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Novay_Z [31]
3 years ago
13

An electron is in a region outside the nucleus

Chemistry
1 answer:
harkovskaia [24]3 years ago
3 0
Is that a question? Doesnt seem like it
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What is the ph of 0.1 m naoh solution??
Rudik [331]
NaOH is a strong base so pH will be around 13 to 12. Whatever number of moles of NaOH you have approximately the pH of NaOH will be around 14 13 or 12  
7 0
3 years ago
What was the weight percent of water in the hydrate before heating?
DedPeter [7]
Data:

weight of water before heating = 0.349

weight of hydrate before heateing = 2.107

Formula:

Weight percent of water = [ (weight of water) / (weight of the hydrate) ] * 100

Solution:

Weight percent of water = [ 0.349 / 2.107] * 100 ≈ 16.6 %

Answer: 16.6%
3 0
3 years ago
Sabendo que A Densidade do Metal De Chumbo e 11,3/cm3
a_sh-v [17]
Sabendo que A Densidade do Metal De Chumbo e 11,3/cm3 means Knowing that the density of lead Metal and 11,3/cm3.
3 0
3 years ago
Why do you require an acid catalyst to make an ester? Why not just mix acid and alcohol? Describe an alternate method of making
djverab [1.8K]

Answer:Acid catalyst is needed to increase the electrophilicity of Carbonyl group of Carboxylic acid as alcohol is a weak nucleophile.

Alternatively esters can be synthesised by converting carboxylic acid into acyl chloride using thionyl chloride(SOCl_{2} and then further treating acyl chloride with alcohol.

Carboxylic acid and esters can be easily distinguished on the basis of IR as carboxylic acid would contain a broad intense peak in 2500-3200cm_{-1} corresponding to OH stretching frequency whereas esters would not contain any such broad intense peak.

Alcohol and esters can also be distinguished using IR as alcohols would contain a broad intense peak at around 3200-3600cm_{-1}

Explanation: For the synthesis of esters using alcohol and carboxylic acid we need to add a little amount of acid in the reaction . The acid used here increases the electrophilicity of carbonyl carbon and hence makes it easier for a weaker nucleophile like alcohol to attack the carbonyl carbon of acid.

The oxygen of the carbonyl group is protonated using the acidic proton which  leads to the generation of positive charge on the oxygen. The positive charge generated is delocalised over the whole acid molecule and hence the electrophilicity of carbonyl group is increased. Kindly refer attachment for the structures.

If we simply mix the acid and alcohol then no appreciable reaction would take place between them and ester formation would not take place because the carboxylic acid in that case is not a good electrophile whereas alcohol is also not a very strong nucleophile which can attack the carbonyl group.

Alternatively we can use thionyl chloride or any other reagent which can convert the carboxylic acid into acyl chloride. Acyl chloride is very elctrophilic and alcohol can very easily attack the acyl chloride and esters could be synthesized.

The carboxylic acid and ester can very easily be distinguished on the basis of broad intense OH stretching frequency peak at around 2500-3200cm_{-1} . The broad intense OH stretching frequency peak is present in carboxylic acids as they contain OH groups and absent in case of esters .

Likewise esters and alcohols can also be distinguished on the basis IR spectra as alcohols will have broad intense spectra  at around 3200-3600cm_{-1}corresponding to OH stretching frequency whereas esters will not have any such peak. Rather esters would be having a Carbonyl stretching frequency at around 1720-1760

4 0
3 years ago
Heat is added to a 200.-gram sample of H2O(s) to melt the sample at 0°C. Then the resulting H2O (image) is heated to a final tem
pochemuha
You are given 200 grams of H2O(s) at an initial temperature of 0°C. you are also given the final temperature of water after heating at 65°C. You are required to get the total amount of heat to melt the sample. The specific heat capacity, cp, of water is 4.186 J/g-°C. Let us say that T1 = 0°C and T2 = 65°C. The equation for heat, Q, is  

Q = m(cp)(T2-T1)
Q = 200g(4.186 J/g-°C )(65°C - 0°C)
<u>Q = 54,418J</u>
3 0
3 years ago
Read 2 more answers
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