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djverab [1.8K]
2 years ago
7

The reaction that you will conduct in this experiment is known as a "haloform" reaction. When a methyl ketone is used as a subst

rate, what is it converted to in this reaction?

Chemistry
1 answer:
RUDIKE [14]2 years ago
6 0

It is converted into a CARBOXYLIC ACID

Haloform reaction mechanism.

First step

The base (hydroxide ion) takes out the alpha hydrogen producing enolate. Then, the reaction between the enolate and the halogen occurs, leading to the formation of the halogenated ketone along with the halogens corresponding anion.

Second step

Step 1 is repeated twice to yield a tri-halogenated ketone. The net reaction till the formation of the tri-halogenated ketone can be written as

The hydroxide ion acts as a nucleophile and attacks the electrophilic carbon which is doubly bonded to oxygen. This carbon-oxygen double bond becomes a single bond making the oxygen atom anionic. This makes the reformation of the carbon oxygen double bond favourable and the carbon attached to three halogens is displaced, leaving us with the carboxylic acid. An acid base reaction ensues, the carboxylic acid donates a proton to the tri-halomethyl anion giving the required haloform product.

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Write a balanced equation showing how this nucleus decays to form an α particle: thorium−230.
Ostrovityanka [42]
When   an  atomic  nucleus  emits  an  alpha  particle  it  decay  into  an atom   with atomic  number  2  less   and   mass number  4  less.  Thus  Thorium   230   decay    as  follows.

230 90Th  -------> 226  88Th  +  4 2 He

thorium  is  in  the  atomic  number   90  thus  it  during   alpha decay  it reduces  to  atomic  number  88  while  its  230   mass  number  reduces  to   226
6 0
2 years ago
I really need help with this question please help me.
oksano4ka [1.4K]
I would say A because it’s the only one that doesn’t conduct electricity. Ionic compounds conduct electricity, and covalent compounds do not.
3 0
2 years ago
C2h4 3 o2 2 co2 2 h2o
Marat540 [252]
Use the question marck Moles of CO2
The the giving = 0.624 mol O2
Find the CF faction = 1 mole=  32.00 of O2

O= 2x16.00= 32.00amu ( writte this in the cf fraction)
SET UP THE CHART
Always start with the giving

0.624 mol O2    /  1mol of CO2
___________  / _____________ = Cancel the queal ( O2)
                       / 32.00c O2
                      /
                     /
Multiply the top and divide by the bottom 
0.624 mol CO x 1mol CO2 = 0.624 divide by 32.00 O2 =0.0195
You should look at the giving number ( how many num u gor ever there) 
Ur answer should have the same # as ur givin so 
= 0.0195 
= .0195 mol of CO2
3 0
3 years ago
Where can I get a oxygen isotope analysis comercialy
miskamm [114]

Answer:

Honestly I don't know.

Explanation:

Haha good luck though

6 0
3 years ago
The solubility of acetanilide in hot water (5.5 g/100 ml at 100∘C) is not very great, and its solubility in cold water (0.53 g/
Olin [163]

Answer:

89.4%

Explanation:

Initially, there is 5.0 of the acetanilide in 100 mL of water, then the solution is chilled at 0ºC. The solubility represents the amount that the solvent (water) can dissolve of the solute (acetanilide). So, at 0ºC, 100 mL of water can dissolve till 0.53 g of the compound, the rest will precipitate and will be recovered.

So, the mass that is recovered is 5.0 - 0.53 = 4.47 g

The percent recovery is:

(4.47/5)x100% = 89.4%

8 0
3 years ago
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