₦airaland Forum

DrBESTJC:
CHEMISTRY.

1 : Calculate the pH of 0.2 mol dm^-3 of Ca(OH)2, assuming that Ca(OH)2 is a strong electrolyte.
(A) 10.6
(B) 12.5
(C) 5.1
(D) 8.0

2 : The concentration of H3O+(oxonium) ions in 1.0 mol dm^-3 solution of C6H5COOH(benzoic acid) is 8.0 * 10^-3 mole per dm^3.Calculate its percentage ionization......
(A) 0.545%
(B) 0.888%
(C) 0.213%
(D) 0.725%

...both question are wrong...
No 1.. Its simple BT wrong.....the most important aspect its just to mutiply the hydroxide concentration by 2 and then find POH.......PH = 14 - POH....

No 2....its also simple....percentage of ionisation
=( hydroxonium concentration/original acid present) × 100%...=( 8 × 10^-3/ 1) × 100...
= 0.8 %....there r some complicated question where hydroxonium ion concentration is not given but Ka is given....

boyejo124:
I'll go for 0.4M.
just WURUWURU TO THE ANSWER
cc:
Gracy3535

..wrong but u tried....there is a small mistake along ur calculation....
Bro,are u interested in the reading group?

A 600ml of water was added to 200ml of a 1.6M CaCl2 solution...what is the concentration of chloride in the new solution....(A) 0.4 (B) 0.8 (C) 1.2 (D) 5.6
PS....if u r in Ibadan,it will be of good help,if you can go to Team tutor...u will be expose to a lot of question like this....

Gracy3535:
Am interested u ve my number ryt?
Pls who is the author of science & college physics

...author of science physics is afolabi....college physics is by Eugene hetch....

Damayogenius18:
Frnds..I need some pple to create a reading group.....you can drop ur number with me...the ethics is we read from 8pm - 5am every day from (may)7 mon to April 7 (thurs).. Just a month ...
U can email me or drop ur number if you are interested...

...membership ethics....
1..no what's app chatting among members during this period...no member shld also be caught on what's app or any social network during this time
2...for science members,we are solely doing physics...each member must finish physics topics during this period....40 topics....you will be given each topics todo for this 30days...
3...Each member must get lamlad physics, science physics,college physics and any ordinary level physics txtbook( just one)...optional,A level physics and calculation in physics...topics to be done per day will be given by me....we have 40 physics topics to cover...
4...Each member must av jamb pq and ui practice question....each member must solve jamb pq everyday and finish it b4 June7 and also the practice question...the admin will communicate how it will be done
5...Each members must read 12hrs per day...8am -5pm is mandatory( that make 9hrs)u av the liberty to choose when you will complete
The remaining 3hrs....
6...their is continuity ..after the 30days( June 7)...we will start chemistry for 30days( June 8 - july8) ....also biology/math for 30days(July 8 - august8)...but let's do it one at a time starting with physics....
8...the membership is also open to art and commercial student....but there won't be overseer.. u choose d subject and topic u want to read.. Thanks
Signed______damayaogenius....supported by Team tutors...Sir caleb and company....

Frnds..I need some pple to create a reading group.....you can drop ur number with me...the ethics is we read from 8pm - 5am every day from (may)7 mon to june 7 (thurs).. Just a month ...
U can email me or drop ur number if you are interested...

dannyboii1:
where can we get such amount of questions on a particular topic..

aside from the exercise.... which are few

...e-mail me ..we will talk about it...av u heard abt team tutors in ui...preparing ppl for post utme..

DrBESTJC:
I saw in HiLearners blog that the probability of Ui conducting their PostUtme earlier(around July--August) this year is very high(as against last year's which was conducted around October).

So being a slow reader,I got thrilled by the news.I bought Understanding chemistry and I have been reading it for a month now without being able to finish it.Though I understood 80% of all that I have read.
Also I bought Idodo Umeh biology but the textbook is damn big!

