Originally posted by UltimaOnline:
Bronsted-Lowry acidification or protonation.
Thanks !!!
What does proticity mean? Can't find the meaning online
Is it mean how many H it can donate?
U.O. told me that formula of salt formed is based on proticity, based on the discussion of george chong's inorganic book having a error at page 40
[IGNORE this epic fail post. Found my mistake, written below]
Omg wtf, ultima hope u can help on this
Page 40 george chong's inorganic book
Oxides of phosphorus with alkalis :
(After making corrections) product when p4o6 react with NaOH is Na2HPO3 . I calculated o.s. of P is +3 (can help check?)
Product when p4o10 react with NaOH is Na3PO4. I calculated o.s. of P is +5 (can help check? )
!? Wait. Update : for george chong's inorganic book, i have serious anomalous results compared to his writing
On both pages 40 and 41, specifically Oxides of P and oxides of S parts , he wrote Four times, that when oxides dissolve in water/neutralised by alkalis , oxidn number/state do not change ! Neutralisation, dissociation are not redox reactions
But I've calculated o.s. if all the P and S compounds in page 40 and 41, the o.s. aren't the same(e.g. the top portion of this post) !!!
Please please help !!
Shit la. Actually i think i made a serious blunder
He meant o.s. of p in p4o10 and os in na3po4 is the same, +5. SOOORRRRYRYYY !!
Originally posted by Flying grenade:What does proticity mean? Can't find the meaning online
Is it mean how many H it can donate?
U.O. told me that formula of salt formed is based on proticity, based on the discussion of george chong's inorganic book having a error at page 40
Since H3PO3 (Latin name phosphorous acid, Stock name phosphoric(III) acid, IUPAC name phosphonic acid) is a diprotic acid, hence upon complete deprotonation using excess NaOH(aq) in a Bronsted-Lowry acid-base reaction, the salt generated is Na2HPO3 (Latin name disodium phosphite, Stock name disodium phosphate(III), IUPAC name disodium phosphonate).
Thanks alot !!
Is it possible to synthesise methanal ? As it will further oxidise to co2 and h2o
Use methanol, k2cr2o7, heat with immediate distillation, can get methanal?
Hi Ultima, hope you can help me, please !
https://www.dropbox.com/s/5kz1e3dc81gswep/20160527_125045-1.jpg?dl=0
Originally posted by Flying grenade:Is it possible to synthesise methanal ? As it will further oxidise to co2 and h2o
Use methanol, k2cr2o7, heat with immediate distillation, can get methanal?
https://en.wikipedia.org/wiki/Formaldehyde#Industry
Using methods within H2 syllabus will have very low yield. Cambridge will understand this and either not ask, or will accept the within-syllabus method, etc.
Originally posted by Flying grenade:Hi Ultima, hope you can help me, please !
https://www.dropbox.com/s/5kz1e3dc81gswep/20160527_125045-1.jpg?dl=0
Originally posted by UltimaOnline:
This problem you need to ask your school teacher or private tutor face-to-face, because it's not a simple balancing of equation using molecular formulae, as it involves molecular structures and mechanisms, cannot efficiently settle on forum.
Ultima, can help me check my answers and teacher's answer at the left column? For the 2nd product, i thought its methanoic acid, but my cher say it's co2, idk why
Originally posted by Flying grenade:Ultima, can help me check my answers and teacher's answer at the left column? For the 2nd product, i thought its methanoic acid, but my cher say it's co2, idk why
THANKSSSSSSS ULTIMAA!!!
Hi ultima, can help page 10 chan kim seng planning book?
Don't understand double titration method to find the actual amount of diprotic K2CO3
Originally posted by Flying grenade:Hi ultima, can help page 10 chan kim seng planning book?
Don't understand double titration method to find the actual amount of diprotic K2CO3
Can sodium beryllium hydride exist? NaBeH
Originally posted by Flying grenade:Can sodium beryllium hydride exist? NaBeH
Yeah it was meant to be a joke
Originally posted by Flying grenade:Yeah it was meant to be a joke
Originally posted by UltimaOnline:
Nah beh?!? Eh your formula wrong lah!
EHHH What's the correct formula ?????
Now after attempting to check oxidation state of elements
E.g. LiAlH4
O.S. : Li=+1 , Al=+3 , H- = -1 , hence overall ionic charge of molecule =0
O.S. : Na=+1 , Be= +2 , H- = -1
Hence NaBeH doesn't exist?
Maybe NaBeH3 exists?
Or Na2BeH4
Originally posted by Flying grenade:Or Na2BeH4
Originally posted by UltimaOnline:
Correct. Because Be wants to have a stable octet.
I dun get it ! I tot Be has to lose 2 e- to gain a stable duplet structure?
If stable octet it has to gain 6 e- ! How is it possible here?
Then does NaBeH3 exist?
Thanksss ultima!!!
Originally posted by Flying grenade:
I dun get it ! I tot Be has to lose 2 e- to gain a stable duplet structure?
If stable octet it has to gain 6 e- ! How is it possible here?
Then does NaBeH3 exist?
Thanksss ultima!!!
Can exist, but won't be stable, coz Be will lack a stable octet and still kena a uninegative formal charge. Either BeH2 (lack stable octet, but at least no formal charge), or Na2BeH4 (has stable octet, but kena dinegative formal charge) are both more stable than your NaBeH3, which kena the worst of both worlds (in terms of stabilities).
But of course, you still won't fully understand, you still have many questions, but that's enough for the forum, go ask your school teacher or private tutor.