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Dovator [93]
3 years ago
14

Write sentences that completely describe each of the chemical reactions shown in these skeleton equations. 11.1

Chemistry
1 answer:
Natasha2012 [34]3 years ago
6 0
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1. Separation of amino acids by ion-exchange Chromatography. Mixtures of amino acids can be analyzed by first separating the mix
mart [117]

a.) Asp and Lys

Asp will elute first from the column because it has less positively charged functional groups than Lys.

 

b.) Arg and Met

Met will elute first from the column because it has less positively charged functional groups than Lys.

c.) Glu and Val

Glu will elute first from the column because it has more negativity functional groups than Lys and will be not be much retained by the -SO₃⁻ groups from the ion-exchange coloumn.

d.) Gly and Val

 Gly will elute first from the column because Lys have a longer alkyl chain which will be attracted by the strongly hydrophobic backbone for the resin.

e.) Ser and Ala

Ser will be eluted first from the column because Ala alkyl chain will be more attracted by the strongly hydrophobic backbone for the resin. Ser have an -OH group which will decrease the hydrophobicity of the alkyl chain and will not be so much retained on the column.

7 0
3 years ago
In a 1.0x10^-4 M solution of HClO(aq), identify the relative molar amounts of these species:HClO, OH-, H3O+, OCl-, H2O
yarga [219]
HClO is a weak acid, which means the ions do not fully dissociate. The hydrolysis reaction for the hypochlorous acid is:

HClO + H2O ⇄ H3O+ +OCl-

Then the equilibrium constant, Ka, of dilute HClO would be:

K_{a} = \frac{[ H_{3}  O^{+} ][O Cl^{-} ]}{HClO}

Then we do the ICE table. I is for the initial concentration, C for the change and E for the excess.
      
          HClO       + H2O   ⇄   H3O+ +  OCl-
I     1.0x10^-4                          0             0
C        -x                                 +x           +x 
E  (1.0x10^-4 - x)                     x             x

Substituting the excess (E) concentration to the Ka equation:

K_{a} = \frac{[x ][x]}{1.0 \ x \  10^{-4} - x }

Simplifying the equation would yield a quadratic equation:

x^{2} + K_{a}x-(1.0 \ x \ 10^{-4}) K_{a}=0

The Ka for HClO is an experimental data which was determined to be 2.9 x 10^-8. Substitute this to the equation, determine the roots, then you get the value for x, which is the concentration of H3O+ and ClO-. Just use your calculator feature Shift-Solve.

x = 1.688 x 10^-6 M = [H3O+] = [ClO-]

Then, you can determine the conc of [OH-] through pH.

pH = -log {H3O+] = -log [1.688 x 10^-6] = 5.77
pOH = 14 - pH = 14 - 5.77 = 8.23
pOH = 8.23 = -log [OH-]
[OH-] = 5.89 x 10^-9 M

Also, since HClO is (1.0x10^-4 - x), then it's concentration would be:
[HClO] = 1.0x10^-4 - 1.688 x 10^-6 = 9.83 x10^-5 M

Let's summarize all concentrations:
[HClO] = 9.83 x10^-5 M
[OH-] = 5.89 x 10^-9 M
[H3O+] = [ClO-] = 1.688 x 10^-6 M
Since the solution is dilute, H2O is relatively higher in concentration.

Thus in relative amounts, the order would be

H2O >>> HClO > H3O+ = ClO- > OH-


6 0
3 years ago
Read 2 more answers
_ S8 +_02 +_ SO3
BigorU [14]
My educated guess should be the 3rd one
3 0
3 years ago
2 HBr(g)+O2(g)—>H2O2(g)+Br2(g)
MrRissso [65]

Answer:

b

Explanation:

8 0
3 years ago
Which are balanced plus if they aren’t please correct them
Veronika [31]

b,f,h are already balanced

7 0
3 years ago
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