Answered

IDNLearn.com makes it easy to find the right answers to your questions. Get accurate and comprehensive answers to your questions from our community of knowledgeable professionals.

Based on the activity series provided, which reactants will form products?

[tex]\[ F \ \textgreater \ Cl \ \textgreater \ Br \ \textgreater \ I \][/tex]

1. [tex]\(\text{CuI}_2 + \text{Br}_2 \rightarrow\)[/tex]
2. [tex]\(\text{Cl}_2 + \text{AlF}_3 \rightarrow\)[/tex]
3. [tex]\(\text{Br}_2 + \text{Na} \rightarrow\)[/tex]
4. [tex]\(\text{CuF}_2 + \text{I}_2 \rightarrow\)[/tex]

Sagot :

GinnyAnsweridn
Sure, let's analyze each reaction based on the activity series:

1. [tex]\( CuI_2 + Br_2 \rightarrow \)[/tex]
- According to the activity series [tex]\( F > Cl > Br > I \)[/tex], bromine (Br) is more reactive than iodine (I).
- However, since Cu is part of the compound as CuI2, we need to compare the reactivities of Br and I directly.
- Since Br is higher in the activity series than I, theoretically Br can displace I.
- But considering the actual reactivity of the compound CuI2, where Cu and I have formed a stable compound, Br2 will not react with CuI2.
- So, no reaction occurs here.

2. [tex]\( Cl_2 + AlF_3 \rightarrow \)[/tex]
- Chlorine (Cl) is more reactive than fluoride (F) as indicated by the activity series.
- But in the compound AlF3, fluorine is already in a highly stable state bounded with aluminum.
- Thus, chlorine will not displace fluorine from this stable compound.
- Therefore, no reaction occurs here.

3. [tex]\( Br_2 + Na \rightarrow \)[/tex]
- Sodium (Na) is highly reactive, much more than bromine (Br).
- When sodium is introduced to bromine, it will react to form sodium bromide (NaBr).
- Therefore, a reaction will occur producing NaBr.

4. [tex]\( CuF_2 + I_2 \rightarrow \)[/tex]
- Fluorine (F) is the most reactive halogen according to the activity series.
- Iodine (I) is the least reactive; it will not be able to displace fluorine from its compounds.
- Thus, iodine will not displace fluorine from CuF2.
- Hence, no reaction takes place here.

Summary of reactions:
1. [tex]\( CuI_2 + Br_2 \rightarrow \)[/tex] No reaction.
2. [tex]\( Cl_2 + AlF_3 \rightarrow \)[/tex] No reaction.
3. [tex]\( Br_2 + Na \rightarrow \)[/tex] Reaction occurs, forming NaBr.
4. [tex]\( CuF_2 + I_2 \rightarrow \)[/tex] No reaction.

Therefore, the answer is: [tex]\( (False, False, True, False) \)[/tex].
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Trust IDNLearn.com for all your queries. We appreciate your visit and hope to assist you again soon.