Hey! I wanna share with you about my final-first-semester-exam, and I hope this article will be useful for you :D

I’m sorry before because my English is not good. :P

1. X atom has atomic number 19, that element in the periodic table is situated on…

a. Period 2, Group IA

b. Period 3, Group IA

c. Period 4, Group IA

d. Period I, Group IVA

e. Period II, Group IVA

*Answer: 19 -> [Ar] 4s ^{1} (Electron configuration which based on quantum number’s
theory) or (2,8,8,1) (Electron configuration which based on Niels Bohr).
It means, this element is located on Period 4 and Group IA. So the answer is
C.*

2. Which of the following sets of quantum numbers is not allowed?

a. n=4, l=2, m=0, s=-1/2

b. n=4, l=1, m=3, s=+1/2

c. n=3, l=2, m=-2, s=-1/2

d. n=2, l=1, m=0, s=+1/2

e. n=4, l=2, m=-1, s=-1/2

*Answer: You know the formulas of Quantum Number? Okay, I will explain you
one by one:
n : It states the amount of the shells of the atom.
l : It states the shape of an orbital, the amount of the sub-shell. (We
call it as Azimuthal). The formula of this Quantum Number is
(n-1)
m : It states the amount of orbital of an atom. (We call it as Magnetic).
The formula is –l till +l.
s : It states the direction of electron in its orbital. (We call it as Spin).
And the formula of this Quantum Number is +1/2 or -1/2.*

* So, the answer is B, because of the (m). Why?
Because the (l) value is 1 so, the value of (m) which allowed is -1 until +1*

3. The following compound pair that has covalent bond is…

a. KCl and HCl

b. H_{2}S and K_{2}S

c. PCl_{3} and FeCl_{3}

d. CH_{4} and NH_{3}

e. H_{2}O and Na_{2}O

*Answer: You have to remember this: Ionic bond is a bond that occurs between Metal
Element and Non-metal element; Covalent bond is a bond that occurs between
2 or more Non-metal element; Metallic bond is a bond that occurs between 2 or
more Metal Element.
So:
KCl -> Ionic Bond*

* HCl -> Covalent Bond*

* H _{2}S -> Covalent Bond*

* K _{2}S -> Ionic Bond
PCl_{3} -> Covalent Bond
FeCl_{3} -> Ionic Bond*

* CH _{4} -> Covalent Bond*

* NH _{3} -> Covalent Bond*

* H _{2}O -> Covalent Bond*

* Na _{2}O -> Ionic Bond.
So, the answer is D*

4. Which groups that consist of polar compounds?

a. HCl, HBr, NH_{3}, H_{2}O

b. CO_{2}, Cl_{2}, Br_{2}, H_{2}

c. H_{2}, O_{2}, CO, CO_{2}

d. MgO, NH_{3}, CO, CO_{2}

e. SO_{2}, Cl_{2}, N_{2}, NH_{3}

*Answer: Intramolecular force is a force between atoms in a molecule. Nah, A molecule will be stated as a polar molecule if it has one or more lone pair.
Mmm would you like to learn about Intermolecular force by yourself? Because
the answer is A.*

5. The following pairs of compounds which both have hydrogen bond are…

a. HF and H_{2}O

b. HF and HI

c. HBr and H_{2}O

d. HCl and HBr

e. NH_{3} and HI

*Answer: In the X grade, we know that the molecules that have hydrogen bond are HF, H _{2}O, and NH_{3}. So the answer is A*

6. The compound that have sp^{3} hybrid orbital has the shape of…

a. Trigonal planar

b. Tetrahedral

c. Trigonal bipyramidal

d. Octahedral

e. Linear

*Answer: sp -> Linear; sp ^{2}-> Trigonal planar; sp^{3} -> Tetrahedral; sp^{3}d^{ } -> Trigonal bipyramidal; sp^{3}d^{2} -> Outer Octahedral; d^{2}sp^{3} -> Inner Octahedral.*

