Answer: if you have this app then you’r not smart
Explanation: because cheating is not allowed
Answer:
It has a negative charge and is located in orbitals around the nucleus
Explanation:
In ionic bonding, atoms transfer electrons to each other. Ionic bonds require at least one electron donor and one electron acceptor.
Answer:
The question is incomplete and confusing.
- In the complete ionic equation you write all the ions that are formed. Those are: Pb²⁺, NO₃⁻, K⁺, and I⁻. They all are present in the complete ionic equation.
- In the net ionic equation, the spectator ions do not appear. They are: NO₃⁻ and K⁺. They would not be present in the net ionic equation, but they do in the complete ionic equation.
See below the details.
Explanation:
Which compound will not form ions?
<u />
<u>1. Write the balanced molecular equation:</u>
- Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq)
<u />
<u>2. Write the ionizations for the ionic aqueous compounds:</u>
<u />
- Pb(NO₃)₂(aq) → Pb⁺²(aq) + 2NO₃⁻(aq)
- 2KI(aq) → 2K⁺(aq) + 2I⁻(aq)
- 2KNO₃(aq) → 2K⁺(aq) + 2NO₃⁻(aq)
<u />
<u>3. Write the complete ionic equation:</u>
Pb⁺²(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + 2I⁻(aq) → PbI₂(s) + 2K⁺(aq) + 2NO₃⁻(aq)
Hence, since PbI₂(s) does not ionize, but stays in solid form, it will not form ions.
All, Pb⁺², NO₃⁻, K⁺, and I⁻ will be present in the total ionic equation.
It is in the net ionic equation that the spectator ions are removed. Those, are NO₃⁻ and K⁺, because they are on both sides of the complete ionic equation.
Many weight charts have 10 pound increments in each frame size; small, medium, and large. ... Being big-boned or small boned doesn't justify a 20-25 pounds differential from a normal size person. If you are truly convinced you are not overweight just big boned.
