Chidorx60:

I suggest you change that method of reading one subject once in two weeks that's academic suicide
try reading each subject at least three times a week,I presume that you are sitting for the jamb exam,so sharing four subjects into a week should not be too hard.

SufferHeadAjebo:
How do everyone tackle Non-metals and its compound, metals and its compd for jamb so bulky

mhizcutie:

How about organic chemistry?the whole thing is bulky
But metals are similar to each other

mhizcutie:
.

SufferHeadAjebo:
so how we go about it ?

WinningEleven:


Its impossible to summarize organic chem to mere pages

SufferHeadAjebo:
Thanks!!! That means I should go with a reading timetable then Thanks anyways!!!
What uni are you applying ?

mhizcutie:

With me! Even the word impossible is spelt I'm possible

Chidorx60:

You wont need to summarize if you had a photographic memory.
But summarizing is quite stressful and tireing,
But i do it once in a while.

WinningEleven:
What is the function of myoglobin in muscles?

Chidorx60:

You wont need to summarize if you had a photographic memory.
But summarizing is quite stressful and tireing,
But i do it once in a while.

Odiaze:
My Name is ODIAZE, SUBJECT COMBO; Mathemactics, Physics, Chemistry & English. COURSE; Chemical Engineering. SCHOOL; University Of Benin (UNIBEN). Am i welcome?.

Quick Revision Questions in CHEMISTRY. 

To be answered in 20minutes !

I will post the CORRECT ANSWERS with EXPLANATIONS to the questions in 24 hours!



1 : The knowledge of half-life can be used to ___
A ) create an element.
B ) detect an element.
C ) split an element.
D ) irradiate an element.

The Answer is B  because half-life of a radioactive isotope is the time taken for a half of the original quantity to change or decay.It is used by geologists in dating geological remains .For instance; the quantities of uranium -238 in rocks are used to determine their ages. Similarly, in archaeological dating, the half-life of carbon-14(an isotope of carbon-12) enables the age of the remains of an organism such as a plant, to be deduced by comparing carbon-14 content of a dead plant with that of a similar living plant.


2 : Sieving is a technique used to separate mixtures containing solid particles of _______
A ) small sizes.
B ) large sizes.
C ) different sizes.
D ) the same size.

The Answer is C  because sieving is a separation technique based on the differences in the particle sizes of the components of the mixture. Examples are: soil sample, and iron filings of various sizes.


3 : A mixture of iodine and sulphur crystals can be separated by treatment with _________
A ) water to filter off sulphur.
B ) carbon (IV) sulphide to filter off iodine.
C ) ethanoic acid to filter of sulphur.
D ) methanol to filter off iodine.

The Answer is B  because sulphur is insoluble in water, but soluble in carbon (IV) sulphide.


4 : The ratio of the initial to the final pressure of a given mass of gas is 1 : 1.5 . Calculate the final volume of the gas if the initial volume was 300cm³ at the same temperature _______
A ) 750cm³
B ) 120cm³
C ) 200cm³
D ) 450cm³

The Answer is C because,using Boyle's law, P₁ x V₁ = P₂ x V₂ , where P₁ = 1, P₂ = 1.5, V₁ = 300cm³, but V₂ =?...Making V₂ the subject of the formula, V₂ = P₁ x V₁ /P₂ = 1 x 300/1.5 = 200cm³.
 

5 : The fundamental difference between the three states of matter is the ________
A ) shape of their particles.
B ) number of particles in each state.
C ) shape of the container they occupy.
D ) degree of movement of their particles.

The Answer is D because the forces of attraction or cohesion between the particles of solid matter are stronger than that in the liquid matter which is in turn stronger than the cohesive forces of the gaseous matter. This implies that gaseous particles can move faster and further than liquid particles which can in turn move faster and furter than solid particles .This accounts for the fundamental disparity between the three states of matter.


