In 1st orbit 2
2nd 8
3rd 10
f orbital has 16
Answer:
Following two compounds have Hydrogen Bond Interactions;
1) CH₃(CH₂)₂NH₂ (Propan-1-amine)<span>
2) </span>CH₃(CH₂)₂NH(CH₂)₄CH₃ (N-propylpentan-1-amine)
Explanation:
Hydrogen Bond Interactions are formed between those molecules which has hydrogen atoms covalently bonded to most electronegative atoms like Fluorine, Oxygen and Nitrogen. This direct attachment of Hydrogen to electronegative atom makes it partial positive resulting in hydrogen bonding with neighbor's partial negative most electronegative atom. So, in above selected compounds it can be seen that both compounds contain hydrogen atoms directly attached to Nitrogen atoms, Therefore, allowing them to form Hydrogen Bonding Interactions.
Answer:
B. To Identify the half reactions for the equation
Noble gases react very unwillingly, because the outermost shell of electrons orbiting the nucleus is full, giving these gases no incentive to swap electrons with other elements. As a result, there are very few compounds made with noble gases. Like its noble gas comrades, neon is odorless and colorless.
Answer:
Fe²⁺(aq) + S²⁻(aq )⟶ FeS(s)
Step-by-step explanation:
Molecular Equation:
(NH₄)₂S(aq) + FeCl₂(aq) ⟶ 2NH₄Cl(aq) + FeS(s)
Ionic equation
:
2NH₄⁺(aq) + S²⁻(aq) + Fe²⁺(aq) + 2Cl⁻(aq) ⟶ 2NH₄⁺(aq) + 2Cl⁻(aq) + FeS(s)
Net ionic equation
:
Cancel all ions that appear on both sides of the reaction arrow (underlined).
<u>2NH₄⁺(aq)</u> + S²⁻(aq) + Fe²⁺(aq) + <u>2Cl⁻(aq)</u> ⟶ <u>2NH₄⁺(aq) </u>+ 2<u>Cl⁻(aq) </u>+ FeS(s)
Fe²⁺(aq) + S²⁻(aq )⟶ FeS(s)
