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andre [41]
3 years ago
12

Which of the following would represent a single displacement reaction between Potassium Bromide (K^+Br^-) and Iodine (I), which

forms iodide (I^-)?
options:
KBr + I -> K + IBr
KBr + I -> KBr + I
KBr + I -> KI + Br
Chemistry
1 answer:
ra1l [238]3 years ago
5 0

Answer:

Option : KBr + I -> KBr+I

Explanation:

Single-replacement reaction or single displacement reactions are a type of chemical reactions in which a whole compound reacts with an element in such a way that the element takes place of one of the compound's own elements and sets it free.

If we talk about KBr and I displacement reaction is not possible among these because Iodine is less reactive than Bromine that is why it will not react with KBr or replace Br.

                       KBr + I -> KBr+ I


Potassium Bromide + Iodine -> Potassium bromide + Iodine


Hope it help!

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3 years ago
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Which of the halide ions (f−, cl−, br−, and i−) is the most stable base? which is the least stable base?
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You should take note that the question is about stability. A compound is stable if it does not easily react with other elements. Hence, its reactivity must be low. As you move down the group, reactivity decreases. So, the halide at the very bottom is the least reactive. It would then be logical that the most stable conjugate base is I⁻ and the least stable conjugate base is the most reactive which is F⁻.
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3 years ago
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Balance the following half reaction in basic conditions. Then, indicate the coefficients for H2O and OH– for the balanced half r
Ugo [173]

Answer:

The ballance half reactions are:

Mg²⁺  + 2e⁻ → Mg

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

Coefficients for H2O and OH– are 3 for H₂O (in products side) and 6 for OH⁻ (in reactants side)

Explanation:

Si (s) + Mg(OH)₂ (s) → Mg (s) + SiO₃²⁻ (aq)

Let's see the oxidations number.

As any element in ground state, we know that oxidation state is 0, so Si in reactants and Mg in products, have 0.

Mg in reactants, acts with +2, so the oxidation number has decreased.

This is the reduction, so it has gained electrons.

Si in reactants acts with 0 so in products we find it with +4. The oxidation number increased it, so this is oxidation. The element has lost electrons.

Let's take a look to half reactions:

Mg²⁺  + 2e⁻ → Mg

Si  → SiO₃²⁻ + 4e⁻

In basic medium, we have to add water, as the same amount of oxygen we have, IN THE SAME SIDE. We have 3 oxygens in products, so we add 3 H₂O and in the opposite site we can add OH⁻, to balance the hydrogen. The half reaciton will be:

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

If we want to ballance the main reaction we have to multiply (x2) the half reaction of oxidation. So the electrons can be ballanced.

2Mg²⁺  + 4e⁻ → 2Mg

Now, that they are ballanced we can sum the half reactions:

2Mg²⁺  + 4e⁻ → 2Mg

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

2Mg²⁺  + 4e⁻  + 6OH⁻ + Si  → 2Mg  +  SiO₃²⁻ + 4e⁻ + 3 H₂O

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Answer:

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Explanation:

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2 years ago
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