I read from morning till night but at the end of the day I feel as though I have not covered much.The next day,I brace up and study even harder......still the same experience(especially with chemistry and physics)....but if I read other books like novels,I could finish it at most within two days and understand everything no matter how voluminous it is.

Bosses here I need your help.Even on the fact that I am going for a very competitive course,I want to inculcate and internalise in me,....THE HABIT OF SWIFT AND EFFECTIVE STUDIES....

Let us take it that the examination is coming up in August...,how do I finish and at the same time absorb everything I need to pass this PostUtme using these textbooks within this short time interval...
BIOLOGY:Explicit,Modern and Idodo Ume.
CHEMISTRY:Understanding,Lamlad,New school
PHYSICS:lamlad,Principles of physics,A-level physics and New school physics.

English is not much of my problems as I can finish it within a week.


Cc..Pes13,Damayogenius18,...and other bosses here...

...the key is DAT,u shld try to understand wateva u r reading...don't muddle things up....u should also solve a lot( like hundreds)of question on each topics u read...

Eben331:
(A) 4:1 The bosses should explain.

...Gud morn bro...y r u not always interested in solving...its time consuming abi...or u dey preserve ur battery..I will check it now..

DrBESTJC:
CHEMISTRY.

Two radioactive elements P and Q initially have the same mass.If their half-lives are 10 and 5 years, the ratio of their masses(ie. P:Q) after 20 years is.....
(A) 4:1
(B) 2:1
(C) 1:2
(D) 1:4


This question is somehow ambiguous to me.Please,I need explanations to the right answer.

..OK bro...the nuclei has d same mass but there half life is different..N/No = 1/2 (t/t1/2)...No is constant for both...
Np/No= 1/2^ 20/10..= {1/2}^ 2..Np/No= 1/4...Nq /No..= {1/2}^20/5...= {1/2}^4....Nq/No =1/16...
No cancel cos its d same for both....so
Np/Nq = {1/4 } ÷{1/16}...open d bracket..
Np/Nq= 1/4 × 16/1...
Np/Nq = 4/1 or 4 : 1....you can use this formula for any radioactive calculation..you just need to understand d question..

Gracy3535:
08094258273 pls edit wen u ve seen it

..u can modify it and delete it

Gracy3535:
Hy am nw on watsapp shuld i send my number

....OK...send it....

Eben331:
since n=(t1/2)/t where t1/2=half-life, t=time to disintegrate. if n=3 and we were askd to find t and not t1/2, then it should be: 3=10/t t=10/3 Correct yourself there. I knw it might be a slip of mind. No offence meant bro.

....yea...n = t/(t1/2)...

Eben331:
C

....bro, how r u...long time..

DrBESTJC:
CHEMISTRY.


1: If 4g of a radioactive material of half-life 10 days is spilled on a laboratory floor, how long would it take to disintegrate 3.5g of the material?
(A) One and 1/4 days.
(B) Eight and 3/4 days.
(C) 30 days.
(D) 80 days.

2: If 1.20 * 10^5 atoms of a radioactive nuclide decays such that only 1.50 * 10^4 remained after 81 days.The half-life of the radioactive nuclide is....
(A) 21 days
(B) 27 days
(C) 54 days
(D) 81 days


Abeg how do I solve questions like these without calculator


Oga Pes13, Damayogenius18,......and other bosses here help out please....

