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Fashion / Re: Senator Style For Men by Qudusthetutor(m): 5:20pm On Jun 09 |
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Fashion / Re: Senator Style For Men by Qudusthetutor(m): 5:19pm On Jun 09 |
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Fashion / Re: Senator Style For Men by Qudusthetutor(m): 1:25pm On May 09 |
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Education / Re: Chemistry by Qudusthetutor(m): 1:23pm On May 09 |
WAEC PAST QUESTION AND ANSWER State Faraday's first law of electrolysis. (ii) Distinguish between a strong electrolyte and a weak electrolyte (b) State one chemical property of ethyne. (c)( i) What is meant by the tern unsaturated hydrocarbon? (ii) Complete the following reaction equation: CH3 + CH3 OH-> (iii) Name the major product formed in the cation stated in 1(c)(ii). (d) State one way by which the rate of esterification could be increased. (e) Consider the reaction represented by the following equation: Zn + H2 SO4 → ZnOS4 + H2 . If 3.75g of Zn dust was added to excess H2 SO4 . Calculate the number of molecules of hydrogen gas produced. [ Zn = 65.0, Na = 6.02 X1023 ]. (f) State one effect of global warming. (g) Consider the following reaction equation: A. Pb(NO3 ) +H2 S --> PbS + 2HNO3 ; B. H2 + C2 H4 → C2 H6 . C. Zn(OH)2 + 2OH → [ Zn(OH)4 ]2 . (i) Which of the equations represent(s) redox process? (ii) State the change in Oxidation number of the species that are oxidized or reduced. (h)(i) State two of the main concepts of Bohr's model of the atom. (ii) State the limitations of Bohr's model. (i) List three factors that could influence the equilibrium position of a reversible reaction. (j) Calcium trioxocarbonate(iv) powder is added to separate equimolar solutions of hydrochloric acid and ethanoic acid. State one: (i) similarity in the observation in both reactions: (ii) difference in the observation in both reactions. ANSWERS (i) Faraday's first law of electrolysis states that the amount of a substance discharged at/dissolved from an electrode is directly proportional to the quantity of electricity that passes through the electrolyte. OR - The mass of an element deposited at the electrode is directly proportional to the quantity of electricity passed. (ii) Strong electrolytes dissociate/ionize completely in aqueous solutions while weak electrolytes dissociate/ionize only slightly/partially in aqueous solution. (b) -it burns in air/oxygen; - it undergoes addition reaction;-it undergoes polymerization, - it undergoes substitution/precipitation reaction. (c)i) hydrocarbon with double/triple bond/multiple bonds between at least a pair of carbon atoms. (ii) CH3 COOH + CH3 OH ⇌ CH3 COOCH3 + H2 (ii) Methylethanoate (d) addition of concentrated H2 SO4 ; - removal of the ester as soon as it forms; - increase the concentration of one of the reactants. (e) 65.0g of Zn2+ will produce 6.02 x 1023 molecules of H2 , OR 65.0g Zn = I molecule of H2 ∴ 3.75g of Zn = frac3.7565 X 6.02 x 1023 = 3.4 X 10 molecules (f) Sea level rising: -severe climate changes,depletion of coral reefs; -alters ecosystem balance/setting animal on the move; - severe drought; flooding or( areas); - polar ice melting (g)(i) equation B, H2 +C2 H4 --> C2 H6 (ii) In equation B, H changes from zero to-1; C changes from +2 to +3 as Concentration (i) (h)(i) an electron in an atom exists/revolves a circular orbit; - energy of an electron is quantified / has a fixed value; - an electron emits energy in the form of radiation when it moves from a higher energy state to a lower energy State. (ii) Bohr's model cannot explain the more complicated spectra lines observed in spectra Other than that of hydrogen. (i) temperature; pressure/volume; -concentration (j) (i) produces a gas/effervescence/bubbles- given off;- HCI will produce a more vigorous reaction/CH3 COOH will produce a less vigorous reacton. |
Education / Re: Chemistry by Qudusthetutor(m): 1:16pm On May 09 |
Rare gases are suitable because they Options A) are monoatomic B) form ions easily C) have duplet or octet electronic configuration in the outermost shells in their atoms D) are volatile gases The answer is C |
Education / Re: Chemistry by Qudusthetutor(m): 1:15pm On May 09 |
Calculate the relative molecular mass of Limestone CaCO3, (Ca = 40, C =12, O = 16). Calcium=40 Carbon=12 Oxygen=16 For caco3 There is 3 Moles of Oxygen ; 40+12+(3×16) 52+48 100 |
Education / Re: Chemistry by Qudusthetutor(m): 1:10pm On May 09 |
Question 1 Which of the following pairs of compounds belongs to the same homologous series? Options A) C3H8 and C3H6 B) C4H10 and C5H10 C) C2H4 and C4H10 D) C2H6 and C4H10 Answer: C Solution C2H4 is ethane while C4H10 is butane Organic Chemistry 1 Methane (carbon=1) Ethane ( carbon=2) Propane (carbon=3) Butane ( carbon=4) Alkanes formula for homologous series ( CnH2n+2) Where n is the number of carbon present For Ethane with carbon 2 CnH2n+2 C2H2(2) +2 C2H4+2 C2H6 For Butane with carbon 4 CnH2n+2 C4H2(4) +2 C4H8+2 C4H10 |
Education / Re: Chemistry by Qudusthetutor(m): 12:16pm On May 09 |
Draw the mechanism of the reaction make sure to include all arrows, lone pairs electron, charges and intermediates
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Education / Chemistry by Qudusthetutor(m): 12:00pm On May 09 |
The most effective method of decontamination of polluted industrial wastewater is its treatment with substances that transform harmful impurities into insoluble compounds. At the same time, the precipitating reagent can be in excess or in short supply. The enterprise supplies wastewater containing 3.31 kg of lead (II) nitrate to the water treatment site. Precipitation of lead cations is carried out by adding 1.27 kg of sodium carbonate to the solution. Determine which reagent and in what quantity is taken in excess. Calculate the mass of the PbCO3 precipitate formed. Solution: The precipitation reaction can be represented as follows- Pb(NO₃)₂ (aq) + Na₂CO₃ (aq) -- PbCO₃ (s) + 2 NaNO₃ (aq) The limiting reagent can be determined as follows- From the above reaction, 1 mol of Pb(NO₃)₂ reacts with 1 mol of Na₂CO₃ in order to form 1 mol of PbCO₃. So, 331.2 g of Pb(NO₃)₂ reacts with 106 g of Na₂CO₃ 1 g of Pb(NO₃)₂ reacts with 106 / 331.2 g of Na₂CO₃ 3310 g of Pb(NO₃)₂ reacts with (106 / 331.2) × 3310 g of Na₂CO₃ Hence, the amount of Na₂CO₃ actually required = 1059.36 g Amount of Na₂CO₃ used in the reaction = 1270 g So, the amount of Na₂CO₃ is in excess, this suggests that Pb(NO₃)₂ is the limiting reagent. Now, The amount of PbCO₃ formed can be calculated as follows- 331.2 g of Pb(NO₃)₂ reacts to give 267.21 g of PbCO₃ 1 g of Pb(NO₃)₂ gives 267.21 / 331.2 g of PbCO₃ So, 3310 g of Pb(NO₃)₂ gives ( 267.21 / 331.2) × 3310 g of PbCO₃ The amount of PbCO₃ precipitate formed = 2670.5 g = 2.67 kg |
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