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3 years ago

At pH ___, the pentapeptide Ala-Glu-His-Val-Cys would contain two positively charged groups and one negatively charged group.

Chemistry

1 answer:

REY [17]3 years ago

3 0

Answer:

Explanation:

At pH 3, the pentapeptide Ala-Glu-His-Val-Cys would contain two positively charged groups and one negatively charged group.

This is evident in that peptides have both carboxylic and amino groups.

Given that peptides have a certain amount of H⁺ and OH⁻ ions, it converts to zwitterions.

This conversion process is based on pKₐ and pKb of the acid and the base accordingly.

Therefore, they have two positively charged groups and one negatively charged grou

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Mole ratios how can the coefficients in a chemical equation be interpreted

navik [9.2K]

Answer:

The coefficient in a balanced chemical equation indicates the mole ratio of both reactants and products.

Explanation:

For example lets consider the reation between Hydrogen and Oxygen to form water:

2H2 + O2 ----------------------- 2H2O

In this reaction, the coefficients of the balanced reaction can be transformed to Mole ratio according to Avogadro's Law which states that at standard temperature and pressure, equal volume of gases contain the same number of moles.

So the mole ratio for the above equation is the ratio of the coefficient:

2moles : 1 mole : 2 moles

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3 years ago

Ni4+ + _____e- LaTeX: \longrightarrow⟶ Nix+

stich3 [128]

Kdd was he e shxiuw r thai ed

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3 years ago

How much heat is required to increase the temperature of 198.5 grams of water from 25.0

velikii [3]

5.27*10^4 (52,700.)

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4 years ago

Methanol, CH3OH, is the simplest of the alcohols. It is synthesized by the reaction of hydrogen and carbon monoxide. Write chemi

Serhud [2]

Answer:

The limiting reactant is H₂

Explanation:

The reaction of hydrogen (H₂) and carbon monoxide (CO) to produce methanol (CH₃OH) is the following:

2H₂(g) + CO(g) → CH₃OH(g)

From the balanced chemical equation, we can see that 1 mol of CO reacts wIth 2 moles of H₂. So, the stoichiometric ratio is:

2 mol H₂/1 mol CO = 2.0

We have 500 mol of CO and 750 mol of H₂, so we calculate the ratio to establish a comparison:

750 mol H₂/500 mol CO = 1.5

Since 2.0 > 1.5, we have fewer moles of H₂ than are needed to completely react with 500 moles of CO. In fact, we need 1000 moles of H₂ and we have 750 moles. So, the limiting reactant is H₂.

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3 years ago

Write the charge and full ground-state electron configuration of the monatomic ion most likely to be formed by each:

Maru [420]

Answer :

(a) The charge and full ground-state electron configuration of the monatomic ion is, (+1) and $1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

(b) The charge and full ground-state electron configuration of the monatomic ion is, (-3) and $1s^22s^22p^6$

(c) The charge and full ground-state electron configuration of the monatomic ion is, (-1) and $1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

Explanation :

For the neutral atom, the number of protons and electrons are equal. But, they are unequal when the atoms present in the form of ions or the atom has some charges.

When an unequal number of electrons and protons then it leads to the formation of ionic species.

Ion : An ion is formed when an atom looses or gains electron.

When an atom looses electrons, it will form a positive ion known as cation.

When an atom gains electrons, it will form a negative ion known as anion.

(a) The given element is, Rb (Rubidium)

As we know that the rubidium element belongs to group 1 and the atomic number is, 37

The ground-state electron configuration of Rb is:

$1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^1$

This element will easily loose 1 electron and form $Rb^+$ ion which attain stable noble gas electronic configuration.

The full ground-state electron configuration of Rb ion is:

$1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

(b) The given element is, N (Nitrogen)

As we know that the nitrogen element belongs to group 15 and the atomic number is, 7

The ground-state electron configuration of N is:

$1s^22s^22p^3$

This element will easily gain 3 electrons and form $N^{3-}$ ion which attain stable noble gas electronic configuration.

The full ground-state electron configuration of N ion is:

$1s^22s^22p^6$

(c) The given element is, Br (Bromine)

As we know that the bromine element belongs to group 17 and the atomic number is, 35

The ground-state electron configuration of Rb is:

$1s^22s^22p^63s^23p^64s^23d^{10}4p^5$

This element will easily gain 1 electron and form $Br^-$ ion which attain stable noble gas electronic configuration.

The full ground-state electron configuration of Br ion is:

$1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

4 0

3 years ago