Chemistry questions and answers

Chemistry Questions and Answers

Learn more about the properties, composition, and structure of substances (elements and compounds) with these Chemistry questions and answers. This Test can be used by students preparing for Chemistry in JAMB, WAEC, NECO or Post UTME.

31.

Arrange the following compounds in decreasing order of their boiling points: \(NH_3\), HF, \(SiH_4\), \(CH_4\)

A.

HF,\(NH_3\), \(SiH_4\), \(CH_4\)

B.

\(CH_4\), \(SiH_4\), HF, \(NH_3\),

C.

\(NH_3\),\(CH_4\), \(SiH_4\), HF

D.

\(CH_4\), \(SiH_4\), \(NH_3\), HF

Correct answer is A

The correct order of boiling points of the compounds \(NH_3\), HF, \(SiH_4\), and \(CH_4\) is HF, \(NH_3\), \(SiH_4\), \(CH_4\).
The correct order of boiling points of the compounds \(NH_3\), HF, \(SiH_4\), and \(CH_4\) is HF, \(NH_3\), \(SiH_4\), \(CH_4\).

The boiling point of a compound is the temperature at which the vapor pressure of the liquid is equal to the atmospheric pressure. The stronger the intermolecular forces, the higher the boiling point.

32.

Consider the following reaction equation:
\(2SO_3(g) → 2SO_2(g) + O_2\)(g)
∆H = +198 kJ/mol
Which of the following statements about the reaction is correct?

A.

198 kJ of energy is given off

B.

The reaction is exothermic

C.

The reaction container would feel warm.

D.

198 kJ of energy is absorbed

Correct answer is D

A positive value of ∆H (+198 kJ mol^-1) indicates that the reaction is endothermic, meaning it requires an input of energy to proceed. In this case, the reaction absorbs 198 kJ of energy per mole of reactants to convert 2 moles of \(SO_3\) into 2 moles of \(SO_2\) and 1 mole of \(O_2\). This means that the reaction container would need to be supplied with energy, and it would feel cooler rather than warmer.

33.

The separation technique that is suitable for separating iodine from tetrachloromethane is

A.

solvent extraction

B.

decantation

C.

fractional distillation

D.

filtration

Correct answer is A

Solvent extraction, also known as liquid-liquid extraction, is a technique used to separate substances based on their differential solubilities in two immiscible liquids. In this case, since iodine and tetrachloromethane have different solubilities in certain solvents, you can use solvent extraction to selectively dissolve one of the substances while leaving the other behind. Iodine is a non-polar molecule, while tetrachloromethane is a polar molecule. This means that iodine will dissolve in a non-polar solvent, while tetrachloromethane will dissolve in a polar solvent

34.

The salt solution formed firm a reaction between ethanoic acid and sodium hydroxide solution would be

A.

amphoteric

B.

neutral

C.

basic

D.

acidic

Correct answer is C

Ethanoic acid is a weak acid, while sodium hydroxide is a strong base. When these two substances react, they neutralize each other, forming salt and water. The salt is a compound of the cation from the base and the anion from the acid. In this case, the salt is sodium ethanoate, which is a basic salt.

35.

Which of the following equations does not illustrate correctly one of the reactions of chlorine?

A.

\(Ca(OH)_2 + Cl_2 → CaOCl_2 + H_2O\)

B.

\(Cl_2 + 2NaF → 2NaCl + F_2\)

C.

\(Cl_2 + 2NaOH → NaCl + NaClO + H_2O\)

D.

\(H_2S + Cl_2 → 2HCl + S\)

Correct answer is B

The equation \(Cl_2 + 2NaF → 2NaCl + F_2\) does not illustrate correctly one of the reactions of chlorine.

In this reaction, chlorine reacts with sodium fluoride to form sodium chloride and fluorine gas. However, this reaction is not possible because fluorine is a much more reactive element than chlorine. When chlorine and fluorine react, they form a compound called chlorine trifluoride (\(ClF_3)\), which is a very powerful oxidizing agent.

36.

The pair of compounds that belongs to the same homologous series is

A.

\(C_2H_6\) and \(C_4H_{10}\)

B.

\(C_3H_8\) and \(C_3H_6\)

C.

\(C_2H_4\) and \(C_4H_{10}\)

D.

\(C_4H_{10}\) and \(C_5H_{10}\)

Correct answer is A

In the given options, \(C_2H_6\) (ethane) and \(C_4H_{10}\) (butane) both belong to the same homologous series, as they have the same alkane functional group \(C_nH_{2n}+2\) and differ by a \(CH_2\) group:
\(C_2H_6\): Ethane
\(C_4H_{10}\): Butane
Both of these compounds are members of the alkane homologous series.

37.

Which of the following equimolar solutions has the highest conductivity?

A.

\(H_2SO{_4(aq)}\)

B.

\(NaOH{_(aq)}\)

C.

\(H_2CO{_3(aq)}\)

D.

\(CH_3COOH{_4(aq)}\)

Correct answer is A

\(H_2SO_4\)(aq) is a strong acid that dissociates into two hydrogen ions (H+) and one sulfate ion \(SO4^{2-}\) when dissolved in water. This leads to a higher concentration of ions in the solution, resulting in higher conductivity.

38.

How many moles are there in 3.0 g of \(O_2\)? [O = 16.0]

A.

0.0093 moles.

B.

0.0930 moles.

C.

0.0625 moles.

D.

0.6250 moles.

Correct answer is B

Number of moles = Mass / Molar mass
Given that the molar mass of oxygen \(O_2\) is 32.0 g/mol (2 × 16.0 g/mol), you can calculate the number of moles:
Number of moles = 3.0 g / 32.0 g/mol = 0.09375 moles
Rounded to four significant figures, the number of moles in 3.0 g of O2 is approximately 0.0938 moles.
Therefore, the closest answer choice is:
0.0930 moles.

39.

Species that occur in a reaction pathway but not in the overall reaction are known as

A.

Reactants

B.

Inhibitors

C.

Products

D.

Intermediates

Correct answer is D

In a chemical reaction, intermediates are the chemical species that are formed during the reaction but are subsequently consumed in further steps, leading to the final products. They are transient species that exist momentarily during the reaction process and are not present in the initial reactants or the final products of the overall reaction.

40.

A mixture containing two salts of different solubility can be separated by

A.

chromatography

B.

crystallization

C.

distillation

D.

evaporation

Correct answer is B

Crystallization is a separation technique that exploits differences in the solubilities of components in a solution. When a solution is allowed to cool or evaporate under controlled conditions, one of the salts may begin to crystallize out of the solution before the other due to its lower solubility. This process can be used to separate the salts based on their differing solubilities and crystal formation.