* So, the answer is B.*

7. Reaction of Sulfur trioxide preparation takes place at certain temperature and pressure.

If the reaction which occurs is SO_{2}(g) + O_{2}(g) -> SO_{3}(g), then gases volume ratio of

This reaction respectively is…

a. 1 : 1 : 1

b. 1 : 1 : 2

c. 1 : 2 : 3

d. 2 : 2 : 2

e. 2 : 1 : 2

*Answer: SO _{2}(g) + O_{2}(g) -> SO_{3}(g). Okay, we already learned it before on my previous*

* Article. But don’t worry, I will explain that lesson in here :D.
The amount of S in the left side is 1 and the amount of S in the right side is
1 too, so we don’t have to balance it. But how about the O???
The amount of O in the left side is 4 and the amount of O in the right side is 3
so what will we do to balance the O??
Aaahhh You were right!! Time SO*

_{2 }with 2, O

_{2}with 1, and the SO

_{3}with 2. :D

* Now let’s see our equation now:
2SO _{2}(g) + O_{2}(g) -> 2SO_{3}(g)*

* So the ratio is 2 : 1 : 2, the answer is E ! :D*

8. The statement of element mass ratio in a compound is constant was expressed by…

a. Amedeo Avogadro.

b. Proust.

c. John Jacob Berzellius.

d. Lavoisier.

e. John Dalton.

*Answer: Avogadro : Express that the coefficient in a molecule equals to its mol in the
same temperature and pressure.*

* Proust : Element mass ratio in a compound is constant.
Berzellius : He set the old version of periodic table.*

* Lavoisier : The mass before reaction equals to the mass after reaction.
Dalton : He stated that atom is the smallest particle that indestructible and
indivisible.*

* So the answer is B.*

9. The volume of 16 gram Sulfur dioxide at 27^{o}C and 1 atm is…

a. 2.10 liter

b. 3.25 liter

c. 4.20 liter

d. 5.25 liter

e. 6.15 liter

*Answer: PV = nRT*

* The Mr of Sulfur dioxide is 64, The Temperature is 300K and The pressure is
1 atm, So:*

* 1 * V = 16/64 * 0.082 * 300*

* V = 6.15 liter.
So the answer is E*

10. What volume will 2.80 gram of N_{2} occupy at STP ?

a. 2.240 liter

b. 5.600 liter

c. 11.20 liter

d. 22.40 liter

e. 44.80 liter

*Answer: PV = nRT*

* Because the volume is occupied at STP so the Molar Volume is 22.4 liter/mol
and we can change this formula into: V = n*Vm
*

* N = m/Mr*

* 2.80 gram/ 2*14 (Remember, The Ar of N is 14) = 0.1 mol
V = 0.1 * 22.4
= 2.24 liter.*

* So the answer is A.*

11. If Avogadro constant = L, and the molar mass of ammonia gas is M g/mol, then in 10 mol of ammonia gas, the number of ….. molecules were found.

a. 10

b. 10L

c. L/10M

d. 10 ML

e. L/M

*Answer: The amount of particle = The mol of the particle * Avogadro constant*

* X = n * L*

* X = 10L*

* PS: Avogadro Constant = 6.02x10 ^{23}*

* So the answer is B.*

12. The mass of nitrogen in a fertilizer containing 40 gram ammonium nitrate is…

a. 1.4 gram

b. 2.8 gram

c. 14 gram

d. 16.8 gram

e. 28 gram

*Answer: Ammonium nitrate = NH _{4}^{+ } + NO_{3}^{-} = NH_{4}NO_{3}*

* Mr of NH _{4}NO_{3} = 14 + (4*1) + 14 + (3*16)*

* = 80
*

* The mass of N = n * Ar / Mr NH _{4}NO_{3} * The mass of NH_{4}NO_{3}*

* = 2 * 14 / 80 * 40*

* = 14 gram*

* So, the answer is: C.*

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