1128 is the momentum for this. i think
Answer:
See explanation below
Explanation:
The first two pictures show the reagents used in these reactions a) and b). As it was stated, An E2 reaction proceeds with an antiperiplanar stereochemistry, so in the case of reaction a) it fill form a product with the groups in opposite directions. In other words, a Trans product.
In the case of reaction b) we have the same reaction, with the difference that we have changed the CH3 and phenyl group of positions. This will cause that the reaction will proceed the same but the stereochemistry of the final product will be changed too. In this case, and according to the picture 3 attached, we can see that the product formed is a cis product. So we can conclude that the relation of product a) and b) is that they are isomers, the trans and cis isomers respectively. See picture below for mechanism and products
K2Cr2O7 + 14HCl → 2CrCl3 + 2KCl + 3Cl2 + 7H2O
the correct option is :
K2Cr2O7, because the oxidation number of Cr changes from +6 to +3.
<u>Oxidation number of Cr in K2Cr2O7 is:</u>
K2Cr2O7 = 2K + 2 Cr + 7 O
= 2(+1) + 2Cr + 7(-2)
= 2 + 2Cr -14
[total charge on K2Cr2O7 = 0], Hence;
2 + 2Cr -14 = 0
2Cr -12 = 0
2Cr = 12
Cr = 12/2
<u>Cr = +6</u>
<u>Oxidation number of Cr in CrCl3 is:</u>
CrCl3 = Cr + 3Cl = 0
Cr + 3(-1) = 0
Cr -3 = 0
<u>Cr = +3</u>
Hence Cr is changing its oxidation number from
+6 in K2Cr2O7 to +3 in CrCl3.
Since the oxidation number of Cr [ +6 → +3] is decreasing here,
Cr is getting reduced.
so K2Cr2O7 is an oxidizing agent,as it is getting itself reduced and oxidizes others.
Answer:
H₂S(aq)+ 2 LiOH(aq) → Li₂S(aq) + 2 H₂O(l)
6 HI(aq) + 2 Al(s) → 2 AlI₃(aq) + 3 H₂(g)
2 H₂SO₄(aq) + TiO₂(s) → Ti(SO₄)₂(aq) + 2 H₂O(l)
H₂CO₃(aq) + 2 LiOH(aq) → Li₂CO₃(aq) + 2 H₂O(l)
Explanation:
H₂S(aq)+ 2 LiOH(aq) → Li₂S(aq) + 2 H₂O(l)
This is a neutralization reaction. The products are salt and water.
6 HI(aq) + 2 Al(s) → 2 AlI₃(aq) + 3 H₂(g)
This is a single displacement reaction.
2 H₂SO₄(aq) + TiO₂(s) → Ti(SO₄)₂(aq) + 2 H₂O(l)
This is a neutralization reaction. The products are salt and water.
H₂CO₃(aq) + 2 LiOH(aq) → Li₂CO₃(aq) + 2 H₂O(l)
This is a neutralization reaction. The products are salt and water.
