Year :
2019
Title :
Chemistry
Exam :
JAMB Exam
Paper 1 | Objectives
31 - 40 of 80 Questions
| # | Question | Ans |
|---|---|---|
| 31. |
At 27°C, 58.5g of sodium chloride is present in 250cm\(^3\) of a solution. The solubility of sodium chloride at this temperature is? A. 2.0 moldm\(^{-3}\) B. 0.25 moldm\(^{-3}\) C. 1.0 moldm\(^{-3}\) D. 0.5 moldm\(^{-3}\) Detailed SolutionGiven the Mass of the salt = 58.5gVolume = 250 cm\(^3\) = 0.25 dm\(^3\) Mass concentration = \(\frac{Mass}{Volume}\) = \(\frac{58.5}{0.25}\) = 234 gdm\(^{-3}\) Solubility (in moldm\(^{-3}\) = \(\frac{234}{111} \) = 2.11 moldm\(^{-3}\) \(\approxeq\) 2.0 moldm\(^{-3}\) There is an explanation video available below. |
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| 32. |
Which of the following properties increases from left to right along the period but decreases down the group in the Periodic Table? A. ii and iii only B. ii and iv only C. iii and iv only D. i and ii only Detailed SolutionIonization energy and electron affinity increase across a period, and decrease down a group.There is an explanation video available below. |
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| 33. |
The molecular shape and bond angle of water are respectively A. linear, 180° B. bent, 109.5° C. tetrahedral, 109.5° D. bent, 105° Detailed Solution The shape of water molecule = Bent/ V- shaped The bond angle of water = 104.5°/ 105° There is an explanation video available below. |
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| 34. |
Na\(_2\)CO\(_3\) + 2HCl \(\to\) 2NaCl + H\(_2\)O + CO\(_2\)The indicator most suitable for this reaction should have a pH equal to A. 5 B. 7 C. 3 D. 9 Detailed SolutionMethyl orange is the best indicator for the reaction with range 3.1 - 4.4.There is an explanation video available below. |
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| 35. |
A radioactive nucleus has a half-life of 20 years, starting with 100,000 particles, how many particles will be left exactly at the end of 40 years A. 75,000 particles B. 35,000 particles C. 25,000 particles D. 50,000 particles Detailed SolutionGiven:t\(_{\frac{1}{2}\) = 20 years After the first 20 years, half of the substance (\(\frac{1}{2} \times 100,000 = 50,000\)) will have decayed. Hence, we have 100,000 - 50,000 = 50,000 particles left. After the second 20 years (being 40 years in all), half of the remaining substance (\(\frac{1}{2} \times 50,000 = 25,000\)) will have decayed. Remaining particles after 40 years = 50,000 - 25,000 = 25,000 particles. There is an explanation video available below. |
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| 36. |
Which two gases can be used for the demonstration of the fountain experiment? A. SO\(_4\) and NH\(_3\) B. SO\(_3\) and NH\(_3\) C. HCl and NH\(_3\) D. SO\(_2\) and CO |
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| 37. |
Which of the following pollutants will lead to the depletion of ozone layer? A. chlorofluorocarbon B. carbon (ii) oxide C. hydrogen sulphide D. sulphur (iv) oxide |
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| 38. |
Which of the following describes the chemical property of acids? A. None of the above B. Acid + XCO\(_3\) \(\to\) Salt + H\(_2\)O + NH\(_3\) C. Acid + NaOH \(\to\) Normal salt + CO\(_2\) D. Acid + X\(_n\)CO\(_3\) \(\to\) Salt + H\(_2\)O + CO\(_2\) |
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| 39. |
Which of the following pairs cannot be represented with a chemical formula? A. air and bronze B. bronze and sodium chloride C. copper and sodium chloride D. caustic soda and washing soda |
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| 40. |
Consider the equation below:Cr\(_2\)O\(_7 ^{2-}\) + 6Fe\(^{2+}\) + 14H\(^{+}\) \(\to\) 2Cr\(^{3+}\) + 6Fe\(^{3+}\) + 7H\(_2\)O.The oxidation number of chromium changes from A. +5 to +3 B. +6 to +3 C. -2 to +3 D. +7 to +3 Detailed SolutionCr\(_2\)O\(_7 ^{2-}\) + 6Fe\(^{2+}\) + 14H\(^{+}\) \(\to\) 2Cr\(^{3+}\) + 6Fe\(^{3+}\) + 7H\(_2\)OThe oxidation of Cr in Cr\(_2\)O\(_7 ^{2-}\) : Let the oxidation of Cr = x; 2x + (-2 x 7) = -2 \(\implies\) 2x - 14 = -2 2x = 12 ; x = +6 Hence, the change in oxidation of Cr = +6 to +3 There is an explanation video available below. |