.....for no 1...you can do it did way without calculator....
N/No = 1/2 ^n ...where n= half-life/timtaken to disintegrate...I.e
n = t/(t 1/2)...N= number of nuclei dat has disintegrated...No= Number of nuclei originally present....in this question, No=4g while N = 4 - 3.5 = 0.5...be careful here...a lot of ppl make sis mistake...3.5g is the no that has disintegrated while 0.5g is no remaining....so we have
0.5/4 = 1/2^n...0.5= 1/2..so we can say
(1/2 * 1/4)= 1/2^n
1/8 =1/2^n..now we use index form
2^-3 = (2^-1)n...2 on each side cancel remaining the powers..so
-3 = -n...so...n= 3.....
But remember that n = t/(t 1/2)= 10 days...so
t = t 1/2 * n...= 3 * 10days
= 30 days.....
U can do the second question dis way too....but der are some complicates one u cant do dis way
But I dont think UI can set dat in post utme
It is very simple and easy to do dis way...the reason it is too long is because am explaining.....
Hope u found it useful...

Ok

DrBESTJC:
CHEMISTRY

In a nuclear fusion process,four protons each of mass Mp, were fused to produce a nucleus X of mass Mx. Which of the following equations is CORRECT?......
(A) 4Mp > Mx
(B) 4Mp = Mx
(C) 4Mp < Mx
(D) Mp = Mx

...answer should be A.....4mp> MX...when different masses are added,energy is released...E= mc^2...some of the mass of the added proton is converted to energy...the energy is called binding energy.....normally,we know that all nuclei have lesser mass than the sum of individual nucleus that form them...

Eben331:
Damayogenius18, show us the corrections to your questions, we dont knw them.

..I will bro...solve d relative question....

Eben331:
(1) A man can row a boat at 4km/hr in still water. He rows the boat 2km upstream and 2km back to his starting place in 2 hours. How fast is the stream flowing? (2) A swimmer is capable of swimming at 1.4m/s in still water. (a) How far downstream will he land if he swims directly across a 180m wide river? (b) How long will it take him to reach the other side?

...no 2..a is not complete...b is t = 180/1.4..=129s..so it will b easier to calculate "a" if the rate of flow of the river is given....dx= speed of river/ 129s....dx= how far downstream....I could av also calculated the speed of river using pythagoras if you av given me the angle the boat made with the shore...I stand corrected o...I may be wrong..

Eben331:
(1) A man can row a boat at 4km/hr in still water. He rows the boat 2km upstream and 2km back to his starting place in 2 hours. How fast is the stream flowing? (2) A swimmer is capable of swimming at 1.4m/s in still water. (a) How far downstream will he land if he swims directly across a 180m wide river? (b) How long will it take him to reach the other side?

..no 1..Velocity of boat related to shore= 2km/ 2hrs = 1km/hr...Velocity of boat related water=4km/hr...velocity of water related to shore= 4^2 - 1^2...
v^2 = 16 - 1. ..v = √15...=3.87km/hr

Eben331:
(1) A man can row a boat at 4km/hr in still water. He rows the boat 2km upstream and 2km back to his starting place in 2 hours. How fast is the stream flowing? (2) A swimmer is capable of swimming at 1.4m/s in still water. (a) How far downstream will he land if he swims directly across a 180m wide river? (b) How long will it take him to reach the other side?

..no 2...the question is not complete or perhaps I don't understand it....I think u should give the speed of the river....even If u don't give it,u should give the angle the boat made wit the shore..

Gracy3535:
tanx cos i really need ur help. Ve gotten new skul phy. Pls take a pix of science phy

...send ur number to me trough mail..I will send it to u via WhatsApp...no need

vicstiqs95:
A simple pendulum oscillate with an amplitude of 30° if the length of the string is 1.0m. calculate the velocity of the pendulum Bob at its lowest point.

...conservation of energy....
Mgh = 1/2mv^2....m cancel...so gh=1/2v^2....
v=√2gh...but h= l - l cos teta....h= l (1-cos teta)
h=1(1 - 0.866)..h=0.134...
v=√2 * 10 *0.134...
v=1.637m/s

Gracy3535:
when u r through wit them come and collect ur cane

..lol..I miss u guys too.. But from Monday..I will always come online at 2am....

