Answer:
The correct answer is "α-helices only use intrachain hydrogen bonds and β-sheets can use either intrachain or interchain hydrogen bonds".
Explanation:
Hydrogen bonding are essential forces for the formation of the secondary structures of proteins. The alpha helix is a coiled secondary structure that is stabilized only by intrachain hydrogen bonds, which are formed between the NH and CO groups of the main chain. On the other hand, β-sheets can use either intrachain or interchain hydrogen bonds for its stabilization. Adjacent chains in β-sheets can run either in the same or opposite directions, which are stabilized by interchain hydrogen bonds.
The statement regarding hydrogen bonding in secondary structures which is true is: C. α-helices only use intrachain hydrogen bonds and β-sheets can use either intrachain or interchain hydrogen bonds.
Hydrogen bond can be defined as a weak chemical bond existing between a partially positively charged hydrogen atom and an electronegative chemical atom.
Protein is a macromolecule which is formed as a result of the union of many amino acids.
A hydrogen bond is an essential intermolecular force of attraction that is required for the formation of the secondary structures of proteins.
Generally, there are two (2) main types of secondary structure in proteins and these include:
Alpha (α) helix: it's a right-handed coiled strand. It is only stabilized or maintained by intrachain hydrogen bonds, which are typically formed between amide hydrogens (NH) and carbonyl oxygens (CO) of the peptide backbone.Beta (β) sheets: it can be stabilized or maintained by either intrachain or interchain hydrogen bonds.In conclusion, α-helices can only use intrachain hydrogen bonds while β-sheets can use either intrachain or interchain hydrogen bonds.
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Which elements cannot have more than an octet of electrons? Select all that apply
C
S
O
N
Br
Answer:
{ Carbon, Oxygen, Nitrogen }
Explanation:
Elements can only have more than an octet of electrons if they demonstrate an expanded octet. This is if they belong to groups in or beyond the third group. Why? Well these elements have d - orbitals that they can rely on to expand the number of electrons that could otherwise be limited. * Here we are focusing on main group elements, P - block elements more specifically. *
Carbon belongs to the 2 group, and thus doesn't have an empty d - orbital. Thus, it can't have more than an octet of electrons. Sulfur belongs to group 3, hence has an empty d - orbital, and can have more than an octet of electrons. Oxygen belongs to the 2 group, and thus doesn't have an empty d - orbital, so it can't have more than an octet of electrons. Same goes for Nitrogen. Bromine belongs to group 4, thus has empty d - orbitals, and can expand further than Sulfur can - it can have more than an octet of electrons.
Solution = { Carbon, Oxygen, Nitrogen }
Carry out the following operations as if they were calculations of experimental results and express each answer in standard notation with the correct number of significant figures and wtih the correct units. Provide both the answer and the units.
1. 5.6792 m + .6 m + 4.33 m
2. 3.70 g - 2.9133 g
3. 4.51 cm x 3.6666 cm
Answer:
1. [tex]10.6\; \rm m[/tex] (one decimal place.)
2.[tex]0.79\; \rm g[/tex] (two decimal places.)
3. [tex]16.5\;\rm cm^2[/tex] (three significant figures.)
Explanation:
1.The first and second expressions are additions and subtractions. When adding two numbers, the accuracy of the result is given by the number of decimal places in it. The result should have as many decimal places as the input with the least number of decimal places.
For example, in the first expression:
[tex]5.6792\;\rm m[/tex] has four decimal places.[tex]0.6\; \rm m[/tex] has only one decimal place.[tex]4.33\; \rm m[/tex] has two decimal places.Therefore, the result should be rounded to one decimal place. Note that these units are compatible for addition, since they are all the same. The result should have the same unit (that is: [tex]\rm m[/tex].)
Therefore:
[tex]\rm 5.6792\; m + 0.6\; m + 4.33\; m \approx 10.6\; \rm m[/tex]. (Rounded to one decimal place.)
