Balanced equation is
HBr + NaOH ----> NaBr + H2O
Using molar masses
80.912 g HBr reacts with 39.997 g of Naoh to give 18.007 g water
so 1 gram of NaOH reacts with 2.023 g of HBR
and 5.7 reacts with 11.531 g HBr so we have excess HBr in this reaction
Mass of water produced = (5.7 * 18.007 / 39.997 = 2.6 g to 2 sig figs
Answer:
the answer is true
Explanation:
it is the smallest particle in an element that takes part in a chemical reaction
We are told we have an oxyacid of the formula HOFO. We will assume the atoms are in this order and will draw a proper lewis structure for this compound by first drawing bonds between each of the 4 atoms and then place the remaining electron pairs on each atom:
.. .. ..
H - O - F - O:
·· ·· ··
We can calculate the formal charge of an atom using the following formula:
Formal charge = [# of valence electrons] - [# of non-bonded electrons + # of bonds]
H: Formal charge = [1]-[0+1] = 0
O: Formal charge = [6]-[4+2] = 0
F: Formal charge = [7]-[4+2] = +1
O: Formal charge = [6]-[6+1] = -1
As we can see the overall charge of the molecule is neutral since the fluorine as a +1 charge and the oxygen a -1 charge.
Answer:
a. Gly-Lys + Leu-Ala-Cys-Arg + Ala-Phe
b. Glu-Ala-Phe + Gly-Ala-Tyr
Explanation:
In this case, we have to remember which peptidic bonds can break each protease:
-) <u>Trypsin</u>
It breaks selectively the peptidic bond in the carbonyl group of lysine or arginine.
-) <u>Chymotrypsin</u>
It breaks selectively the peptidic bond in the carbonyl group of phenylalanine, tryptophan, or tyrosine.
With this in mind in "peptide a", the peptidic bonds that would be broken are the ones in the <u>"Lis"</u> and <u>"Arg"</u> (See figure 1).
In "peptide b", the peptidic bond that would be broken is the one in the <u>"Phe"</u> (See figure 2). The second amino acid that can be broken is <u>tyrosine</u>, but this amino acid is placed in the <u>C terminal spot</u>, therefore will not be involved in the <u>hydrolysis</u>.