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Chemistry: Post your doubts here!

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sulfuric acid is not present in the first stage :unsure: Im soooooo confused :cry: so NaBr is supposed to react with water
yes thats what i thoght at first but see NaBr reaction with Br is not possible
Na compounds dont undergo hydrolysis
 
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well add Hcl to both side and you will find that they will have same product then subtract them keeping in mind the molar ratio
 
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well add Hcl to both side and you will find that they will have same product then subtract them keeping in mind the molar ratio
I don't get this part :|
I've written the previous equations, and everything.
But I don't get why should I add Hcl to both sides? Can you explain a bit further, maybe write an equation?
Tnx.
 
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well here goes nothing
HCL + 2 KHCO3 = K2CO3+H2O + CO2 +HCl

2KCL +H2O +CO2 here
I just don't get it! Can you explain with a diagram or something?
The addition of HCl to both sides doesn't make sense to me.
 
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Yea, these (i.e. Concentrated H2SO4) are good dehydrating agents.

Thank you, also one quick question. For testing for alkenes, we use bromine water.
Now what I'm not sure is how it tests for them. Does the alkene decolourise or the bromine water? And does it mean it's present or not present? Could you please briefly explain how to test for them with bromine water ?
 
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Thank you, also one quick question. For testing for alkenes, we use bromine water.
Now what I'm not sure is how it tests for them. Does the alkene decolourise or the bromine water? And does it mean it's present or not present? Could you please briefly explain how to test for them with bromine water ?
Bromine water is a test for presence of C=C double bonds in a compound.
What happens is that the bromine water (Br2 molecule - brown color), gets decolourized (it means the solution becomes colorless).
This is because of electrophilic addition reaction taking place, forming a colourless halogenoalkane.
 
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Bromine water is a test for presence of C=C double bonds in a compound.
What happens is that the bromine water (Br2 molecule - brown color), gets decolourized (it means the solution becomes colorless).
This is because of electrophilic addition reaction taking place, forming a colourless halogenoalkane.

Ah all right, gotcha' .
 
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Thank you, also one quick question. For testing for alkenes, we use bromine water.
Now what I'm not sure is how it tests for them. Does the alkene decolourise or the bromine water? And does it mean it's present or not present? Could you please briefly explain how to test for them with bromine water ?
look the colour actually was of bromine so when it reacts with ethene there are no bromine molecules left hence the colour disappears
 
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when you make titration can we start with 50cm3 as an initial reading for each rough titration
in the MS it is written that do not give marks if start with 50cm3
plaese answer i am confused
 

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when you make titration can we start with 50cm3 as an initial reading for each rough titration
in the MS it is written that do not give marks if start with 50cm3
plaese answer i am confused
you cannot start with 50 cm3 as thats the maximum reading a burete can have
 
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sulfuric acid is not present in the first stage :unsure: Im soooooo confused :cry: so NaBr is supposed to react with water

Read this part

The overall reaction may be considered to take place in two stages. In the first stage the
inorganic reagents react together to form HBr. In the second stage, the organic reagent
reacts with the HBr that is formed in the first stage.

meaning that both the stages are carried out .. but it is carried out in 2 reactions ..

Non-Organic reagents are .. NaBr H2O and H2SO4

NaBr can't react with water.. it will just dissolve in it.

Stage I NaBr undergoes Displacement reaction with H2SO4 where SO4^2- displaces Br^-

2NaBr + H2S04 --> Na2SO4 + 2HBr --> Reaction

Then in Stage II HBr reacts with the alcohol to give Bromo-Butane + Water.
 
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no offence meant bro but its very basic
5bi) since carboxylic acid is removed in the first option Na can only react with alcohols hence the group is OH group
5bii use the equation 2Na + 2C3H5OH = 2C3H5O-Na + H2 so calculate moles of alcohol used and find your answer
5biii) it will show that 1 mole of the compond produces one mole of H2 hence it must contain two OH groups
hope it solves the problem :)
 
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when you make titration can we start with 50cm3 as an initial reading for each rough titration
in the MS it is written that do not give marks if start with 50cm3
plaese answer i am confused
What's ur problem!?!!
 
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