What systems are responsible for moving the gases oxygen and carbon dioxide?

Calculated from measurements of distance and time defined

prisoha [69]

Answer:

Any of the six chemical elements that markup group1

of the periodic table.

Explanation:

8 0

3 years ago

An aqueous solution contains 0.23 M potassium hypochlorite.

Arturiano [62]

Answer:

0.22 mol HClO, 0.11mol HBr.

0.25mol NH₄Cl, 0.12 mol HCl

Explanation:

A buffer is defined as a mixture in solution between weak acid and its conjugate base or vice versa.

Potassium hypochlorite (KClO) could be seen as conjugate base of HClO (Weak acid). That means the addition of 0.22 mol HClO will convert the solution in a buffer. HBr reacts with KClO producing HClO, thus, 0.11mol HBr will, also, convert the solution in a buffer. 0.23 mol HBr will react completely with KClO and in the solution you will have only HClO, no a buffering system.

Ammonia (NH₃) is a weak base and its conjugate base is NH₄⁺. That means the addition of 0.25mol NH₄Cl will convert the solution in a buffer. Also, NH₃ reacts with HCl producing NH₄⁺. Thus, addition of 0.12 mol HCl will produce NH₄⁺. 0.25mol HCl consume all NH₃.

5 0

3 years ago

Determine the primary structure of an octapeptide from the following data: acid-catalyzed hydrolysis gives 2 arg, leu, lys, met,

erma4kov [3.2K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The sequence of these amino acid would be

Lue Arg Lys Arg Met Phe Arg Ser

Explanation:

The acid-catalyzed hydrolysis from makes us to understand that the polypeptides contains 8 amino acids

From the question we are told that the Edman's reagent releases Leu it means that the N-terminal amino acid would be Lue(i.e Leucine)

Also from the question we are told that the Carboxypeptidase released Ser this mean the the C-terminal amino acid would be Ser ( i.e Serine)

The Amino acids would in the polypeptides would be arranged like this

Lue __ __ __ __ __ __ Ser

Now from the question we are told that treatment with cranogen bromide form two peptides.

Now generally cranogen bromide divides a peptide on the C- side(i.e the extreme left side ) of Met(Methionine)(This is an amino acid obtained by hydrolysis of most common peptides )

So this means that any peptides containing Met(Methionine) must have Methionine as a C- terminal amino acid(i.e at extreme left) and for peptides that does not contain Met must be C - terminal peptides

From the question we see that it is the second peptide that contain Met and it is a penta peptide(i.e it contains 5 amino acid)

Thus the fifth amino acid is Met

So the sequence of these amino acid would now be

Lue __ __ __ Met __ __ Ser

From the question we are told that the the Trypsin-catalyzed hydrolysis forms two amino acid and two peptides

Now generally Trypsin divides a peptide on the C- side(i.e the extreme left) of Arg(Arginine) and Lys (lysine) and any peptide that holds Arg or Lys must have them as their C- terminal amino acids

From the first peptide in the two peptide formed we see that Arg would be the Seventh amino acid of the octapeptide because commonly the trypsin that sticks to the C-side of Arg would for Ser

and Phe would be the sixth amino acid of the octapeptide

So the sequence of these amino acid would be

Lue __ __ __ Met Phe Arg Ser

Looking at the first amino acid formed from the Trypsin-catalyzed hydrolysis we see that Arg would be the fourth amino acid of octapeptide as Trypsin divides a peptide on the C- side(i.e the extreme left) of Arg(Arginine).

From the second peptide of the Trypsin-catalyzed hydrolysis we see that Lys would be the third amino acid of the octapeptide as trypsin divides on the C- side of Lys (lysine) and Tyr would be the second amino acid of the octapeptide

So the sequence of these amino acid would be

Lue Arg Lys Arg Met Phe Arg Ser

4 0

4 years ago

What are the half-reactions for a galvanic cell with Zn and Mg electrodes?

Alona [7]

the half-reactions

cathode : Zn²⁺ (aq) + 2e⁻ ---> Zn (s)

anode : Mg (s) → Mg²⁺ (aq) + 2e−

a balanced cell reaction

Zn²⁺(aq) + Mg(s)→ Zn(s) + Mg²⁺ (aq)

<h3 /><h3>Further explanation</h3>

Given

Zn and Mg electrodes

Required

The half-reactions for a galvanic cell

Solution

To determine the reaction of a voltaic cell, we must determine the metal that serves as the anode and the metal that serves as the cathode.

To determine this, we can either know from the standard potential value of the cell or use the voltaic series

1. voltaic series

Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au

The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent

So the metal on the left will easily undergo oxidation and function as anode

Since Mg is located to the left of Zn, then Mg functions as anode and Zn as a cathode

2. Standard potentials cell of Mg and Zn metals :

Mg2+ + 2e– → Mg E° = -2,35 V

Zn2+ + 2e– → Zn E° = -0,78 V

The anode has a smaller E°, then Mg is the anode and Zn is the cathode.

7 0

3 years ago

Type of bond between na+1 ion and Cl-1 ion is ion

makvit [3.9K]

A neutral sodium atom, for example, contains 11 protons and 11 electrons. By removing an electron from this atom we get a positively charged Na+ ion that has a net charge of +1. Atoms that gain extra electrons become negatively charged. A neutral chlorine atom, for example, contains 17 protons and 17 electrons.

6 0

3 years ago