Paper 1  Objectives  27 Questions
WASSCE/WAEC MAY/JUNE
Year: 2006
Level: SHS
Time:
Type: Question Paper
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#  Question  Ans 

1. 
How many orbitals are in the dsub shell? A. 1 B. 3 C. 5 D. 7 
C 
2. 
An element X has isotopic masses of 6 and 7. if the relative abundance is 1 to 12.5 respectively, what is the relative atomic mass of X? A. 6.0 B. 6.1 C. 6.9 D. 7.0
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Detailed SolutionContribution by isotope 6 = 6 x 1 = 6contribution by isotope 7 = 7 x 12.5 = 87.5 This contribution is made by 1 + 12.5 = 13.5 atoms Relative atomic mass of = 87.5 + 6 = 93.5/13.5 = 6.9 

3. 
An atom ^{238}_{92}X decays by alpha particle emission to give an atom Y. The atomic number and mass number of Y are A. 90 and 234 respectively B. 91 and 238 respectively C. 92 and 236 respectively D. 93 and 238 respectively
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Detailed Solution238_{92} X 234 _{90}Y + ^{4 2 He Atomic number is 90 while mass number is 234 } 

4. 
If 1 mole of sodium contains 6 x 10^{23} atoms, how many atoms are contained in 0.6 g of sodium? [Na = 23] A. 1.56 x 10^{23} B. 1.56 x 10^{22} C. 3.6 x 10^{23} D. 3.6 x 10^{22}
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Detailed Solution23g of sodium contain 6 x 10_{23} atoms :. 0.6g of sodium will contain \(\frac{0.6 x 6 x 10^{23}}{23}/) = 1.56 x 10^{22} 

5. 
If 20 cm^{3} of distilled water is added to 80cm ^{3} of 0. 50 mol dm^{3} hydrochloric acid, the new concentration of the acid will be A. 0.10mol dm^{3} B. 0.20mol dm^{3} C. 0.40mol dm^{3} D. 2.00mol dm^{3}
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Detailed SolutionC_{1} x 100 = 0.5 x 80C_{1} = \(\frac{0.5 x 80}{100}/) = \(\frac{4}{10}/) = 0 . 40mol/dm^{3} 

6. 
Consider the reaction represented by the equation. 2NaHCO_{3(s)} \(\rightarrow\) 2Na_{2}CO _{3(S)} + CO _{2(g)} + H_{2}O_{(g)} what volume of carbon (IV)oxide at s.t.p is evolved when 0. 5 moles of NaHCO_{3} is heated? [Molar volume = 22.4 dm^{3} at s.t. p] A. 1.12dm^{3} B. 2.24dm^{3} C. 5.6dm^{3} D. 56.0dm^{3}
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Detailed Solution2NaHCO_{3(s)} \(\rightarrow\) 2Na_{2}CO _{3(S)} + CO _{2(g)} + H_{2}O_{(g)}2 moles of 2NaHCO_{3(s)} produces 22.4dm^{3} of CO_{2} 0.5 moles of 2NaHCO_{3} will produce \(\frac{0.5 \times 22.4}{2}\) = \(\frac{11.2}{2}\) = 5.6 dm^{3} 
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