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Chemistry
Paper 1 | Objectives | 49 Questions
JAMB Exam
Year: 1999
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper has been added to Cart!
Paper 1 | Objectives | 49 Questions
JAMB Exam
Year: 1999
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
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| # | Question | Ans |
|---|---|---|
| 1. |
200cm3 each of 0.1M solutions of lead (II) trioxonitrate (V) and hydrochloric acid were mixed. Assuming that lead (II) chloride is completely insoluble, calculate the mass of lead (II) chloride that will be precipitated. A. 2.78g B. 5.56g C. 8.34g D. 11.12g
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Detailed SolutionThe balanced equation of the reaction is:Pb(NO3)2(aq) + 2HCl(aq) ---> PbCl2(s) + 2HNO3(aq) From the balanced equation, 1 mole of Pb(NO3)2 gave 1 mole of PbCl2 using, molar concentration = no of moles/volume in dm3 therefore, no of moles = molar concentration x volume in dm3 where, molar concentration of Pb(NO3)2 = 0.1M volume of Pb(NO3)2 in dm3 = 0.2 dm3 therefore, no of moles of Pb(N |
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| 2. |
56.00cm3 of a gas at S.T.P. weighed 0.11g. What is the vapour density of the gas? A. 11.00 B. 22.00 C. 33.00 D. 44.00
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Detailed Solution56.00cm3 of a gas at S.T.P weighed 0.11gMolar volume of the gas = 22.4dm3 = 22400cm3 molar mass = (22400 x 0.11)/56 molar mass = 2464/56 = 44 Vapour density = Molar mass/2 = 44/2 = 22 |
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| 3. |
Which of the following gases will diffuse fastest when passed through a porous plug? A. Propane B. Oxygen C. Methane D. Ammonia |
C |
| 4. |
Which is the temperature of a given mass of a gas initially at 0°C and 9 atm, if the pressure is reduced to 3 atm at constant volume? A. 91K B. 182K C. 273K D. 819K
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Detailed SolutionFrom P1/T1 = P2/T2 = 9/273 = 3/T2T2 x 9 = 273 x 3 T2 = (273 x 3)/9 T2 = 91K |
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| 5. |
Which of the following will have its mass increased when heated in air? A. Helium B. Magnesium C. Copper pyrites D. Glass |
B |
| 6. |
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) A. 0.3g B. 1.5g C. 2.4g D. 3.0g
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Detailed SolutionMolar conc. =no.of moles／Vol in dm3Vol=250cm3= 0.25dm3 Molar conc=0.5M No of moles=molar conc *Vol in dm3 =0.5*0.25 =0.125mole of HCl 2moles of HCl → １Mole of Mg 0.125 mole →(0.125*1)/2 =0.0625mole of Mg Reacting mass=no of moles* molar mass Mg=24 Mass=0.0625*24 =1.5g |
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| 7. |
A gaseous metallic chloride MClx consists of 20.22% of M by mass. The formula of the chloride is? A. MCl B. MCl2 C. MCl3 D. M2Cl6
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Detailed SolutionPercentage by mass of M = 20.22%Percentage by mass of Cl = ? 100 - 20.22 = 79.78% Mass given ration of M M = 20.22/27 = 0.75 Chlorine = 79.78/35.5 = 2.24 Divide by the smallest M = 0.75/0.75 = 1 Cl = 2.24/0.75 = 3 ∴ The formula = MCl3 |
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| 8. |
In which of the following are water molecules in the most disorderly arrangement? A. Ice at -10°C B. Ice at 0°C C. Water at 100°C D. Steam at 100°C
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Detailed SolutionThe gaseous state of matter has the most disorderly arrangement of particles. Hence, steam (water in gaseous form) is the most disorderly arrangement on the list. |
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| 9. |
In order to remove one electron from a 3S orbital of gaseous sodium atom, about 496KJMol-1 of energy is required. This energy is referred to as A. electron affinity B. ionization energy C. activation evergy D. electronegativity |
B |
| 10. |
Nitrogen obtained from the liquefaction of air has a higher density than that obtained from nitrogen containing compounds because the former contains A. water vapour B. oxygen C. carbon (IV) oxide D. rare gases |
D |
Preview displays only 10 out of the 49 Questions