Chemistry, 2019, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

2019

Title :

Chemistry

Exam :

JAMB Exam

Paper 1 | Objectives

61 - 70 of 80 Questions

#	Question	Ans
61.	The oxidation state(s) of nitrogen in ammonium nitrite is/are A. -3, +3 B. +5 C. -3, +5 D. -3 Show Content Detailed Solution Ammonium nitrite = NH\(_4\)NO\(_2\) NH\(_4 ^+\): Let the oxidation number of Nitrogen = x x + 4 = 1 \(\implies\) x = 1 - 4 x = -3 NO\(_2 ^-\): x - 4 = -1 x = -1 + 4 \(\implies\) x = +3. The oxidation numbers for Nitrogen in Ammonium Nitrite = -3, +3. There is an explanation video available below.	A
62.	Hydrogen bond is a sort of A. van der waals force B. dative bond C. ionic bond D. a covalent bond Show Content Detailed Solution Hydrogen bond is a covalent intermolecular bond that exists between hydrogen and highly electronegative elements like nitrogen, oxygen and fluorine. There is an explanation video available below.	D
63.	Methane is prepared in the laboratory by heating a mixture of sodium ethanoate with soda lime. The chemical constituent(s) of soda lime is/are A. Ca(OH)\(_2\) and NaOH B. Na\(_2\)CO\(_3\) C. NaCl D. NaOH Show Content Detailed Solution Soda lime is a mixture of caustic soda (NaOH) and lime water (Ca(OH)\(_2\)). There is an explanation video available below.	A
64.	Which of the following sets of operation will completely separate a mixture of sodium chloride, sand and iodine? A. addition of water, filtration, evaporation to dryness and sublimation B. filtration, evaporation to dryness, addition of water and sublimation C. sublimation, filtration, evaporation and addition of water D. sublimation, addition of water, filtration and evaporation to dryness Show Content Detailed Solution Process: Add water to the mixture and then filter. The resultant is a salt solution as your filtrate and a mixture of iodine and sand as your residue. Evaporate the filtrate to dryness to get your sodium chloride (salt). Heat the residue to sublime the iodine from the sand. Addition of water \(\to\) Filtration \(\to\) Evaporation \(\to\) Sublimation. There is an explanation video available below.	A
65.	200cm\(^3\) of 0.50mol/dm\(^3\) solution of calcium hydrogen trioxocarbonate (IV) is heated. The maximum weight of solid precipitated is A. 10g B. 25g C. 20g D. 15g Show Content Detailed Solution Equation: Ca(HCO\(_3\))\(_2\) \(\to\) CaCO\(_3\) (precipitate) + H\(_2\)O + CO\(_2\) (in the presence of heat) V = 200 cm\(^3\) = 0.2 dm\(^3\) C = 0.5 mol/dm\(^3\) = 0.5M N = CV \(\implies\) N = 0.5 \(\times\) 0.2 = 0.1 mole From the equation 1 mole of Ca(HCO\(_3\))\(_2\) gives 100g of CaCO\(_3\) (1 mole CaCO\(_3\) = 40 + 12 + (3 x 16) = 100g\) 0.1 mole of Ca(HCO\(_3\))\(_2\) gives x g of CaCO\(_3\). \(\frac{1}{0.10} = \frac{100}{x}\) \(x = 100 \times 0.10 = 10 g\) There is an explanation video available below.	A
66.	Which of the following conditions will most enhance the spontaneity of a reaction? A. \(\Delta\)H is negative and greater than T\(\Delta\)S B. \(\Delta\)H is negative and \(\Delta\)S = 0 C. \(\Delta\)H is positive and less than T\(\Delta\)S D. \(\Delta\)H is positive and equal to T\(\Delta\)S Show Content Detailed Solution There is an explanation video available below.	A
67.	X is a substance which liberates CO\(_2\) on treatment with concentrated H\(_2\)SO\(_4\). A warm solution of X can decolorize acidified KMnO\(_4\). X is A. HCOOH B. NaHCO\(_3\) C. FeCO\(_3\) D. H\(_2\)C\(_2\)O\(_4\) Show Content Detailed Solution It should be noted that for X to liberate CO\(_2\), X must be a carbonate or an oxalate. Since X decolorizes KMnO\(_4\), X must be an oxalate. Therefore, X is H\(_2\)C\(_2\)O\(_4\). There is an explanation video available below.	D
68.	The velocity, V of a gas is related to its mass, M by (k = proportionality constant) A. V = \(\frac{k}{M}\) B. V = \(\frac{k}{M^{\frac{1}{2}}}\) C. V = \(kM^2\) D. V = \((\frac{k}{M})^{\frac{1}{2}}\) Show Content Detailed Solution Recall: V = \(\sqrt{\frac{3RT}{M}}\) \(\therefore V \propto \frac{1}{\sqrt{M}}\) \(V = \frac{k}{\sqrt{M}}\) V = \(\frac{k}{M^{\frac{1}{2}}}\) There is an explanation video available below.	B
69.	The emission of two successive beta particles from the nucleus \(^{32} _{15} P\) will produce A. \(^{30} _{14}\)Si B. \(^{32} _{13}\)Al C. \(^{32} _{17}\)Cl D. \(^{32} _{16}\)S Show Content Detailed Solution \(^{32} _{15}\)P \(\to\) 2\(^{0} _{-1}\)e + \(^{a} _{b}\)X 32 = 2(0) + a a = 32 15 = 2(-1) + b b = 15 + 2 b = 17 \(\therefore\) \(^{a} _{b}\)X = \(^{32} _{17}\)Cl There is an explanation video available below.	C
70.	Hydrogen diffused through a porous plug A. thrice as fast as oxygen B. at the same speed as oxygen C. four times as fast as oxygen D. twice as fast as oxygen Show Content Detailed Solution Using Graham's Law of Diffusion, \(\frac{R_1}{R_2} = \sqrt{\frac{m_2}{m_1}}\) \(\frac{R_H}{R_O} = \sqrt{\frac{m_O}{m_H}}\) \(\frac{R_H}{R_O} = \sqrt{\frac{32}{2}}\) \(\frac{R_H}{R_O} = \sqrt{16} = 4\) \(\therefore R_H = 4 \times R_O\) Therefore, hydrogen diffuses 4 times faster than oxygen. There is an explanation video available below.	C