2.Similarly:
[tex]\rm 3.70\; \rm g[/tex] has two decimal places.[tex]2.9133\; \rm g[/tex] has four decimal places.Therefore, the result should be rounded to two decimal places. Its unit should be [tex]\rm g[/tex] (same as the unit of the two inputs.)
[tex]\rm 3.70\; g - 2.9133\; g \approx 0.79\; \rm g[/tex]. (Rounded to two decimal places.)
3.When multiplying two numbers, the accuracy of the result should be based on the number of significant figures in it. The result should have as many significant figures as the input with the least number of significant figures. In this expression:
[tex]4.51\; \rm cm[/tex] has three significant figures.[tex]3.6666\; \rm cm[/tex] has five significant figures.Therefore, the result should have only three significant figures.
The unit of the result is supposed to be the product of the units of the input. In this expression, that unit will be [tex]\rm cm \cdot cm[/tex], which is occasionally written as [tex]\rm cm^2[/tex].
[tex]\rm 4.51 \; cm \times 3.6666 \; cm \approx 16.5\; \rm cm^2[/tex]. (Rounded to three significant figures.)
Temperature on Reaction Rate Use the drop-down menus to answer the questions. Which form of the sodium bicarbonate tablet has the most surface area? As the surface area increases, what happens to the average time required for the reaction?
Answer:Crushed, decreased
Explanation:
Just got it right
A gaseous mixture of O2 and N2 contains 37.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 525 mmHg?
PLEASE HELP, will mark brainliest!!!
Answer: The partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg
Explanation:
mass of nitrogen = 37.8 g
mass of oxygen = (100-37.8) g = 62.2 g
Using the equation given by Raoult's law, we get:
[tex]p_A=\chi_A\times P_T[/tex]
[tex]p_{O_2}[/tex] = partial pressure of [tex]O_2[/tex] = ?
[tex]\chi_{O_2} = mole fraction of O_2=\frac{\text{Moles of }O_2}{\text{Total moles}}[/tex]
[tex]P_{T}[/tex] = total pressure of mixture = 525 mmHg
[tex]{\text{Moles of }O_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{62.2g}{32g/mol}=1.94moles[/tex]
[tex]{\text{Moles of }N_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{37.8g}{28g/mol}=1.35moles[/tex]
Total moles = 1.94 + 1.35 = 3.29 moles
[tex]\chi_{O_2}=\frac{1.94}{3.29}=0.59[/tex]
[tex]p_{O_2}=\chi_{O_2}\times P_T=0.59\times 525=310mmHg[/tex]
Thus the partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg
A gaseous mixture of O₂ and N₂ that contains 37.8% nitrogen by mass, and whose total pressure is 525 mmHg, has a partial pressure of oxygen of 310 mmHg.
A gaseous mixture of O₂ and N₂ contains 37.8% nitrogen by mass, that is, in 100 g of the mixture, there are 37.8 g of N₂. The mass of O₂ in 100 g of the mixture is:
[tex]mO_2 = 100 g - 37.8 g = 62.2 g[/tex]
We will convert both masses to moles using their molar masses.
[tex]N_2: 37.8 g \times 1 mol/28.00 g = 1.35 mol\\\\O_2: 62.2 g \times 1 mol/32.00 g = 1.94 mol[/tex]
The mole fraction of O₂ is:
[tex]\chi(O_2) = \frac{nO_2}{nN_2+nO_2} = \frac{1.94mol}{1.35mol+1.94mol} = 0.590[/tex]
Given the total pressure (P) is 525 mmHg, we can calculate the partial pressure of oxygen using the following expression.
[tex]pO_2 = P \times \chi(O_2) = 525 mmHg \times 0.590 = 310 mmHg[/tex]
A gaseous mixture of O₂ and N₂ that contains 37.8% nitrogen by mass, and whose total pressure is 525 mmHg, has a partial pressure of oxygen of 310 mmHg.
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