Answer:
I > III > II
Explanation:
I) A disulfide bond between two cystines is created when a sulfur atom from one cystine forms a strong, single covalent bond with a sulfur atom from a second cystine. When a disulfide bond is created, each cystine loses one hydrogen atom. The atom count is 11 for a cystine in mid-chain, but changes to 10 if the cystine joins with another in a disulfide bond. This lead to a much more stable intermolecular interaction.
III) Hydrogen Bonding in water
These hydrogen bonds are at best an interaction, inducing slight positive and negative charges in the Hydrogen and Oxygen/Nitrogen atoms.
The Hydrophilic amino acids have O & N atoms, which form hydrogen bonds with water. These atoms have an uneven distribution of electrons, creating a polar molecule that can interact and form hydrogen bonds with water.
The hydrogen bonds aren't as strong as the covalent bonds in disulfides.
II) Hydrophobic interactions between two leucines
A hydrophobic interaction is formed between two nonpolar molecules.
It describes the preference of nonpolar molecular surfaces to interact with other nonpolar molecular surfaces, thereby displacing water molecules from the interacting surfaces.
Answer: The molarity of 8 grams of an aqueous solution of sodium hydroxide in 2 liters of solute is 0.1 M
Explanation:
Given: Mass of solute = 8 g
Volume of solution = 2 L
Molar mass of NaOH is 40 g/mol.
Number of moles of NaOH are calculated as follows.
As molarity is the number of moles of solute present in a liter of solution. Therefore, molarity of given solution is calculated as follows.
Thus, we can conclude that the molarity of 8 grams of an aqueous solution of sodium hydroxide in 2 liters of solute is 0.1 M.
The fire rearranges the molecules of a log while it is burning making it a chemical change.
Your answer is C
Atomic radius, as the higher atomic numbers in 2A will have more electron energy levels, making them bigger.
Answer:
270. mL
General Formulas and Concepts:
<u>Acid-Base Titrations</u>
Dilution: M₁V₁ = M₂V₂
- M₁ is stock molarity
- V₁ is stock volume
- M₂ is new molarity
- V₂ is new volume
Explanation:
<u>Step 1: Define</u>
<em>Identify</em>
[Given] V₂ = 629.1 mL
[Given] M₂ = 11.2 M
[Given] M₁ = 26.1 M
[Solve] V₁
<u>Step 2: Find Stock Volume</u>
- Substitute in variables [Dilution]: (26.1 M)V₁ = (11.2 M)(629.1 mL)
- Multiply: (26.1 M)V₁ = 7045.92 M · mL
- Isolate V₁ [Cancel out units]: V₁ = 269.959 mL
<u>Step 3: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs as our lowest.</em>
269.959 mL ≈ 270. mL