Eben331:
1 A ball of mass 1kg is dropped from a height of 5m onto a floor and rebounds to a height of 1m. Find the loss of kinetic energy due to impact. (g=9.8m/s2) 2 An electron of mass 9.4*10^-31kg has a wavelength of 1.3*10^-10m. Calculate the velocity of the electron. (h=6.6*10^-34Js) 3 A compound microscope, with lenses of focal lengths of 3cm and 5cm for the objective and eyepiece respectively, has an object placed 2cm from the objective lens. If the first image formed 3cm from the eyepiece lens, calculate the distance of the final image from the eyepiece. Happy Weekend Bosses.

....no 3....fe= 5cm..Ue = 3cm...Ve=?...so lens formula...
1/f=1/u + 1/v....1/5=1/3 + 1/v...1/v= 1/5 - 1/3 ...Ve = -15/2 = -7.5m....
D way I think it is DAT the image of the objective lens form the object distance of the eyepiece ...

dolapo74:
How can I calculate pH without using calculator

..u just need smartness..pH of 0.001= 10^-3..since.pH= - log[H^+]..pH =3....but there are some compicacated one...for dat u av to b smart too...like pH of 3.4 x 10-4....u can just treat it as 1.0 x 10^-4....u get 4.....u will now say....4 - log 3.4...I don't knw log 3.4..but I knw is "0.something".. I knw my answer will be BTW 4 and 3...like 3.1 - 3.9...actually the answer is 3.5......so its just smartness

Gracy3535:
Guys pls ve been trying to understand hybrization of orbital but it not going any simple defination of it.

..it is the mixing or blending of orbital to av orbital of the same energy or to av the same degeneracy...p orbital has slightly higher energy than s orbital..but when hybridize,they will both have the same energy...a new orbital is form different from the original s
and p orbital..the new name of the orbital is now called sp orbital...we have sp,sp^2( called sp two not sp square) also sp^3....

No 2...i misread the question...am suppose to use debrogile equation... Lamda= h/mv...mistake on my part....

Eben331:
1 A ball of mass 1kg is dropped from a height of 5m onto a floor and rebounds to a height of 1m. Find the loss of kinetic energy due to impact. (g=9.8m/s2) 2 An electron of mass 9.4*10^-31kg has a wavelength of 1.3*10^-10m. Calculate the velocity of the electron. (h=6.6*10^-34Js) 3 A compound microscope, with lenses of focal lengths of 3cm and 5cm for the objective and eyepiece respectively, has an object placed 2cm from the objective lens. If the first image formed 3cm from the eyepiece lens, calculate the distance of the final image from the eyepiece. Happy Weekend Bosses.

...1..we can calculate it in many....easiest way...1/2mvi^2- 1/2mvf^2....vi=√2gh..√2 *10*5....vi= 10ms^-1.
Vf=√2 *1*10.. Vf=4.47.....1/2m(vi^2-vf^2)....
0.5 * 1(100-22)..=0.5 * 1 * 78...=39J...another method...mghinitiall=mghfinal +energy lost...energy lost = mghinitial - mghfinal...
Mg(hinitial- hfinal)...
1 * 10(5-1)...=10 *4= 40J...got 39j base or rounding up in first method...I use g= 10 for convenience...should be= 9.81

Eben331:
come back bro

...OK bro...only back to solve ur ques

Whenever I solve question which I consider little bit technical,I post it so DAT u guys can learn by trying to solve it.....b4,u guys wld atleast try to solve it bt now,I guess d enthusiasm as faded..our ogas are d one solving...m sure some days or a week to the exam,the thread will b lively...I will b back then...thanks guys...

A 2kg body moving on an level surface collide and compress an horizontal spring of force constant K= 2 N/m through 2m..The coefficient of kinetic friction between the body and the surface is 0.25..Calculate the velocity of the body while colliding....(A) 4.28..(B) 5.49...(C)1.89..(D)3.74...

U guys are not answering nor posting question...it is the newly admitted DAT r answering...wats wrong?..