Answer:
(a) Ala; (b) Tyr; (c) Ser; (d) His
Explanation:
The general formula for an amino acid is NH₂-CH(R)-COOH
The differences among the different amino acids is caused by the side groups R.
(a) Ala, Leu
Ala: R = CH₃-
Leu: R = (CH₃)₂CHCH₂-
Leucine has a larger hydrophobic side group. It should be the less soluble of the pair,
Alanine is the more soluble amino acid.
(b) Tyr, Phe
Tyr: R = HOC₆H₄CH₂-
Phe: R = C₆H₅CH₂-
Tyrosine has a polar OH group. It can form hydrogen bonds with water.
Tyrosine is the more soluble amino acid.
(c) Ser, Ala
Ser: R = HOCH₂-
Ala: R = CH₃-
Serine has a polar OH group. It can form hydrogen bonds with water.
Serine is the more soluble amino acid.
(d) Trp, His
Trp: R = Indole-CH₂-
His: R = Imidazole-CH₂-
Tryptophan has a large aromatic hydrophobic side chain.
Histidine is a basic amino acid. At pH 7 it exists mainly as the anion NH₂CHRCOO⁻, which is hydrophilic.
Histidine is the more soluble amino acid.
Answer:
-Grignard reagents are strong bases (third choice)
-Grignard reagents are strong nucleophiles ( second choice)
Answer:
Zero order
Explanation:
Looking at the data we can note a linear dependence between concentration and time.
Time Conc.
0 2
15 1.82
30 1.64
48 1.42
75 1.10
In the first 15 min it was consumed 2-1.82=0.18. So the rate is 
From 15 to 30 min (it has passed 15 min) is consumed 1.82-1.64=0.18, so as in the previous calculation the rate is 
.
From 30 to 48 (it has passed 18 min)the rate is 
From 48 to 75 (it has passed 27 min) the rate is 
So these results suggest that despite of the ever minor concentration of the reactant the rate is ever the same. Hence the reaction rate could be expressed as 
that is, the reaction is the zero order respect to C2H5I since it is not depending on concentration of C2H5I.
<span>This question asksyou to apply Hess's law.
You have to look for how to add up all the reaction so that you get the net equation as the combustion for benzene. The net reaction should look something like C6H6(l)+ O2 (g)-->CO2(g) +H2O(l). So, you need to add up the reaction in a way so that you can cancel H2 and C.
multiply 2 H2(g) + O2 (g) --> 2H2O(l) delta H= -572 kJ by 3
multiply C(s) + O2(g) --> CO2(g) delta H= -394 kJ by 12
multiply 6C(s) + 3 H2(g) --> C6H6(l) delta H= +49 kJ by 2 after reversing the equation.
Then,
6 H2(g) + 3O2 (g) --> 6H2O(l) delta H= -1716 kJ
12C(s) + 12O2(g) --> 12CO2(g) delta H= -4728 kJ
2C6H6(l) --> 12 C(s) + 6 H2(g) delta H= - 98 kJ
______________________________________...
2C6H6(l) + 16O2 (g)-->12CO2(g) + 6H2O(l) delta H= - 6542 kJ
I hope this helps and my answer is right.</span>
