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3 years ago

A protein has the amino acid sequence

Chemistry

1 answer:

kherson [118]3 years ago

6 0

Answer:

Number of peptide fragments resulting from cleaving with cyanogen bromide? A: Three peptide fragments

Number of peptide fragments resulting from cleaving with trypsin? A: Four peptide fragments

Which of these reagents gives the smallest single fragment (in number of amino acid residues)? A: CnBr, a dipeptide fragment consisting of AL (Alanine-Leucine)

Explanation:

Cyanogen bromide cleaves the methionine C-terminus, then we have a first fragment of 8 amino acids: DSRLSKTM, a second fragment of 15 aas YSIEAPAKLDWEQNM, and a last fragment of only 2 aas is produced, AL

Trypsin cuts the C-terminus of Arginine and Lysine, then we'll have a first fragment of 3 aas DSR, a second fragment consisting of also 3 aas LSK, a third fragment of 10 aas TMYSIEAPAK, and a last fragment of 9 aas LDWEQNMAL. All produced in three cut sites.

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Analyze: The metallic character of an element is determined by how readily it loses electrons. Elements that lose electrons most

GenaCL600 [577]

Answer:

Group 1 or akali metals have the greatest metallic property.

Group 17 has the lowest metallic character.

C. As you move from right to lefton the periodic table, metallic character increases which is the ability to lose electrons. Ionization energy decrease as we move from right to left on the periodic table.

Explanation:

Akali metals in group 1 have the greatest metallic property and they are the most reactive metals. Francium metal on the group has the most metallic characteristics. It is rare and very radioactive. Group 17 has the lowest metallic character. This is because while moving across the period, the number of electrons in the outermost shell increases. This make it difficult for atoms to leave see electrons and become electropositive . Group 17 has the highest tendency of accepting electrons.

Ionization energy is the energy use to remove electron from an atom in gaseous stage. Ionization energy decrease as we move from right to left on the periodic table and metallic character increases as we move from right to left on the periodic table.

5 0

3 years ago

How were weapons and utensils fashioned during the Stone Age

Alenkasestr [34]

They were created with stone and wood and sometimes even bones, I hope this helped!

4 0

3 years ago

Classify each of the following compounds as a strong acid, weak acid, strong base, or weak base, and write the Ka expression for

Katyanochek1 [597]

Answer:

See explanation below

Explanation:

There are several ways to know if an acid or base is strong. One method is calculating the pH. If the pH is really low, is a strong acid, and if it's really high is a strong base.

However we do not have a pH value here.

The other method is using bronsted - lowry theory. If an acid is strong, then his conjugate base is weak. Same thing with the bases.

Now, Looking at the 4 compounds, we can say that only two of them is weak and the other two are strong compounds. Let's see:

LiOH ---> Strong. If you try to dissociate :

LiOH ------> Li⁺ + OH⁻ The Li⁺ is a weak conjugate acid.

HF -----> Weak

HF --------> H⁺ + F⁻ The Fluorine is a relatively strong conjugate base.

HCl -----> Strong

This is actually one of the strongest acid.

NH₃ ------> Weak

Now writting the Ka and Kb expressions:

Ka = [H⁺] [F⁻] / [HF]

Kb = [NH₄⁺] [OH⁻] / [NH₃]

Finally, to calculate the [OH⁻] we need to use the following expression:

Kw = [H⁻] [OH⁻]

Solving for [OH⁻] we have:

[OH⁻] = Kw / [H⁺]

Remember that the value of Kw is 1x10⁻¹⁴. So replacing:

[OH⁻] = 1x10⁻¹⁴ / 7x10⁻⁶

[OH⁻] = 1.43x10⁻⁹ M

And now, multiplying by 10¹⁰ we have:

[OH⁻] = 1.429x10⁻⁹ * 1x10¹⁰

Hope this helps

4 0

2 years ago

100.0 g of 4.0°C water is heated until its temperature is 37.0°C. If the specific heat of water is 4.184 J/g°C, calculate the am

Lady bird [3.3K]

Answer:

13.8072 kj

Explanation:

Given data:

Mass of water = 100.0 g

Initial temperature = 4.0 °C

Final temperature = 37.0°C

Specific heat capacity = 4.184 j/g.°C

Heat absorbed = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 37.0°C - 4.0 °C

ΔT = 33.0°C

Q = 100.0 g ×4.184 j/g.°C × 33.0°C

Q = 13807.2 j

Joule to KJ:

13807.2 j × 1kj /1000 j

13.8072 kj

5 0

3 years ago

Find the length of time required for the total pressure in a system containing N2O5 at an initial pressure of 0.110 atm to rise

USPshnik [31]

Answer:

t = 37.1 s

Explanation:

The equation for the reaction is given as;

2 N2O5(g) --> 4 NO2 + O2

Initial: 0.110 - -

change: -2x +4x +x

Final: 0.110 - 2x +4x +x

But final = 0.150atm;

0.110 - 2x + 4x + x = 0.150 atm

3x = 0.150 - 0.110

x = 0.0133 atm

Pressure in reactant side;

0.110 - 2x

0.110 - 2 (0.0133) = 0.0834 atm

The integral rate law expression is given as;

ln ( [A] / [Ao] ) = -kt

k = rate constant = 7.48*10^-3*s-1

ln (0.0834/0.11) = (7.48*10^-3) t

upon solving, t = 37.1 s

3 0

3 years ago