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Alex_Xolod [135]
3 years ago
11

What two parts of an amino acid are involved in a peptide bond?

Chemistry
1 answer:
Lemur [1.5K]3 years ago
6 0

Answer:

The answer to your question is below.

Explanation:

Amino acids are composed by one amino group, one carboxyl group and one  chain.

The parts of the amino acid that are involved in a peptide group are the amino group (- NH₂) and the carboxyl group (-COOH).

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Consider the hypothetical reaction 4A + 3B → C + 2D Over an interval of 3.00 s the average rate of change of the concentration o
pogonyaev

Answer:

Final concentration of C at the end of the interval of 3s if its initial concentration was 3.0 M, is 3.06 M and if the initial concentration was 3.960 M, the concentration at the end of the interval is 4.02 M

Explanation:

4A + 3B ------> C + 2D

In the 3s interval, the rate of change of the reactant A is given as -0.08 M/s

The amount of A that has reacted at the end of 3 seconds will be

0.08 × 3 = 0.24 M

Assuming the volume of reacting vessel is constant, we can use number of moles and concentration in mol/L interchangeably in the stoichiometric balance.

From the chemical reaction,

4 moles of A gives 1 mole of C

0.24 M of reacted A will form (0.24 × 1)/4 M of C

Amount of C formed at the end of the 3s interval = 0.06 M

If the initial concentration of C was 3 M, the new concentration of C would be (3 + 0.06) = 3.06 M.

If the initial concentration of C was 3.96 M, the new concentration of C would be (3.96 + 0.06) = 4.02 M

3 0
3 years ago
A women steps in front of a child to keep him from running off which term best describes this example
valina [46]
Negative force

The woman is applying force to stop the child or to prevent him from running.

I hope this helps...
4 0
3 years ago
An element crystallizes in a face-centered cubic lattice. If the length of an edge of the unit cell is 0.408 nm, and the density
V125BC [204]

Answer:

Au

Explanation:

For the density of a face-centered cubic:

Density = \dfrac{4 \times M_w}{N_A \times a^3}

where

M_w = molar mass of the compound

N_A= avogadro's constant

a^3 = the volume of a unit cell

Given that:

Density (\rho) = 19.30 g/cm³

a = 0.408 nm

a = 0.408 \times 10^{-9} \times 10^{2} \ cm

a = 4.08 \times 10^ {-8} \ cm

∴

19.3 = \dfrac{4 \times M_w}{(6.023 \tmes 10^{23})\times (4.08 \times 10^{-8})^3}

M_w= \dfrac{19.3\times (6.023 \times 10^{23})\times (4.08 \times 10^{-8})^3}{4}

M_w=197.37 \ g/mol

Thus, the molar mass of 197.37 g/mol element is Gold (Au).

4 0
2 years ago
Consider the reaction of magnesium metal with hydrochloric acid to produce magnesium chloride and hydrogen gas. If 3.29 mol of m
gregori [183]

Answer:

1.645 moles of excess reactant that is of magnesium metal are left over.

Explanation:

Moles of magnesium metal = 3.29 mol

Moles of HCl = 3.29 mol

Mg(s)+2HCl(aq)\rightarrow MgCl_2(aq)+H_2(g)

According to recation, 2 moles of HCl reacts with 1 mol of magnesium metal, then 3.29 moles of HCl will react with :

\frac{1}{2}\times 3.29 mol=1.645 mol of magnesium metal

Moles of HCl left = 3.29mol - 3.29 mol = 0

Moles of magnesium metal left = 3.29 mol - 1.645 mol = 1.645 mol

1.645 moles of excess reactant that is of magnesium metal are left over.

7 0
2 years ago
How many solutions are there in the following equation.
givi [52]
When they ask you for the solution they are usually asking for x. So solve for X.
X=-12  so you only have one solution. so the answer is B
6 0
3 years ago
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