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Sindrei [870]
4 years ago
10

Is the formation of NaCL and other compound a physical or chemical change?

Chemistry
1 answer:
Mnenie [13.5K]4 years ago
5 0
NaCl is a molecule formed from 1 Sodium (Na) atom and 1 Chlorine (Cl) atom.  Because the formation of NaCl deals with the chemical structure of the compound, it is a chemical change.
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Which of these equations are balanced
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3 years ago
What is the difference between the crest and the trough of a wave?
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Trough is the lower bound of wave, while the crest is the upper bound of a wave.

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3 0
3 years ago
When ethyl acetoacetate (CH3COCH2CO2CH2CH3) is treated with one equivalent of CH3MgBr, a gas is evolved from the reaction mixtur
grigory [225]

Answer:Methane gas is evolved from the reaction mixture.

Explanation:

When ethyl acetoacetate is treated with grignard reagent a carbanion is generated.

There are two acidic hydrogens which are present on the carbon which is in between the ester and the ketone group in ethylacetoacetate.

These two protons are also called active methylene protons and they are very acidic in nature due to the presence of two electron withdrawing substituents that is an ester and ketone.

CH₃MgBr is grignard reagent and it is an organo-metallic copmpound  . Carbon here in CH₃MgBr exists as carbanion  CH3⁻ which is basic enough to abstract the acidic protons present on ethylacetoacetate.

As CH3⁻ abstracts an acidic  proton from ethylacetoacetate it become CH₄ which is methane. As methane is a gas so it is methane gas which is evolved from the reaction mixture.

As  the acidic proton is abstracted from ethylacetoacetate which leads to generation of carbanion and this carbanion is very stable as it can be delocalized on to the two carbonyl groups . As we add aqueous acid to the reaction mixture the carbanion can again be protonated and its protonation would lead to the generation of ethylacetoacetate again.

8 0
4 years ago
Calculate the frequency of visible light having a wavelength of 486 nm
hjlf
Ν= c/λ

λ= 486 nm * ( 1 m / 1x10^9 nm) = 4.86 x10^-7 m 

v= (3.00 x 10^8 m/s) /  (4.86 x10^-7 m) 
v= 6.1728395x 10^14 s^-1
= 6.14 x10^-14 Hz
8 0
3 years ago
How many moles of caco3 are there in an antacid tablet containing 0.515 g caco3?
MaRussiya [10]
The  moles  of  CaCO3  which  are  there  in  antacid  tablet  that  contain  0.515g  CaCO3  is  calculated  as  follows

moles  =mass/molar  mass

the  molar  mass  of  CaCO3 =  ( 40  x1)+  (12  x  1)  + (16 x  3)=  100  g/mol

moles  is  therefore=  0.515g/100 g/mol=  5.15  x10^-3  moles


8 0
3 years ago
Read 2 more answers
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