In order for a solute to dissolve in a solvent,
the attractive forces between solute particles and the solvent particles must
be stronger than the attractive forces between solute-solute and
solvent-solvent particles. This is important so that the solute will remain in
solution.
Definitely their jawline… just kidding, it’s how much they dissociate
Answer:
According to Bohr, the amount of energy needed to move an electron from one zone to another is a fixed, finite amount. ... The electron with its extra packet of energy becomes excited, and promptly moves out of its lower energy level and takes up a position in a higher energy level. This situation is unstable, however.
The statement is False.
The molecular (chemical) formulas are same in case of isomers but the arrangement of atoms is different in them. The physical properties are determined by the arrangement of atoms within the molecule.
Since the arrangement of atoms is different in case of isomers, the Physical Properties will also be different.
Answer:
0.00370 g
Explanation:
From the given information:
To determine the amount of acid remaining using the formula:
where;
v_1 = volume of organic solvent = 20-mL
n = numbers of extractions = 4
v_2 = actual volume of water = 100-mL
k_d = distribution coefficient = 10
∴




Thus, the final amount of acid left in the water = 0.012345 * 0.30
= 0.00370 g