61.	The oxidation state(s) of nitrogen in ammonium nitrite is/are A. -3, +3 B. +5 C. -3, +5 D. -3 Show Content Detailed Solution Ammonium nitrite = NH\(_4\)NO\(_2\) NH\(_4 ^+\): Let the oxidation number of Nitrogen = x x + 4 = 1 \(\implies\) x = 1 - 4 x = -3 NO\(_2 ^-\): x - 4 = -1 x = -1 + 4 \(\implies\) x = +3. The oxidation numbers for Nitrogen in Ammonium Nitrite = -3, +3. There is an explanation video available below.	A
62.	Hydrogen bond is a sort of A. van der waals force B. dative bond C. ionic bond D. a covalent bond Show Content Detailed Solution Hydrogen bond is a covalent intermolecular bond that exists between hydrogen and highly electronegative elements like nitrogen, oxygen and fluorine. There is an explanation video available below.	D
63.	Methane is prepared in the laboratory by heating a mixture of sodium ethanoate with soda lime. The chemical constituent(s) of soda lime is/are A. Ca(OH)\(_2\) and NaOH B. Na\(_2\)CO\(_3\) C. NaCl D. NaOH Show Content Detailed Solution Soda lime is a mixture of caustic soda (NaOH) and lime water (Ca(OH)\(_2\)). There is an explanation video available below.	A
64.	Which of the following sets of operation will completely separate a mixture of sodium chloride, sand and iodine? A. addition of water, filtration, evaporation to dryness and sublimation B. filtration, evaporation to dryness, addition of water and sublimation C. sublimation, filtration, evaporation and addition of water D. sublimation, addition of water, filtration and evaporation to dryness Show Content Detailed Solution Process: Add water to the mixture and then filter. The resultant is a salt solution as your filtrate and a mixture of iodine and sand as your residue. Evaporate the filtrate to dryness to get your sodium chloride (salt). Heat the residue to sublime the iodine from the sand. Addition of water \(\to\) Filtration \(\to\) Evaporation \(\to\) Sublimation. There is an explanation video available below.	A
65.	200cm\(^3\) of 0.50mol/dm\(^3\) solution of calcium hydrogen trioxocarbonate (IV) is heated. The maximum weight of solid precipitated is A. 10g B. 25g C. 20g D. 15g Show Content Detailed Solution Equation: Ca(HCO\(_3\))\(_2\) \(\to\) CaCO\(_3\) (precipitate) + H\(_2\)O + CO\(_2\) (in the presence of heat) V = 200 cm\(^3\) = 0.2 dm\(^3\) C = 0.5 mol/dm\(^3\) = 0.5M N = CV \(\implies\) N = 0.5 \(\times\) 0.2 = 0.1 mole From the equation 1 mole of Ca(HCO\(_3\))\(_2\) gives 100g of CaCO\(_3\) (1 mole CaCO\(_3\) = 40 + 12 + (3 x 16) = 100g\) 0.1 mole of Ca(HCO\(_3\))\(_2\) gives x g of CaCO\(_3\). \(\frac{1}{0.10} = \frac{100}{x}\) \(x = 100 \times 0.10 = 10 g\) There is an explanation video available below.	A

