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Nairaland Forum / Nairaland / General / Education / University Of Ibadan 2015/16 Applicants (1098965 Views)
University Of Ibadan 2015/16 Applicants Call 07037300634 / Federal University Of Petroleum Resources {fupre} 2015/16 Applicants / University Of Ibadan 2014/15 Admission Process.... (2) (3) (4)
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Re: University Of Ibadan 2015/16 Applicants by Holamarncy: 1:33am On Feb 10, 2015 |
Pr0ton:e suppose decrease i guess |
Re: University Of Ibadan 2015/16 Applicants by Holamarncy: 1:36am On Feb 10, 2015 |
[i][/i]jamb z still very far... ![]() |
Re: University Of Ibadan 2015/16 Applicants by FunkySilver(f): 3:36am On Feb 10, 2015 |
Holamarncy: still very far? |
Re: University Of Ibadan 2015/16 Applicants by ogunsinamayowa(m): 4:42am On Feb 10, 2015 |
estijaz: hahaha! bad market really. but no one wishes such situation. |
Re: University Of Ibadan 2015/16 Applicants by ogunsinamayowa(m): 5:02am On Feb 10, 2015 |
cassyrooy: i think all your comments are jokes,right? and for allegation of telling you the thread to be: did i told u that? i beg recheck my initial post.... i'm also a social science aspirant..... let continue to exhibit our initial civility. |
Re: University Of Ibadan 2015/16 Applicants by DrHost: 6:10am On Feb 10, 2015 |
Holamarncy:yes it decreases there is supply of excess electron..... The larger d r d less acidic...... Ur calc use 2 mol of na= 2 mole of naoh 0.1 mol =x o.1 is get by dividing 2.3 with 23 cross multiply that gives 0.1 multiply that naoh mm and u have ur ans 4...... U can also solv with mm directly.... Organic get calcs ooo |
Re: University Of Ibadan 2015/16 Applicants by DrHost: 6:18am On Feb 10, 2015 |
FunkySilver:24hrs is even enough to pass depending on some factors |
Re: University Of Ibadan 2015/16 Applicants by Nobody: 6:54am On Feb 10, 2015 |
DrHost: You're starting to sound cocky. What kind of factors will make you prepare and pass an exam that is only 24 hours away ![]() You all should read for jamb and forget the post ume for now. 1 Like |
Re: University Of Ibadan 2015/16 Applicants by MockinJay(m): 8:31am On Feb 10, 2015 |
Ogas and Madams...I greet una morning o! |
Re: University Of Ibadan 2015/16 Applicants by Pr0ton: 8:36am On Feb 10, 2015 |
Holamarncy: Correct! Good morning bro |
Re: University Of Ibadan 2015/16 Applicants by MockinJay(m): 8:40am On Feb 10, 2015 |
Pr0ton: 4g. The only elements/compounds to include is Na and NaOH. Since they both have the same mole(2). Cancel out so it'd be 1. Na : NaOH 1 : 1 23 : 40 2.3 : xg X= 40 * 2.3/ 23 X= 4g NaOH= 4g when Na is 2.3g |
Re: University Of Ibadan 2015/16 Applicants by Pr0ton: 8:42am On Feb 10, 2015 |
DrHost: Where did you get 0.1 mol from ![]() |
Re: University Of Ibadan 2015/16 Applicants by Holamarncy: 8:42am On Feb 10, 2015 |
XavierG::-D u c doctor for person name u know know say na boss. |
Re: University Of Ibadan 2015/16 Applicants by Holamarncy: 8:44am On Feb 10, 2015 |
DrHost:its nt organic calculation its jxt common stoichiometry |
Re: University Of Ibadan 2015/16 Applicants by Pr0ton: 8:49am On Feb 10, 2015 |
MockinJay: Please if I may ask: 1. Why not use 2:2 mole ratio instead of 1:1? 2. I noticed you didn't use the mole ratio in the calculation, then why should we be concerned with the mole ratio in the first place? 3. Could you tell me the exact formular you used for the calculation? |
