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1aii) Strong electrolytes ionize completely (100%), while weak electrolytes ionize only partially
[9/2, 12:56 PM] Fatai: *1ai* Faraday’s first law of electrolysis states that the amount of element deposited at the electrodes is directly proportional to the quantity of electricity applied in the electrolysis
*1ai* Faraday’s first law of electrolysis states that the amount of element deposited at the electrodes is directly proportional to the quantity of electricity applied in the electrolysis
1aii) Strong electrolytes ionize completely (100%), while weak electrolytes ionize only partially
1ai) Faraday’s First Law of Electrolysis states that the chemical deposition due to the flow of current through an electrolyte is directly proportional to the quantity of electricity (coulombs) passed through it.
1aii) Strong electrolytes ionize completely (100%), while weak electrolytes ionize only partially
1b) Properties of Ethyne (choose one)
Ethyne is a colorless gas with a sweet smell when pure.
It is slightly less dense than air
It is almost insoluble in water.
It burns with a sooty flame due to high proportion of unburnt carbon.
1c) Unsaturated hydrocarbons are hydrocarbons that have double or triple covalent bonds between adjacent carbon atoms. ..
3. (a) Explain briefly the observations in each of the following processes:
(i) when carbon(IV) oxide is bubbled through lime water, it turns milky but the milkiness disappears when the gas is bubbled for a long time;
(ii) a precipitate of calcium hydroxide is insoluble in excess sodium hydroxide solution whereas that of lead.
(II) hydroxide is soluble.
(b) (i) What is a primary standard solution?
(ii) Calculate the mass of sodium trioxocarbonate(IV) required to prepare 250 cm3 of 0.15 moldm-3 solution.
[Na = 23.0; O = 16.0; C = 12.0]
(c) Name one gas that can be collected by:
(i) upward displacement of air;
(ii) downward displacement of air.
ANS: (a) (i): Insoluble CaCO3 formation is responsible for the milkness produce when CO2 is bubbled through lime water while the disappearance of milkness is due to the formation of soluble Ca(HCO3)2.
Lime water turns milky with CO2 because CaCO3/ CaCO3(s) is formed. Milkiness disappear when excess CO2 reacts with CaCO3 in water medium forming the soluble Ca(HCO3)2/ Ca(HCO3)2 (aq).
(ii) Calcium hydroxide is not amphoteric. Does not react with an alkali NaOH whereas lead (II) hydroxide is amphoteric so reacts with excess NaOH.
(b) (i) Primary standard solution is one whose concentration is known and can be used to standardize another solution.
(ii) M (Na2CO3) = 106 gmol-1
m(Na2CO3) = C x M x V
= 0. 15 x 106 x 0. 25
= 3. 98 g
(c) (i) Carbon(IV) oxide, sulphur (IV) oxide, hydrogen chloride, oxygen, nitrogen (IV) oxide, chlorine, hydrogen sulphide.
(ii) Ammonia, oxygen, hydrogen, methane.
[8/18, 7:53 AM] +234 906 354 8595: 1. A is 0.100 mol dm-3 solution of an acid. B is a solution KOH containing 2.8 g per 500 cm3.
(a) Put A into the burette and titrate it against 25.0 cm3 portions B using methyl orange as indicator. Repeat the titration to obtain consistent titres. Tabulate your readings and calculate the average volume of A used.
(b) From your results and the information provided above, calculate the:
(i) number of moles of acid in the average titre;
(ii) number of moles of KOH in the volume of B pipetted;
(iii) mole ratio of acid to base in the reaction
[H = 1.00, O = 16.0, K = 39.0]
[8/18, 7:56 AM] +234 906 354 8595: ANS: (a) titration answer soon.
.
(b) (i) number of moles of acid = 0.100 x VA
1000
= X mole(s) [3sig. Fig to score]
*SIR GNOMIC ✍️*
1000cm3 contains 0.100 mole(s)
VA will contain 0.100 x VA
1000
= X moles [3 Sig. Fig. to score]
.
(ii) Number of moles of KOH in B
500cm3 of B contains 2.8g of KOH
1000cm3 of B will contain 2.8 x 1000 = 5.6 KOH
500
Molar mass of KOH = 39 + 16 + 1 or 56 gmol-1
Conc of B = 5.6 = 0.100 mol dm-3
56
(iii) Mole ratio of acid to base = X: Y to nearest whole number ratio.
.
2. C is a mixture of two salts. Carry out the following exercises on C. Record your observations and identify any gas(es) evolved. State the conclusion drawn from the result of each test.
.
(a) Put all of C into a boiling tube and add about 5cm3 of distilled water. Stir thoroughly and filter. Keep both the residue and the filtrate.
.
(b) To about 2 cm3 of the filtrate, add few drops of Pb(NO3)2(aq)‑Boil the mixture and then allow to cool.
.
(c) (i) Put the residue in a test tube and add dilute HNO3. Shake the mixture and divide the solution into two portions.
.
(ii) To the first portion from (c)(i), add NaOH(aq) in drops and then in excess.
(iii) To the second portion from (c)(i), add aqueous ammonia in drops and thein in excess.
