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ra1l [238]
3 years ago
12

When the phenol shown below is treated with KOH, it forms a product whose IR spectrum does not show an absorption in the 3200-36

00 cm-1 region. Propose a structure for the product.?

Chemistry
1 answer:
AlladinOne [14]3 years ago
7 0

Answer:

Explanation:

The first attached diagram shows the proposed phenol that is being treated with KOH to form Oxonine (CₐHₐO).

Reaction of Arylhalide after treatment with KOH lead to the displacement of Bromine atom with the hydrogen from the hydroxyl (OH) substituent , then  forming HBr  , leaving the oxy (-O) atom  alone to form an Oxonine compound.

Therefore the propose structure for the product (Oxonine) CₐHₐO is shown in the second diagram attached below.

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Write the complete balanced equation for the neutralization reaction that occurs when aqueous hydroiodic acid, HI, and sodium hy
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Answer:

H+ ( aq ) + HCO3- ( aq ) ------> H2O( l ) + CO2 ( g )

Explanation:

The complete reaction when hydroiodic acid and sodium hydrogen carbonate combine, would be as follows -

HI + NaHCO3 ----> NaI + H2O + CO2

net reaction

H2CO3 is highly unstable, and thus decomposes into the water and carbon dioxide you see present as the reactants. If you didn't know already, H2CO3 is also reffered to as carbonic acid. The rest of the elements present on the reactant side are Iodine and Sodium, which is why they are present on the product side as NaI.

Let me include the " physical states " in this reaction as well -

HI ( aq ) + NaHCO3 ( aq ) ----> NaI ( aq ) + H2O ( l ) + CO2 ( g )

Now the complete ionic equation would simply be each compound present as ions in an aqueous solution, so there is no need for an explanation on this step -

H+ ( aq ) + I- ( aq ) + Na+ ( aq ) + HCO3- ( aq ) -------> Na+ ( aq ) + I- ( aq ) + H2O( l ) + CO2 ( g )

The spectator ions in this reaction are I- and Na+, so canceling them out, you would receive the following net ionic equation -

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4 0
3 years ago
How many grams of silver chromate will precipitate when 150. mL of 0.500 M silver nitrate are added to 100. mL of 0.400 M potass
sergiy2304 [10]

The amount of silver chromate that precipitates after addition of solutions is 12.44 g.

Number of moles:

The number of moles is the product of molarity of the solution and its volume. The formula is expressed as:

Moles = Molarity x Volume

Calculations:

Step 1:

The molecular formula of silver nitrate is AgNO3. The number of moles of silver nitrate is calculated as:

Moles of AgNO3 = 0.500 M x (150/1000) L

= 0.075 mol

Step 2:

The molecular formula potassium chromate is K2CrO4. The number of moles of potassium chromate is calculated as:

Moles of K2CrO4 = 0.400 M x (100/1000) L

= 0.04 mol

Step 3:

The balanced chemical reaction between AgNO3 and K2CrO4 is:

2AgNO3 + K2CrO4 -----> Ag2CrO4 + 2KNO3

The required number of moles of K2CrO4 = 0.075 mol/2 = 0.0375 mol

The given number of moles of K2CrO4 (0.04 mol) is more than the required number of moles (0.0375 mol). Therefore, AgNO3 is the limiting reagent.

Step 4:

According to the reaction, the molar ratio between AgNO3 and Ag2CrO4 is 2:1. Hence, the number of moles of Ag2CrO4 formed is 0.0375 mol.

The molar mass of Ag2CrO4 is 331.74 g/mol.

The mass of Ag2CrO4 is calculated as:

Mass = 0.0375 mol x 331.74 g/mol

= 12.44 g

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brainly.com/question/13859041

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