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dedylja [7]
3 years ago
8

In the peptide GLVIW, the C-terminal end is ________.

Chemistry
1 answer:
zheka24 [161]3 years ago
5 0
The peptide given above is made up of the following amino acids: glycine [G], leucine [L], valine [V], isoleucine [I] and tryptophan [W]. These amino acids are joined together by amide bond to form peptide. Peptides usually have two terminals, the N terminal and the C terminal. For GLVIW, the C terminal end amino acid is tryptophan, that is the last amino acid on the peptide chain. The N terminal amino acid is glycine, that is, the first amino acid on the peptide chain.
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Explain why the ionization energy to remove a second electron from potassium is higher than the ionization energy to remove four
never [62]

Explanation:

It is more difficult to remove electrons from the second shell or energy level because of the imbalance between the positive nuclear charge and the remaining electrons.

  • The amount of energy required to remove electrons in ground state of an atom is the ionization energy.
  • The first ionization energy is the energy needed to remove the most loosely bound electron of an atom in the gas phase in ground state.
  • The second energy has a greater nuclear pull as it is closer to the nucleus.
  • Both potassium and silicon have the same number of energy levels.
6 0
3 years ago
How does the law of conservation of mass apply to this reaction C2H4 + O2 > 2H2O + 2CO2
Sindrei [870]
Coefficients are used here to balance the number of elements in the products and reactants. Coefficients are numbers that are placed in front of a chemical formulas in an equation. Hope it's useful.
7 0
3 years ago
Read 2 more answers
When the nuclide iron-59 undergoes beta decay: The name of the product nuclide is . The symbol for the product nuclide is .
Travka [436]

Answer:

Explanation:

⁵⁹₂₆Fe --------- ⁰₋₁e + ⁵⁹₂₇Co

Co- 59 is known as Cobalt

its symbol is ⁵⁹₂₇Co

5 0
3 years ago
How much energy must be removed from a 125 g sample of benzene (molar mass= 78.11 g/mol) at 425.0 K to liquify the sample and lo
riadik2000 [5.3K]

Answer : The energy removed must be, -67.7 kJ

Solution :

The process involved in this problem are :

(1):C_6H_6(g)(425.0K)\rightarrow C_6H_6(g)(353.0K)\\\\(2):C_6H_6(g)(353.0K)\rightarrow C_6H_6(l)(353.0K)\\\\(3):C_6H_6(l)(353.0K)\rightarrow C_6H_6(l)(335.0K)

The expression used will be:

\Delta H=[m\times c_{p,g}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,l}\times (T_{final}-T_{initial})]

where,

\Delta H = heat released by the reaction = ?

m = mass of benzene = 125 g

c_{p,g} = specific heat of gaseous benzene = 1.06J/g^oC

c_{p,l} = specific heat of liquid benzene = 1.73J/g^oC

\Delta H_{vap} = enthalpy change for vaporization = 33.9kJ/mole=33900J/mole=\frac{33900J/mole}{78.11g/mole}J/g=434.0J/g

Molar mass of benzene = 78.11 g/mole

Now put all the given values in the above expression, we get:

\Delta H=[125g\times 1.06J/g.K\times (353.0-(425.0))K]+125g\times -434.0J/g+[125g\times 1.73J/g.K\times (335.0-353.0)K]

\Delta H=-67682.5J=-67.7kJ

Therefore, the energy removed must be, -67.7 kJ

4 0
3 years ago
If a solution conducts electricity, it is positive evidence that
BlackZzzverrR [31]

Answer:

If a solution conducts electricity, it is positive evidence that solute dissolved in solvent is electrolyte.

3 0
3 years ago
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