3s it is what it says sorry if this is wrong but I just did a qu like this and that's what it was
Answer: The correct answer is option B)
in the synthesis of cryolite for aluminum production.
Explanation:
Hydrogen fluoride (HF) is used in synthesis of cryolite for aluminum production.
HF is a gas that dissolves in water to form hydrofluoric acid. However, HF also react with halogenated alkanes like trichloromethane CHCl3 to form organofluorine compounds that serves as precursors in the synthesis of cryolite, an Ore from which aluminum is produced
Answer:
The number and the kinds of atoms in the compound
Explanation:
- A chemical formula of a compound contains symbols o the atoms of the elements present in the compound as well as the number of atoms of each element in the form of subscripts.
- Therefore it helps us to know the elements in a compound and the number of atoms in the compounds.
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?
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<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>
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- 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.
