₦airaland Forum

A man stands one-third of d way between two walls and claps his hands. He hears two distinct echoes 1 sec apart,If the speed of sound is 330ms^-1,d distance btween the walls is?

emekakelvin: Baba I am loyal ooo, but what's up with that Doctor before my name?

smiles*....I wil soon expose maself to you bro!

Dr smart aw far na

Swagalord18: [color=grey]remember the charge on 12H⁺ cancels that on 12e^- ...making the LHS net charge = 2^- from the 2ClO₃^-
....the 6H₂O has no charge, ... So the total charge on the RHS is also 2^- from the 2Cl^-
so RHS net charge = LHS net charge ...
Making it correct
i hope you're getting me[/color]

i can grab,......thanx boss!

great.david:
P1 = 720 mmHg and not 722 mmHg......typographical error

thanx 4 d observation!!...I ddnt evn boda 2 luk tru d workings wella

great.david:
balance the charge at both sides.....
Total no of charge in the RHS=> +14 + (-2)= +12
total no of charge in the LHS=> 2 *(+3)= +6
: . RHS must be equal to LHS
i.e +12 + 6e^- = +6
x=6
thus, we have the equation as:
Cr2O7^2- + 14H^+ + 6e^-=2cr^3+ + 7H2O

if dats d matter,...den what abt dis
2ClO3^- +12H^+ +12e^-=2Cl^- + 6H2O...does it mean iT is nt correct!....

Swagalord18: according to Dalton's law,
total pressure = sum of partial pressures
so 752(total pressure) = 32(water pressure) + x(oxygen pressure)
x = 752 - 32 = 722mmHg(oxygen pressure)
^p1V1/^T1 =
^p2V2/^T2 (gas equation)
p1 = 722 mmHg
V1 = 500 cm³
T1 = (30 + 273)K = 303K
p2 = s.p = 760 mmHg
V2 =
T2 = s.t = 273K

v2 = ^p1V1T2/_p2T1
= ^{722 x 500 x 273}/_{760 x 303}
=
426.78 cm³

dats gr8

In d below reaction,.what is d value of x?Cr2O7^2- + 14H^+ + xe^-=2cr^3+ + 7H2O
pls waht I need most is aw u get dat value,...xo ur workings is all I need

labodinho: Guys are boiling here,let's go there...

1) 2H2(g) + O2(g)------>2H2O(g)

In the reaction above,what volume of hydrogen would be left over when 300cm^3 of oxygen and 1000cm^3 of hydrogen are exploded in a sealed tube?

NB.-The Numbers In front of the equation are subscript.

2)A piece of filter paper moistened with Lead(ii)ethanoate solution turns black when the paper is dropped into a gas jar containing an unknown gas.The gas is likely to be

A)Sulphur(iv)oxide
B)Hydrogen chloride
C)Sulphur(vi)oxide
D)Hydrogen sulphide

3)What number of moles of Cu^2+ will be deposited by 360 coulombs of electricity?

A)5.36 * 10^-4 mole
B)1.87 * 10^-3 mole
C)9.35 * 10^-4 mole
D)3.73 * 10^-3 mole




Tho they are all simple,just manage it ma gurus.Show workings where necessary.

2H2+O2=2H2O ratio of d volumes of d gases

2:1:2
pickin d least value,300cm3 2 multiply tru
we av,
300*2:300*1:300*2
600 : 300 :600
subtract d new vol from d initials
H O
1000 : 300
600:300
we av 400cm3 for hydrogen and Ocm3 for oxygen
soderefor d ans is 400cm3
2 D,...H2S
3 CU^2+ + 2e^-=CU
1 mol of Cu^2+ is deposited by 2*96500c
x mole mole of Cu^2+ is deposited by 360*1/2*96500=1.87*10^-3 moles

500cm3 of oxygen was collected over water at 30 degCelcius and 752mmHg pressure. What is the volume of dry Oxygen at S.t.p
(vapour pressure of water at 30 degcelcius=32mmHg)

Dr. Laz'Devitan:
Once the value has exceeded it,you don't have to it again...u use the one lesser....If you observe my working....u shall discover,i did not make use of it!

thanx a bunch sir...!

Dr. Laz'Devitan:
This a matter of throwing trials to the calculation to knw which is excess or not,
1mol NH(3) + 1mol HCl=> 1mol of NH(4)Cl
Mass before sparking:
19.04g NH(3) + 31.10g HCl=>........
using
36.5g HCl => 17g NH(3)
31.10g HCl => (31.10 * 17)/(36.5)
so we have 14.48g of NH(3)
when i tried the other......I got
40.58g of HCl.....this means that HCl was just enough while the NH(3) was in excess
a) So NH(3) was in excess and by an amount of (19.04 - 14.48)=4.56g

b) so we then can use 14.48g of NH(3) or 31.10g of HCl to calculate that of the AMMONIUM ION.....
36.5g of HCl=> 53.5g of NH(4)Cl
31.10g of HCl=>(31.1 * 53.5)/(36.5)
so, the Mass amount of the ammonium ion=45.58g

c)To get the other amount to react with the excess NH(3).....You do it as follow!
17g of NH(3) => 36.5g of HCl
4.56g of NH(3) => (4.56 * 36.5)/(17)
=9.79g of HCl
So the mass just need is
9.79g of HCl..!

I xo much respect d workings u gave bro,buht am confused abt smfn....dou I got dat 40.58g of d HCl,bu I was contemplatin mayb I was wrong because d value has already exceed d initial value,.sir my Q now is dat weneva we found it as such,do we av 2 consider it 2 b enof 4 d reaction...nd bsyds thanx for d disintegration

Pls d gurus in d aus,as u take mr Labodinho's Q,U can as well break dis down 4 us!

