<u>Q</u><u>U</u><u>E</u><u>S</u><u>T</u><u>I</u><u>O</u><u>N</u><u> </u><u>:</u>
- highest energy level for lithium
<u>A</u><u>N</u><u>S</u><u>W</u><u>E</u><u>R</u><u> </u><u>:</u>
- There are 2 electrons on the first energy level and 1 electron on the second. Explain that the first energy level can only have 2 electrons so the next electron in lithium is on the next (second) level. Explain that neon has 10 protons and 10 electrons.
Answer:
a and b are correct
Explanation:
This because both are aqueous solutions,therefore, identity of solvent is same that is water.
And because both solutions are non electrolyte they would not ionize in solution, and for the same concentration, the freezing point of both solution would also be same. Since depression in freezing point is a colligative property this means that it depends on number of solute particles not nature of particles
.
Hence answer is that their freezing points and Identity of the solvent shall remain the same.
Explanation:
The different subatomic particles in an atom are the:
- Protons which are the positively charged particles.
- Electrons which are the negatively charged particles.
- Neutrons which do not carry any charges.
Protons and neutrons are located in the nucleus of an atom which is the tiny center of the atom.
Electrons orbits around the nucleus and fill the rest of the volume of atom.
Answer : The electrons in a bond between two iodine atoms are shared equally and non-polar.
Explanation :
Non-polar covalent bond : It is a type of bond that is present between the two similar and different atoms. This type of bond is formed when the electrons are shared equally between the atoms.
This type of bond is termed as non-polar because of the electronegativity difference between the atoms is negligible.
The given molecule
is formed by the equal sharing of electrons and the bond formed between the iodine-iodine is non-polar due to negligible electronegativity difference.
Therefore, the electrons in a bond between two iodine atoms are shared equally and non-polar.