## Chemistry 2018 | Study Mode

Question 31
A
< 7
B
> 7
C
= 7
D
Explanation.

$$Cl_{2} + H_{2}O \to HCl + HOCl$$

The two compounds in the product end are acidic and hence, their mixture is acidic. Hence, the pH is < 7

Question 32
A
$$1.20 \times 10^{23}$$
B
$$2.41 \times 10^{23}$$
C
$$3.62 \times 10^{23}$$
D
$$4.82 \times 10^{23}$$
Explanation.

$$1 mole = 6.02 \times 10^{23} atoms$$

$$0.2 moles = 6.02 \times 10^{23} \times 0.2$$ atoms

= $$1.204 \times 10^{23}$$

$$\approxeq 1.2 \times 10^{23}$$

Question 33
A
charge density of the atoms
B
ductility of the metal
C
number of valence electrons
D
total number of electrons in the atom
Explanation.

It is determined by the difference in the electron structures of the atoms. These primarily are: The number of outer electrons, i.e the number of valence shell electrons released during a metallic bond.

Question 34
A
oxidised
B
reduced
C
decomposed
D
dissociated
Explanation.

Reaction: $$2AgNO_{3} + Zn \to Zn(NO_{3})_{2} + 2Ag$$

The oxidation number of Zn in the reactant is 0, and in the product is +2. Therefor, it is oxidized.

Question 35
A
$$Cu_{2}O$$
B
$$CuO$$
C
$$CuO_{2}$$
D
$$CuO_{4}$$
Explanation.

Cu : O

0.067 : 0.066

mole ratio $$\frac{0.067}{0.066}$$ : $$\frac{0.066}{0.066}$$

= 1 : 1

= CuO

Question 36
A
+2 to +3
B
0 to +2
C
+3 to +2
D
+3 to 0
Explanation.

The oxidation number of Iron in $$FeCl_{3}$$ = +3. The oxidation number in $$FeCl_{2}$$ = +2.

Question 37
A
Centrifugation
B
Filtration
C
Chromatography
D
Distillation
Explanation.

No official Explanation yet!

Question 38
A
increases as nuclear charge increases
B
decreases as nuclear charge increases
C
decreases as nuclear charge decreases
D
remains constant as nuclear charge increases
Explanation.

No official Explanation yet!

Question 39
A
1:1
B
1:2
C
2:1
D
2:3
Explanation.

1g of Hydrocarbon = 0.93g of Carbon, therefore, Hydrogen = (1 - 0.93)g = 0.07g

C : H

0.93 : 0.07

$$\frac{0.93}{12}$$ : $$\frac{0.07}{1}$$

0.078 : 0.07

$$\frac{0.078}{0.07} : \frac{0.07}{0.07}$$

$$\approxeq 1:1$$

Question 40
A
pure samples of the same compound contain the same elements combined in the same proportion by mass
B
pure samples of substances are in the same proportion by mass
C
chemical compounds are pure because they contain the same elements
D
matter can neither be created nor destroyed.
Explanation.

The Law of Definite Proportions states that all pure samples of a particular chemical compound contain similar elements combined in the same proportion by mass.

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