19.04g of ammonia were mixed with 31.10g of hydrogenchloride gas in a closed container,
a.whch of the reactants was in excess and by howmuch
b.how much ammoniumchloride was formed
c.how much more of d insufficient reactant would b needed to completely react with excess of other reactant?
(N=14,H=1, Cl=35.5)

pls,I need it worked out!

Just passin by*hyy...*

Dr. Laz'Devitan:
The name of the compound is

3-CHLOROPROPAN-2-ONE!

upon d whole dat dis Q is nt properly set,..stl u bombard it....hmmn,..sm ppl ar just naturally guru...kudos bro!

H CL
| |
H-C-C ---C=O
| | |
H H H-C-H
|
H

what is d IUPAC name?

H CL
| |
H-C-C ---C=O
| | |
H H H-C-H
|
H

Mismak: A big/special shout out to evry1 in dis forum.....

#odun yii a san wa sire, a san sayo.....HNY

Luff y'all

xame to u maam,.missed ya xo much!

great.david:
it has been some time now i've been contemplating over this question since you posted it initially.......but within my own knowledge, i resulted into this:
let exterior angle be x and interior angle be 4x.......then x + 4x = 180.....
X=36 degree
thus,
1)size of exterior angle=36
2) 2n-4)90/n=144
n = 10
pls my answer is subject to any correction.....

bro,without much emphasy.* are very correct

aysuccess99: This is my own questions for Physics, they are jamb based question.
1. The magnitude of the resultant of two mutually perpendicular forces, F1, and F2 is 13N. If the magnitude of F1 is 5N, what is the magnitude of F2?
A2.6N B8.0N C12.0N D18.0N
2. A vibrator of frequency 60Hz is used in generating transverse stationery waves in a long thin wire. If the average distance between successive nodes on the wire is 45cm. Find the speed of the transverse waves in the wire.
A.27m/s B 54m/s C90m/s D108m/s
*workings would be highly appreciated* Thanks a lot.

usin pythagoras theorem, F^2+f^2=R^2
5^2+f^2=13^2
25+f^2=169
f^2=169-25
f^2=144
f=root144
f=12N opt C is d coret ans
2.F=60Hz averageNN=45cm NN=45*2=90cm V=?
,\=2L
,\=2*90
,\=180cm
V=F,\
v=60*180/100
v=108ms^-1
opt D is d ans

emekakelvin: Abeg modify ur post, no let nairalander's see am b'cos I am still a learner.

lolz,...d question is a lil bit awkward!..

And bsyds,...d ans wil stl b dsame if I take dsme step u re talkin abt..cos it is * dats in btwn dem

aysuccess99: okay, i get it bro, that's nice. kudos to you.
this is the workings:
1.25*0.025/0.05
using BODMAS, we first deal with the division rule,
1.25*[0.025/0.05]
We now get:
1.25*0.05=(125/100)*(5\100)=625/10000=0.625, and to 1decimal place, it's 0.63.....
What a tricky maths question. Welldone and thanks oga emakelvin for making that clarification. God Bless.
@oga Ai-bi-ke, am on the road, when I get home, I will tackle your question.

what made me went anoda way was because I tot emeka question is dsame as sayin (a*b)/c
most ppl nake d mistake of takin dem 2 b dsame mostly wen typin on a social media lyk ds ,meanyl it is dfrnt frm a*b/c
xo mr emeka dats waht pop up d problem....I am sorry dou!

emekakelvin: A)36
B)10
Tongue out.

guy u b mathematics bokoharam o!

U can as well use dis as a shewing stick,.....The sides of a regular polygon are produced so that evry pair of exterior angle and interior angle is ratio 1:4. Calculate
a.the size of an exterior angle
b.number of sides of d polygon

1.25*0.025/0.05
(125*25*100)/(5*1000*100)
5/8=0.625 ans is A
2..optin C
3..opt C


mr aysuccess is very correct

emekakelvin: 3 in 1 question.

Jamb 2013.
Mathematic question4 type 1

.Evaluate 1.25*0.025/0.05, correct to 1 decimal place.
A)0.6
B)6.2
C)6.3
D)0.5

This maths question do me strong thing 4 hall.

Chemistry Jamb 2013 Type D Number 31
Which of the following compounds is a neutral oxide?
A)sulphur(VI)oxide
B)sulphur(IV)oxide
C) carbon(II)oxide
D)carbon(IV)oxide.

Physic Jamb 2014 Type B, number 4
The pair of physical quantities that are scalar only are
A) length and displacement
B)impulse and time
c)volume and area
D) moment and momentum.

Happy New Year

1.25*0.025/0.05
(125*25*100)/(5*1000*100)
5/8=0.625 ans is A
2..optin C
3..opt C

aysuccess99: good afternoon boss. Happy New Year to you. hmmmn, when it's almost time for jamb exam, i know that people/jambites would pop in here, so i know that this thread is very useful and I hope all the jambites here will gain admission.
@all, let the asking and answering of questions continue. Best of God and best of luck to us all.

same to you sire,but I fink u re already admitted!

Ifyx: Nope,aint,,2nd choiceakcpt wen u c Ifyx16

yes boss....

Una don bid here abi?,...

Ifyx: kul boss,drop ur 2go_id,nid to add up more uites

bro are u coc candidate?
.
.
,
¡and here is my 2go Id,..Afolayan696

Hmn!....sir taylor uv actually done a great work,..u ar mouthed bro..*make I just fold my hands dey follow d thread jeje..*