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nlexa [21]
2 years ago
6

Which electron configuration is impossible?

Chemistry
1 answer:
Verizon [17]2 years ago
7 0

Answer:

MP hippopotamus

Explanation: I'm in the fifth grade so yeah I really don't know this so I'm just going to say some random stuff and white cheddar mac and cheese warm TV perfect back MD 11443 to 2885 eleven 12:20 to 11 132 0.24 answer

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What reaction conditions most effectively conver a cabocxylic acid to a methly ester?
VLD [36.1K]

Answer:

Esterification reaction

Explanation:

When we have to go from an acid to an ester we can use the <u>esterification reaction</u>. On this reaction, an alcohol reacts with a carboxylic acid on acid medium to produce an ester and water. (See figure).  

In this case, we need the <u>methyl ester</u>, therefore we have to choose the <u>appropriate alcohol</u>, so we have to use the <u>methanol</u> as reactive if we have to produce the methyl ester.

7 0
3 years ago
I NEED HELP ASAP!! What is the difference between the experimental group and a control group?​
dexar [7]

I believe the control group is what doesn't change in the experiment, and the experimental group is what is being tested / receives the treatment :)

3 0
3 years ago
One peanut M&amp;M weighs approximately 2.33 g.
Sphinxa [80]

Answer:

There are 23076 peanut M&M's in 53.768 kg of M&M's.

Explanation:

First we <u>convert 53.768 kg into g</u>:

  • 53.768 kg * 1000 = 53768 g

Then we <u>divide the total mass of M&M's by the mass of one peanut M&M,</u> in order to calculate the answer:

  • 53768 g / 2.33 g = 23076

So there are 23076 peanut M&M's in 53.768 kg of M&M's.

3 0
3 years ago
An organic molecule is likely to contain all of these elements except: Answer C H O Ne N
eimsori [14]
Organic molecules typically do not contain the noble gases, so they would contain all but Ne
8 0
3 years ago
stbank, Question 075 Get help answering Molecular Drawing questions. Compound A, C6H12 reacts with HBr/ROOR to give compound B,
Law Incorporation [45]

Answer:

Explanation:

In this case we want to know the structures of A (C6H12), B (C6H13Br) and C (C6H14).

A and C reacts with two differents reagents and conditions, however both of them gives the same product.

Let's analyze each reaction.

First, C6H12 has the general formula of an alkene or cycloalkane. However, when we look at the reagents, which are HBr in ROOR, and the final product, we can see that this is an adition reaction where the H and Br were added to a molecule, therefore we can conclude that the initial reactant is an alkene. Now, what happens next? A is reacting with HBr. In general terms when we have an adition of a molecule to a reactant like HBr (Adding electrophyle and nucleophyle) this kind of reactions follows the markonikov's rule that states that the hydrogen will go to the carbon with more hydrogens, and the nucleophyle will go to the carbon with less hydrogen (Atom that can be stabilized with charge). But in this case, we have something else and is the use of the ROOR, this is a peroxide so, instead of follow the markonikov rule, it will do the opposite, the hydrogen to the more substituted carbon and the bromine to the carbon with more hydrogens. This is called the antimarkonikov rule. Picture attached show the possible structure for A. The alkene would have to be the 1-hexene.

Now in the second case we have C, reacting with bromine in light to give also B. C has the formula C6H14 which is the formula for an alkane and once again we are having an adition reaction. In this case, conditions are given to do an adition reaction in an alkane. bromine in presence of light promoves the adition of the bromine to the molecule of alkane. In this case it can go to the carbon with more hydrogen or less hydrogens, but it will prefer the carbon with more hydrogens. In this case would be the terminal hydrogens of the molecules. In this case, it will form product B again. the alkane here would be the hexane. See picture for structures.

8 0
3 years ago
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