| 31. |
At 27°C, 58.5g of sodium chloride is present in 250cm\(^3\) of a solution. The solubility of sodium chloride at this temperature is? A. 2.0 moldm\(^{-3}\) B. 0.25 moldm\(^{-3}\) C. 1.0 moldm\(^{-3}\) D. 0.5 moldm\(^{-3}\) Detailed SolutionGiven the Mass of the salt = 58.5gVolume = 250 cm\(^3\) = 0.25 dm\(^3\) Mass concentration = \(\frac{Mass}{Volume}\) = \(\frac{58.5}{0.25}\) = 234 gdm\(^{-3}\) Solubility (in moldm\(^{-3}\) = \(\frac{234}{111} \) = 2.11 moldm\(^{-3}\) \(\approxeq\) 2.0 moldm\(^{-3}\) There is an explanation video available below. |
|
| 32. |
Which of the following properties increases from left to right along the period but decreases down the group in the Periodic Table? A. ii and iii only B. ii and iv only C. iii and iv only D. i and ii only Detailed SolutionIonization energy and electron affinity increase across a period, and decrease down a group.There is an explanation video available below. |
|
| 33. |
The molecular shape and bond angle of water are respectively A. linear, 180° B. bent, 109.5° C. tetrahedral, 109.5° D. bent, 105° Detailed SolutionThe shape of water molecule = Bent/ V- shaped The bond angle of water = 104.5°/ 105° There is an explanation video available below. |
|
| 34. |
Na\(_2\)CO\(_3\) + 2HCl \(\to\) 2NaCl + H\(_2\)O + CO\(_2\)The indicator most suitable for this reaction should have a pH equal to A. 5 B. 7 C. 3 D. 9 Detailed SolutionMethyl orange is the best indicator for the reaction with range 3.1 - 4.4.There is an explanation video available below. |
|
| 35. |
A radioactive nucleus has a half-life of 20 years, starting with 100,000 particles, how many particles will be left exactly at the end of 40 years A. 75,000 particles B. 35,000 particles C. 25,000 particles D. 50,000 particles Detailed SolutionGiven:t\(_{\frac{1}{2}\) = 20 years After the first 20 years, half of the substance (\(\frac{1}{2} \times 100,000 = 50,000\)) will have decayed. Hence, we have 100,000 - 50,000 = 50,000 particles left. After the second 20 years (being 40 years in all), half of the remaining substance (\(\frac{1}{2} \times 50,000 = 25,000\)) will have decayed. Remaining particles after 40 years = 50,000 - 25,000 = 25,000 particles. There is an explanation video available below. |
| 36. |
Which two gases can be used for the demonstration of the fountain experiment? A. SO\(_4\) and NH\(_3\) B. SO\(_3\) and NH\(_3\) C. HCl and NH\(_3\) D. SO\(_2\) and CO |
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| 37. |
Which of the following pollutants will lead to the depletion of ozone layer? A. chlorofluorocarbon B. carbon (ii) oxide C. hydrogen sulphide D. sulphur (iv) oxide |
|
| 38. |
Which of the following describes the chemical property of acids? A. None of the above B. Acid + XCO\(_3\) \(\to\) Salt + H\(_2\)O + NH\(_3\) C. Acid + NaOH \(\to\) Normal salt + CO\(_2\) D. Acid + X\(_n\)CO\(_3\) \(\to\) Salt + H\(_2\)O + CO\(_2\) |
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| 39. |
Which of the following pairs cannot be represented with a chemical formula? A. air and bronze B. bronze and sodium chloride C. copper and sodium chloride D. caustic soda and washing soda |
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| 40. |
Consider the equation below:Cr\(_2\)O\(_7 ^{2-}\) + 6Fe\(^{2+}\) + 14H\(^{+}\) \(\to\) 2Cr\(^{3+}\) + 6Fe\(^{3+}\) + 7H\(_2\)O.The oxidation number of chromium changes from A. +5 to +3 B. +6 to +3 C. -2 to +3 D. +7 to +3 Detailed SolutionCr\(_2\)O\(_7 ^{2-}\) + 6Fe\(^{2+}\) + 14H\(^{+}\) \(\to\) 2Cr\(^{3+}\) + 6Fe\(^{3+}\) + 7H\(_2\)OThe oxidation of Cr in Cr\(_2\)O\(_7 ^{2-}\) : Let the oxidation of Cr = x; 2x + (-2 x 7) = -2 \(\implies\) 2x - 14 = -2 2x = 12 ; x = +6 Hence, the change in oxidation of Cr = +6 to +3 There is an explanation video available below. |