3725: ANS:
TEST
OBSERVATION
INFERENCE
a) C+water, mixt
ure stirred
and filtered
Partly dissolves/ soluble Colourless filtrate
White residue (do not accept ppt /solid)
C contains soluble and insoluble salts
b) Filtrate + Pb (NO3)2(aq). Boiled then cooled
White precipitate formed
Precipitate dissolves
Precipitate re-appears
Cl-, SO42- or CO32.
Cl- present
Cl- confirmed
c) (i) Residue + dil HNO3
Effervescence /bubbles /gas evolved colourless, odourless gas. Gas turns lime water milky.
CO2 evolved from CO32.
3. (a) Explain briefly the observations in each of the following processes:
(i) when carbon(IV) oxide is bubbled through lime water, it turns milky but the milkiness disappears when the gas is bubbled for a long time;
(ii) a precipitate of calcium hydroxide is insoluble in excess sodium hydroxide solution whereas that of lead.
(II) hydroxide is soluble.
(b) (i) What is a primary standard solution?
(ii) Calculate the mass of sodium trioxocarbonate(IV) required to prepare 250 cm3 of 0.15 moldm-3 solution.
[Na = 23.0; O = 16.0; C = 12.0]
(c) Name one gas that can be collected by:
(i) upward displacement of air;
(ii) downward displacement of air.
ANS: (a) (i): Insoluble CaCO3 formation is responsible for the milkness produce when CO2 is bubbled through lime water while the disappearance of milkness is due to the formation of soluble Ca(HCO3)2.
Lime water turns milky with CO2 because CaCO3/ CaCO3(s) is formed. Milkiness disappear when excess CO2 reacts with CaCO3 in water medium forming the soluble Ca(HCO3)2/ Ca(HCO3)2 (aq).
(ii) Calcium hydroxide is not amphoteric. Does not react with an alkali NaOH whereas lead (II) hydroxide is amphoteric so reacts with excess NaOH.
(b) (i) Primary standard solution is one whose concentration is known and can be used to standardize another solution.
(ii) M (Na2CO3) = 106 gmol-1
m(Na2CO3) = C x M x V
= 0. 15 x 106 x 0. 25
= 3. 98 g
(c) (i) Carbon(IV) oxide, sulphur (IV) oxide, hydrogen chloride, oxygen, nitrogen (IV) oxide, chlorine, hydrogen sulphide.
(ii) Ammonia, oxygen, hydrogen, methane.
[8/18, 7:53 AM] +234 906 354 8595: 1. A is 0.100 mol dm-3 solution of an acid. B is a solution KOH containing 2.8 g per 500 cm3.
(a) Put A into the burette and titrate it against 25.0 cm3 portions B using methyl orange as indicator. Repeat the titration to obtain consistent titres. Tabulate your readings and calculate the average volume of A used.
(b) From your results and the information provided above, calculate the:
(i) number of moles of acid in the average titre;
(ii) number of moles of KOH in the volume of B pipetted;
(iii) mole ratio of acid to base in the reaction
[H = 1.00, O = 16.0, K = 39.0]
[8/18, 7:56 AM] +234 906 354 8595: ANS: (a) titration answer soon.
.
(b) (i) number of moles of acid = 0.100 x VA
1000
= X mole(s) [3sig. Fig to score]
*SIR GNOMIC ✍️*
1000cm3 contains 0.100 mole(s)
VA will contain 0.100 x VA
1000
= X moles [3 Sig. Fig. to score]
.
(ii) Number of moles of KOH in B
500cm3 of B contains 2.8g of KOH
1000cm3 of B will contain 2.8 x 1000 = 5.6 KOH
500
Molar mass of KOH = 39 + 16 + 1 or 56 gmol-1
Conc of B = 5.6 = 0.100 mol dm-3
56
(iii) Mole ratio of acid to base = X: Y to nearest whole number ratio.
.
2. C is a mixture of two salts. Carry out the following exercises on C. Record your observations and identify any gas(es) evolved. State the conclusion drawn from the result of each test.
.
(a) Put all of C into a boiling tube and add about 5cm3 of distilled water. Stir thoroughly and filter. Keep both the residue and the filtrate.
.
(b) To about 2 cm3 of the filtrate, add few drops of Pb(NO3)2(aq)‑Boil the mixture and then allow to cool.
.
(c) (i) Put the residue in a test tube and add dilute HNO3. Shake the mixture and divide the solution into two portions.
.
(ii) To the first portion from (c)(i), add NaOH(aq) in drops and then in excess.
(iii) To the second portion from (c)(i), add aqueous ammonia in drops and thein in excess.
3725: ANS:
TEST
OBSERVATION
INFERENCE
a) C+water, mixt
I-Zac:
(1ai)
Faraday’s first law of electrolysis states that the amount of element deposited at the electrodes is directly proportional to the quantity of electricity applied in the electrolysis.
(1aii)
Strong electrolytes ionize completely (100%), while weak electrolytes ionize only partially.
(1b)
Properties of Ethyne
(Pick one)
(i) Ethyne is a colorless gas with a sweet smell when pure.
(ii) It is slightly less dense than air
(iii) It is almost insoluble in water.
(iv) It burns with a sooty flame due to high proportion of unburnt carbon.
(1c)
Unsaturated hydrocarbons are hydrocarbons that have double or triple covalent bonds between adjacent carbon atoms.
(1cii)
CH3COOH + CH3OH–>CH3COOCH3 + H2O
(1ciii)
methly ethanoate
(1d)
By removing the main product continuously
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