45 m. If each student needs 750 mm of tubing, the teacher should order 45 m of tubing.
a) Find the <em>length in millimetres</em>
Length = 60 students x (750 mm tubing/1 student) = 45 000 mm tubing
b) Convert <em>millimetres to metres
</em>
Length = 45 000 mm tubing x (1 m tubing/1000 mm tubing) = 45 m tubing
Reactivity of non-metals depend on their ability to gain electrons. So, smaller is the size of a non-metal more readily it will attract electrons because then nucleus will be more closer to valence shell. ... Hence, Br is the non-metal which will be more reactive than At.
Talking about ions, a cation has a positive charge. That means a positive charge is gained by an atom that gives out electrons.
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.