Answer:
Option C= A hydrogen bond formed between a polar side chain and a hydrophobic side chain.
Explanation:
All three given options a, b and d have common mechanism to accommodate the polar amino acid.
A= A hydrogen bond forms between two polar side chains.
B= A hydrogen bond from between a polar side chain and protein back bone.
D = hydrogen bond form between polar side chains and a buried water molecules.
All these are use to accommodate the polar amino acid.
While option C is not used. which is:
A hydrogen bond formed between a polar side chain and a hydrophobic side chain.
Answer:
(upper right) corner of the periodic table to the bottom left corner
A) For balanced chemical equation: 2HgO(s) → 2Hg(l) + O₂(g).
1) Mole ratio 1: n(HgO) : n(Hg) = 2 : 2 (1 : 1).
2) Mole ratio 2: n(HgO) : n(O₂) = 2 : 1.
3) Mole ratio 3: n(Hg) : n(O₂) = 2 : 1.
B) Balanced chemical equation: 4NH₃(g) + 6NO(g) → 5N₂(g) + 6H₂O(l).
1) Mole ratio 1: n(NH₃) : n(NO) = 4 : 6 (2 : 3).
2) Mole ratio 2: n(NH₃) : n(N₂) = 4 : 5.
3) Mole ratio 3: n(NH₃) : n(H₂O) = 4 : 6 (2 : 3).
4) Mole ratio 4: n(NO) : n(N₂) = 6 : 5.
5) Mole ratio 5: n(NO) : n(H₂O) = 6 : 6 (1 :1).
6) Mole ratio 6: n(N₂) : n(H₂O) = 5 : 6.
Answer
7665 years
Procedure
Let N₀ be the amount of carbon-14 present in a living organism. According to the radioactive decay law, the number of carbon-14 atoms, N, left in a dead tissue sample after a certain time, t, is given by the exponential equation:
N = N₀e^(-λt)
where λ is the decay constant which is related to half-life (T1/2) by the equation:
Here, ln(2) is the natural logarithm of 2.
The percent of carbon-14 remaining after time t is given by N/N₀.
Using the first equation, we can determine λt.
The half-life of carbon-14 is 5,720 years, thus, we can calculate λ using the second equation, and then find t.
Solving the second equation for t, and using the λ we have just calculated we will have
t= 7665 years
Whether or not they are within the same group (vertical columns) within the periodic table, determines similarity of chemical properties.