66.	Which of the following conditions will most enhance the spontaneity of a reaction? A. \(\Delta\)H is negative and greater than T\(\Delta\)S B. \(\Delta\)H is negative and \(\Delta\)S = 0 C. \(\Delta\)H is positive and less than T\(\Delta\)S D. \(\Delta\)H is positive and equal to T\(\Delta\)S Show Content Detailed Solution There is an explanation video available below.	A
67.	X is a substance which liberates CO\(_2\) on treatment with concentrated H\(_2\)SO\(_4\). A warm solution of X can decolorize acidified KMnO\(_4\). X is A. HCOOH B. NaHCO\(_3\) C. FeCO\(_3\) D. H\(_2\)C\(_2\)O\(_4\) Show Content Detailed Solution It should be noted that for X to liberate CO\(_2\), X must be a carbonate or an oxalate. Since X decolorizes KMnO\(_4\), X must be an oxalate. Therefore, X is H\(_2\)C\(_2\)O\(_4\). There is an explanation video available below.	D
68.	The velocity, V of a gas is related to its mass, M by (k = proportionality constant) A. V = \(\frac{k}{M}\) B. V = \(\frac{k}{M^{\frac{1}{2}}}\) C. V = \(kM^2\) D. V = \((\frac{k}{M})^{\frac{1}{2}}\) Show Content Detailed Solution Recall: V = \(\sqrt{\frac{3RT}{M}}\) \(\therefore V \propto \frac{1}{\sqrt{M}}\) \(V = \frac{k}{\sqrt{M}}\) V = \(\frac{k}{M^{\frac{1}{2}}}\) There is an explanation video available below.	B
69.	The emission of two successive beta particles from the nucleus \(^{32} _{15} P\) will produce A. \(^{30} _{14}\)Si B. \(^{32} _{13}\)Al C. \(^{32} _{17}\)Cl D. \(^{32} _{16}\)S Show Content Detailed Solution \(^{32} _{15}\)P \(\to\) 2\(^{0} _{-1}\)e + \(^{a} _{b}\)X 32 = 2(0) + a a = 32 15 = 2(-1) + b b = 15 + 2 b = 17 \(\therefore\) \(^{a} _{b}\)X = \(^{32} _{17}\)Cl There is an explanation video available below.	C
70.	Hydrogen diffused through a porous plug A. thrice as fast as oxygen B. at the same speed as oxygen C. four times as fast as oxygen D. twice as fast as oxygen Show Content Detailed Solution Using Graham's Law of Diffusion, \(\frac{R_1}{R_2} = \sqrt{\frac{m_2}{m_1}}\) \(\frac{R_H}{R_O} = \sqrt{\frac{m_O}{m_H}}\) \(\frac{R_H}{R_O} = \sqrt{\frac{32}{2}}\) \(\frac{R_H}{R_O} = \sqrt{16} = 4\) \(\therefore R_H = 4 \times R_O\) Therefore, hydrogen diffuses 4 times faster than oxygen. There is an explanation video available below.	C

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