₦airaland Forum

JoyceTed:
Help a sis please.....
0.16g of methane when burnt increases the temperature of 100g of water by 40 degree centigrade, what is the heat of combustion of methane if the heat capacity of the water is 4.2jg/10/c

Ajet001:
T=24days
N1=64g
N2=2g
R=N1/N2
R=64÷2=32
R=2^n
R=2^5
n=5
T½=t/n
T½=24÷5=4.8days




2. The count rate of an alpha particle source is 400 per minute. If the half life of the source is 5 days, what would be the count rate per minut after 15 days?

T½=5days
C1=400
t=15days
T½=t/n
n=15/5
n=3
R=2^n
R=2^3
R=8
R=C1/C2
8=400/C2
C2=50per minute




3. 0.46g of ethanol when burned raised the temperature of 50g of water by 14.3k. Calculate the heat of combustion of ethanol. (O=16, H=1, Specific heat capacity of water = 4.2JgK^-1
Q=MCΦ
Heat loss by Ethanol =Heat gained by water
Q=50×4.2×14.3
Q=3003J
0. 46g of ethanol------> 3003
46g------> (46×3003)/0.46
=300300J
Heat of combustion of ethanol =300.3KJ

SHIJUADE02:
Pls i need the admission office number. #abeg ??

Ajet001:
Good evening everyone.,
pls I need more clarification on this negative marking of a thing.
Thanks.

Any MBBS aspirant in the house.??

#TeamMBBS

Ajet001:
No

samuelizz:
1. KE is converted to PE

½mv² = mgh

v² = 2gh.

h = 5cm = 0.05m

v² = 2(10)0.05

v = √1

v = 1ms-¹


2, the acceleration doubles when the force doubles, since the mass is kept constant.

3. y = vertical component, x = horizontal component

V² = U² + 2gh

30² - 10² = 2gh

h = 40m.

H = ½gt²

40 = ½(10)t²

t = 2√2s. ( time taken for the journey)

Now to the ball thrown horizontally,

Vx = Ux = 10ms-¹

Vy = Uy + gt..

Uy = 0, since the ball was thrown horizontally, the vertical component is zero

Vy = 10(2√2)

Vy = 20√2ms-¹

V = √(Vx² + Vy²)

V = √(10²) + (20√2)²

V = √(100 + 800)

V = √(900)

V = 30ms-¹


4. KE = ½mv² = ½m(20²)

KE = 200m

PE = mgh = m(100)10

PE = 1000m

% = 200m/1000m × 100

m cancels out with annoyance..

% = 0.2 × 100

% = 20%


5. mass of water = 5 - 4.26 = 0.74g

mass of anhydrous salt = 4.26g

Mass of water/mass of anhydrous salt = 18x/molar mass of salt

0.74/4.26 = 18x/208


x ≈ 2( the number of molecules of water)

Jacktheripper:
Where did u see all these questions

DaQueen7:
R.d = upthrust in water/upthrust in water

Upthrust in water= 10-7 =3N

1.5 = Upthrust in liquid/3

4.5 = upthrust in liquid

The weight in the liquid equals the original weigjt minus the upthrust
W =10N -4.5N
W= 5.5N

CondomSir:
Guys, the registration portal has been enabled now. You may now go ahead with your registration.

holuwajobar:
i thought UI sed dah we should upload our bio data by d last week of august buh dy have not open portal