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.
Answer:
the gravitational force is proportional to the mass of both interacting objects, bigger objects will attract each other with a greater gravitational force. So the mass of either object increases, the force of gravitational attraction between them also increases.
Replacement I think, hope this helps ;)
Explanation:
Answer:
Explanation:
Because of the acid-base reaction, as sodium bicarbonate is introduced to the separatory funnel, the additional or unreacted HBr reacts vigorously to yield CO2 gas, which exits the separatory funnel together with any dissolved compound(s) in the ether layer. This is due to a wrong and incorrect selection of the solvent mixture and the addition of sodium bicarbonate to an acidic solution.
Nothing to do with it until it has leaked out of the separatory funnel. Even then, the student may separate the components from the remaining reaction mixture by washing the ether coating layer several times with brine water, then running it into a dry sodium sulfate bed and evaporating solvent ether under decreased pressure.
<h3>Answer:</h3>
221.90 g.mol⁻¹
<h3>Explanation:</h3>
The mass of contained by a molecule is known as molecular mass. It is the sum of atomic weights of the elements contained by the molecule. In given case Iodine Pentafluoride has a chemical formula,
IF₅
The atomic weights of each element are as;
Iodine = 126.90 g.mol⁻¹
Fluorine = 19.00 g.mol⁻¹
As there are five atoms of Fluorine so, we will multiply the atomic weight of Fluorine by five and that of Iodine by one as there is only one Iodine atom.
Therefore,
Molecular Mass of IF₅ = (1 × 126.90 g.mol⁻¹) + (5 × 19.00 g.mol⁻¹)
Molecular Mass of IF₅ = (126.90 g.mol⁻¹) + (95.00 g.mol⁻¹)
Molecular Mass of IF₅ = 221.90 g.mol⁻¹