphatic amino acid: Alanine is an aliphatic, nonpolar and neutral amino acid with a methyl side chain.

5. An aromatic amino acid: Tryptophan is an aromatic, nonpolar and neutral amino acid with an indole side chain.

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What is kind of property is denisty?
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INTENSIVE PROPERTY OF MATTER

Explanation:

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How can melting point and boiling point identify a substance
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8 0
3 years ago
When 40.5 g of Al and 212.7 g of Cl2 combine in the reaction: 2 Al (s) + 3 Cl2 (g) --> 2 AlCl3 (s) c) How many moles of the e
sineoko [7]

Answer:

The answer to your question is 1.49 mol of Cl₂  

Explanation:

Data

mass of Al = 40.5 g

mass of Cl₂ = 212.7 g

moles of excess reactant = ?

- Balanced chemical reaction

                2 Al(s)  +  3Cl₂(g)  ⇒   2AlCl₃

Process

a) Calculate the molar mass of the reactants  

molar mass of Al = 2 x 26.98 = 53.96 g

molar mass of Cl₂ = 6 x 35.45 = 212.7 g

b) Calculate the theoretical proportion  Al/Cl₂ = 53.96/212.7 = 0.254

   Calculate the experimental proportion Al/Cl₂ = 40.5/212.7 = 0.19

As the experimental proportion is lower than the theoretical proportion we conclude that the limiting reactant is Aluminum.

c) Calculate the grams of excess reactant

                    53.96 g of Al ------------------ 212.7 g of Cl₂

                     40.5 g of Al -------------------  x

                      x = (40.5 x 212.7) / 53.96

                      x = 8614.35 / 53.96

                      x = 159.64 g of Cl₂

Excess Cl₂ = 212.7 - 159.64

                  = 53.057 g

d) Calculate the moles of Cl

                       35.45 g of Cl ----------------- 1 mol

                       53.057 g of Cl ---------------  x

                        x = (53.057 x 1)/35.45

                       x = 1.49 mol of Cl₂                    

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Temperature will raise and water will evaporate
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