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Chemistry: Post your doubts here!

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View attachment 59904

Can someone please give me a description on how to do this ?
Take a measuring cylinder. Place it on a balance. Zero the balance. Add distilled water to the measuring cylinder until 30g of water has been poured. Now zero the balance again. Add KCl solid, carefully, so that 2.238g are added. This means you added 0.03mol of KCl.
The molality of this solution (stir it) will be 1mol/kg. Lets call this your Ultra Solution (US)
Now we will do serial dilution.
Take five more measuring cylinders. Add 0g of distilled water to first, 2g to second, 4g to third, 6g to fourth, and 8g to fifth.
Now take your Ultra Solution and add 10g of it to first, 8g to second, 6g to third, 4g to fourth, and 2g to fifth.
Stir them all.
Note that we've used 50g of distilled water so far.
Your first cylinder has molality 1.0mol/kg
Second one is 0.8mol/kg
Fourth is 0.4mol/kg
Last one is 0.2mol/kg

Use the remaining water to clean your apparatus ;)
Nah just kidding you could use it to repeat your experiment or just make larger volumes of the above solutions.
 
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interpret and explain qualitatively the variation in solubility of the
hydroxides and sulfates in terms of relative magnitudes of the
enthalpy change of hydration and the corresponding lattice energy

any notes or explantion? Please HELP :cry:
 
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Option A shows a simple dehydration or elimination reaction. No change in oxidation takes places.

Option C is the hydrolysis of esters. Again no change in oxidation states.

Option D: The reaction does not take place, as ketones do not react with fehling's solution.

Only Option B shows a redox reaction as in this reaction the aldehyde is oxidised to form carboylic acid, and the Ag+1 ions present in the tollen's reagent are reduced to form Ag(s). So as both oxidation and reduction take place, this is the only redox reaction.
agreed with ur statement ;)
 
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Take a measuring cylinder. Place it on a balance. Zero the balance. Add distilled water to be measuring cylinder until 20g of water has been poured. Now zero the balance again. Add KCl solid, carefully, so that 1.492g are added. This means you added 0.02mol of KCl.
The molality of this solution (stir it) will be 1mol/kg. Lets call this your Ultra Solution (US)
Now we will do serial dilution.
Take five more measuring cylinders. Add 0g of distilled water to first, 2g to second, 4g to third, 6g to fourth, and 8g to fifth.
Now take your Ultra Solution and add 10g of it to first, 8g to second, 6g to third, 4g to fourth, and 2g to fifth.
Stir them all.
Note that we've used 40g of distilled water so far.
Your first cylinder has molality 1.0mol/kg
Second one is 0.8mol/kg
Fourth is 0.4mol/kg
Last one is 0.2mol/kg

Use the remaining water to clean your apparatus ;)
Nah just kidding you could use it to repeat your experiment or just make larger volumes of the above solutions.
You added a total of 10+8+6+4+2 which is 30 g but you only made 20 g ?
 
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So is there a way to tell whether carbon atoms in a ring (or any molecule) lie in a plane?View attachment 59908
To see if atoms lie in a plane, we have to think of the type of bonding and the shape it would have.
If the atoms form a part of trigonal planar structure, then of course it's planar. Same goes for C atoms in a benzene ring. Also ethene (because of C=C bond) is planar I think.

In your particular question, most of the carbon atoms involved have undergone sp3 hybridisation and thus are involved in tetrahedral arrangements. There is no way they could all lie on one single plane.
 
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is iodoform test included in organic chemistry 2016 course?
It is there for the AS part.
So if you're doing AS or AL then you need to know it.


http://www.cie.org.uk/images/164502-2016-2018-syllabus.pdf

A Level material moved to AS material
10.4 (c) deduce the presence of a CH3CH(OH)
– group in an alcohol from its reaction with alkaline aqueous iodine to form tri-iodomethane 10.5 (e) describe the reaction of CH3CO– compounds with alkaline aqueous iodine to give tri-iodomethane
 
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It is there for the AS part.
So if you're doing AS or AL then you need to know it.


http://www.cie.org.uk/images/164502-2016-2018-syllabus.pdf

A Level material moved to AS material
10.4 (c) deduce the presence of a CH3CH(OH)
– group in an alcohol from its reaction with alkaline aqueous iodine to form tri-iodomethane 10.5 (e) describe the reaction of CH3CO– compounds with alkaline aqueous iodine to give tri-iodomethane
oh thank u so much :)
 
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Is Manganese(II) sulfate same as manganese sulfate? of course manganese II sulfate has 2+ ions but I have a salt labelled just 'manganese sulfate', it's pale pink, and I need manganese(II) sulfate, could it be the same as that?
 
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Is Manganese(II) sulfate same as manganese sulfate? of course manganese II sulfate has 2+ ions but I have a salt labelled just 'manganese sulfate', it's pale pink, and I need manganese(II) sulfate, could it be the same as that?
Mn +2 salts should be white. Pink salts of manganese contain Mn+7, i guess.
 
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Mn +2 salts should be white. Pink salts of manganese contain Mn+7, i guess.
no, it's very pale pink, not really pink. I just googled and see the one in the image:
https://www.google.com.pk/#q=manganese+sulfate
mine is exactly the same color
but it's labelled magnesium sulfate only and with the formula MnSo2 which is probably wrong MnSO2 is no salt xD, these labels aren't very authentic
do you think I can use it where I need manganese II sulfate?
 
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no, it's very pale pink, not really pink. I just googled and see the one in the image:
https://www.google.com.pk/#q=manganese+sulfate
mine is exactly the same color
but it's labelled magnesium sulfate only and with the formula MnSo2 which is probably wrong MnSO2 is no salt xD, these labels aren't very authentic
do you think I can use it where I need manganese II sulfate?
Lol, give it a try. If the color matches with that, it may be MnSO4. And thanks for correcting me that MnSO4 is pale pink in colour. :)
 
Messages
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It is there for the AS part.
So if you're doing AS or AL then you need to know it.


http://www.cie.org.uk/images/164502-2016-2018-syllabus.pdf

A Level material moved to AS material
10.4 (c) deduce the presence of a CH3CH(OH)
– group in an alcohol from its reaction with alkaline aqueous iodine to form tri-iodomethane 10.5 (e) describe the reaction of CH3CO– compounds with alkaline aqueous iodine to give tri-iodomethane
Hi,

Have you ever come across any question relating to it in P4? If you see any question, please do tell me its year.
 
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