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Irina18 [472]
3 years ago
6

I don’t need the answers to this but u can if u want but I’m just asking what is this is asking for and how to solve it? I hope

u can help!

Chemistry
1 answer:
strojnjashka [21]3 years ago
7 0

Answer:

See explanation

Explanation:

Salts are produced from the reaction of an acid and a base. In general ...

Acid + Base => Salt + Weak Electrolyte

Acids from the 'Arrhenius Definition' contain an 'ionizable' hydrogen (-H). Such as, HCl, HBr, HI, HNO₃, HClO₄, HF, etc.

Bases from the 'Arrhenius Definition' contain an 'ionizable' hydroxide (-OH). Such as, LiOH, NaOH, KOH, CsOH, Ca(OH)₂, etc.

When the acid and base react, they proceed by what is known as a 'Double Replacement Reaction' or 'Metathesis Reaction'. In the process, the ions of the reactant compounds exchange positions such that a 'Driving Force' compound is formed on the product side. The Driving Force compound is <u>always</u> on the product side of the equation and is a compound that takes one of three forms => A precipitating salt, a compound of a weak acid or weak base (weaker than the starting acid or base) or gas decomposition product (~ vinegar + backing soda rxn => CO₂ gas).

For your problem, split the compound into cations and anions. You can usually tell which is which by using this format on formulas like those listed in your question => reading formula from left to right, place an imaginary line after the 1st metal => this metal will be the cation & the remaining formula will be the anion.

KBr => K | Br => K (Potassium) is the cation (K⁺) and Br (Bromide) is the anion (Br⁻)

KBr => K⁺ + Br⁻

Now. Apply this rule => Add 'H' to anion => HBr,  then add 'OH' to cation => KOH.

HBr is the acid and KOH is the base. Therefore ...        

Acid + Base => Salt + Wk Electrolyte

HBr + KOH => KBr + H₂O (note how ions of reactants exchange places to form products).

So, KBr comes from the reaction of acid HBr and base KOH.

HBr + KOH => KBr + H₂O

To determine the acid and base origins of LiCl and NaF, use the same logic. Hope this helps. Doc :-)

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