Answer:
D
Explanation:
This explains how two noble gases molecules can have an attractive force between them.
This force is called as van dar Waals forces.
It plays a fundamental role in fields in as diverse as supramolecular chemistry structural biology .
If no other forces are present, the point at which the force becomes repulsive rather than attractive as two atoms near one another is called the van der Waals contact distance. This results from the electron clouds of two atoms unfavorably coming into contact.[1] It can be shown that van der Waals forces are of the same origin as the Casimir effect, arising from quantum interactions with the zero-point field.[2] The resulting van der Waals forces can be attractive or repulsive.[3] It is also sometimes used loosely as a synonym for the totality of intermolecular forces.[4] The term includes the force between permanent dipoles (Keesom force), the force between a permanent dipole and a corresponding induced dipole (Debye force), and the force between instantaneously induced dipoles
Answer is: concentration of hydrogenium ions is 9,54·10⁻⁵ M.
c(HNO₂) = 0,075 M.
c(NaNO₂) = 0,035 M.
Ka(HNO₂) = 4,5·10⁻⁵.
This is buffer solution, so use <span>Henderson–Hasselbalch equation:
pH = pKa + log(c(</span>NaNO₂) ÷ c(HNO₂)).
pH = -log(4,5·10⁻⁵) + log(0,035 M ÷ 0,075 M).
pH = 4,35 - 0,33.
pH = 4,02.
<span>[H</span>₃O⁺] = 10∧(-4,02).
<span>[H</span>₃O⁺] = 0,0000954 M = 9,54·10⁻⁵ M.
Answer:
500 mL
Explanation:
Step 1: Find conversions
1 mL = 0.0338 oz
Step 2: Use Dimensional Analysis
= 500 mL
Your answer should be C.) +2. "All the elements in Group 2 have two electrons in their valence shells, giving them an oxidation state of +2."
Credits: https://chem.libretexts.org/Core/Inorganic_Chemistry/Descriptive_Chemistry/Elements_Organized_by_Blo...
Hopefully this has helped! :)
Reduction half reaction: Cu²⁺(aq) + 2e⁻ → Cu⁰(s).
Oxidation half reaction: NO₂⁻(aq) + H₂O(l) → NO₃⁻(aq) + 2H⁺(aq) + 2e⁻.
Balanced chemical reaction:
Cu²⁺(aq) + NO₂⁻(aq) + H₂O(l) → Cu(s) + NO₃⁻(aq) + 2H⁺(aq).
Copper is reduced from oxidation number +2 (Cu²⁺) to oxidation number 0 (Cu) and nitrogen is oxidized from oxidation number +3 (in NO₂⁻) to oxidation number +5 (in NO₃⁻).