6 : 50cm³ of carbon (II) oxide was exploded with 150cm³ of air containing 20% of Oxygen by volume. Which of these reactants was in excess? ________
A ) carbon (II) oxide.
B ) carbon (IV) oxide.
C ) Oxygen.
D ) Nitrogen.

The Answer is C because from the equation of the reaction:
2CO + O₂ -------> 2CO₂
and applying Gay-Lussac's Law of Combining Volumes,  2cm³ of CO reacted with 1cm³ of O₂ to give 2cm³ of CO₂ .This suggests that 50cm³ of CO should react with 25cm³ of O₂ to give 50cm³ of CO₂ .Volume of Oxygen given is 20% of 150cm³ of air which is equal to 30cm³.This volume of oxygen (30cm³) is more than the conventional or espected volume of oxygen (25cm³) needed to react with 50cm³ of CO;Thus,Oxygen is in excess by 5cm³.
 

7 : The distance between the nuclei of chlorine atoms in a chlorine is 0.194mm. The atomic radius of chlorine atom is ? ________
A ) 0.097mm
B ) 0.194mm
C ) 0.388mm
D ) 2.388mm

The Answer is A because radius of an atom is the distance between the centre of the nucleus and the outermost electronic shell. In an atom, the orbiting electrons may be best represented by an electron cloud. Since the electron cloud of an atom has no definite limit, the size of an atom cannot be defined easily. However, using X-ray and electron diffraction techniques,it is possible to determine the distance between covalently bonded atoms. Generally, the atomic radius is taken to be one-half the distance of closest approach between the nuclei of atoms in the elemental substance. So, in this case, the radius of a chlorine atom is 0.194mm/2 =0.097mm.


8 : The pOH of a solution of 0.25 mol per dm³ of hydrochloric acid is? _________
A ) 12.40
B ) 13.40
C ) 14.40
D ) 14.60

The Answer is B because according to Sørensen Peder Lauritz Sørensen in 1909, the pH of a solution is the negative logarithm of the hydrogen ion concentration to the base 10. That is : pH= -Log[H⁺] . The hydrogen ion concentration in hydrochloric acid has been given to be 0.25mole per dm³ . Therefore, pH= -Log[0.25] = 0.6021.But,
pOH = 14 - pH.  Therefore, pOH = 14th - 0.6021 = 13.3979 ≈ 13.4 .


9 : ½Zn²⁺_(aq) + e-  ------> ½Zn_(s)
In the reaction above, calculate the quantity of electricity required to discharge Zinc. ______
A ) 0.965 x 10⁴ C
B ) 4.820 x 10⁴ C
C ) 9.650 x 10⁴ C
D ) 48.200 x 10⁴ C
[F = 96500 Coulomb per mol]

The Answer is C because the equation for the discharge of one mole of Zinc is : Zn²⁺_(aq) +  2e-  ------>  Zn_(s) . From the equation of the reaction, 2Faradays of electricity are required to discharge 1mole of Zinc. This means that 1mole of Zinc requires 2 x 96500 Coulombs of electricity to be discharged. Therefore, ½ mole of Zink will reuqire ½ x 2 x 96500 = 96500 Coulombs = 9.650 x 10⁴ C


10 : 0.46g of ethanol when burned raised the temperature of 50g of water by 14.3K. Calculate the heat of combustion of ethanol. _______
A ) +3000kJ per mol
B ) +300kJ per mol
C ) -300kJ per mol
D ) -3000kJ per mol
[C = 12, O = 16, H = 1, Specific heat capacity of water = 4.2 Joule per gram per Kelvin]

The answer is C because, calculating the heat accumulated by the 50g of water, using, H = MC∅ = 50 x 4.2 x 14.3 = 3003Joules of heat. This means that 0.46g of ethanol produced 3003 Joules of heat. Therefore 46g of ethanol will produce 46 x 3003/0.46 = 300300J ≈ -300kJ
(because the process is exothermic)

WinningEleven:
Where did everyone go?