Re: University Of Ibadan 2015/16 Applicants by Pr0ton: 8:54am On Feb 10, 2015 |
Kumpharm.. What's up nw? |
Re: University Of Ibadan 2015/16 Applicants by Omoyele96: 8:58am On Feb 10, 2015 |
Holamarncy:. Please if I may ask,which course are you going for.... |
Re: University Of Ibadan 2015/16 Applicants by Holamarncy: 9:02am On Feb 10, 2015 |
Omoyele96:mbbs ..u? |
Re: University Of Ibadan 2015/16 Applicants by Holamarncy: 9:04am On Feb 10, 2015 |
Pr0ton:dat z method z tym consuming since dey av d same mole 23 = 40 2.3 = X of course its jxt 4 |
Re: University Of Ibadan 2015/16 Applicants by MockinJay(m): 9:04am On Feb 10, 2015 |
Pr0ton: 1. For two reasons. First, they are both carrying the same mole. So one could easily cancel them out. But why? Cause it's easier and faster.(reason 2) 2. I made use of it. Since it's 1,it won't show just like HCl is also 1HCl. Now, if I were to use the 2mole instead of the 1mole. The molar masses of Na and NaOH would be 46 and 80g/mol respectively. 3. The "I don't have a calculator but want to be fast" formula ![]() Whether I use the 1:1 mole ratio or 2:2 mole ratio, I would still get the same answer. 2Na : 2NaOH 2* 23 : 2* 40 46 : 80 2.3 : x X= 80 * 2.3/46 X= 4g |
Re: University Of Ibadan 2015/16 Applicants by Pr0ton: 9:08am On Feb 10, 2015 |
Holamarncy: Yours is faster but not comprehensive. |
Re: University Of Ibadan 2015/16 Applicants by MockinJay(m): 9:08am On Feb 10, 2015 |
Just imagine an equation of this nature having a 3mole of Cl and 3mole of say, NaOH. Instead of using a calculator or stressing yourself to multiple 3 by 35.5 for Cl. Why not just cancel both mole ratio out? So it'd be 35.5 of Cl and 40 of NaOH. Faster..ryt? 1 Like |
Re: University Of Ibadan 2015/16 Applicants by MockinJay(m): 9:09am On Feb 10, 2015 |
Holamarncy: That's the same formula I used but I needed to explain so the guy could easily grab. |
Re: University Of Ibadan 2015/16 Applicants by Pr0ton: 9:12am On Feb 10, 2015 |
MockinJay: Got it! Thanks Pr0ton: Could you answer that? Meanwhile, more calculations coming up. @Holamarncy, MockinJay and DrHost. |
Re: University Of Ibadan 2015/16 Applicants by kumpharm(f): 9:20am On Feb 10, 2015 |
Re: University Of Ibadan 2015/16 Applicants by MockinJay(m): 9:27am On Feb 10, 2015 |
.. |
Re: University Of Ibadan 2015/16 Applicants by MockinJay(m): 9:29am On Feb 10, 2015 |
Pr0ton: Compounds present: Phenol and Toluene(but not attached to benzene) My guess (Hydroxymethyl)benzene |
Re: University Of Ibadan 2015/16 Applicants by Pr0ton: 9:35am On Feb 10, 2015 |
kumpharm: Good morning! How about your admission?? |
Re: University Of Ibadan 2015/16 Applicants by Nobody: 9:42am On Feb 10, 2015 |
Holamarncy:Lol |
Re: University Of Ibadan 2015/16 Applicants by kumpharm(f): 9:46am On Feb 10, 2015 |
Pr0ton: Been admitted already.. |
Re: University Of Ibadan 2015/16 Applicants by Pr0ton: 9:51am On Feb 10, 2015 |
MockinJay: Compounds present are actually Phenyl (C6H 5) and Menthanol (CH3OH) It's Phenylmenthanol. Are you an aspirant? |
Re: University Of Ibadan 2015/16 Applicants by Pr0ton: 9:52am On Feb 10, 2015 |
kumpharm: Oh..(Never knew) What course? |
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