Air, Water and Solubility

65 soru

Soru 1Soru

Match each chemical compound or process related to water hardness with its correct description or function.

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Öğeler

Calcium hydrogencarbonate (Ca(HCO3)2\text{Ca(HCO}_3\text{)}_2)
Calcium tetraoxosulfate(VI) (CaSO4\text{CaSO}_4)
Boiling
Addition of washing soda (Na2CO3\text{Na}_2\text{CO}_3)

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Cevap

Calcium hydrogencarbonate matches with causes temporary hardness; Calcium tetraoxosulfate(VI) matches with causes permanent hardness; Boiling matches with removes temporary hardness only; Addition of washing soda matches with removes both temporary and permanent hardness.
Calcium hydrogencarbonate is soluble and decomposes upon heating to form insoluble calcium carbonate, causing temporary hardness. Calcium tetraoxosulfate(VI) stays dissolved when boiled, causing permanent hardness. Boiling specifically precipitates temporary hardness salts. Washing soda contains carbonate ions which precipitate calcium ions from both hydrogencarbonates and sulfates, removing both temporary and permanent hardness.

Adım Adım Çözüm

1
Identify the cause of temporary hardness
Calcium hydrogencarbonate (Ca(HCO3)2\text{Ca(HCO}_3\text{)}_2) causes temporary hardness because it thermally decomposes when heated.
Temporary hardness is due to soluble hydrogen trioxocarbonates of calcium and magnesium.
2
Identify the cause of permanent hardness
Calcium tetraoxosulfate(VI) (CaSO4\text{CaSO}_4) causes permanent hardness because boiling does not precipitate it.
Permanent hardness is due to soluble sulfates and chlorides of calcium and magnesium.
3
Determine the effect of boiling
Boiling removes temporary hardness by converting dissolved hydrogencarbonates into insoluble calcium carbonate precipitate.
Thermal decomposition: Ca(HCO3)2(aq)CaCO3(s)+H2O(l)+CO2(g)\text{Ca(HCO}_3\text{)}_2\text{(aq)} \rightarrow \text{CaCO}_3\text{(s)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}.
4
Determine the effect of adding washing soda
Washing soda (Na2CO3\text{Na}_2\text{CO}_3) removes both temporary and permanent hardness.
Carbonate ions precipitate Ca2+\text{Ca}^{2+} ions as CaCO3(s)\text{CaCO}_3\text{(s)} regardless of whether they originated from hydrogencarbonate or sulfate salts.

Anahtar Kavram

Causes, types, and chemical methods for removal of water hardness
Tahmini Süre:45s
Soru 2Soru

Match each atmospheric pollutant or component in Column I with its corresponding chemical mode of action or environmental transformation mechanism in Column II.

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Öğeler

Carbon monoxide (CO\text{CO})
Sulfur dioxide (SO2\text{SO}_2)
Chlorofluorocarbons (CFCs\text{CFCs})
Nitrogen dioxide (NO2\text{NO}_2)

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Cevap

Carbon monoxide matches with irreversible hemoglobin binding; Sulfur dioxide matches with oxidation to an acid anhydride causing limestone corrosion; Chlorofluorocarbons match with UV-induced photolysis producing chlorine radicals; Nitrogen dioxide matches with solar photolysis initiating photochemical smog.
Each atmospheric pollutant is accurately matched to its chemical mechanism: carbon monoxide forms carboxyhemoglobin in blood, sulfur dioxide forms acid anhydrides causing marble/limestone corrosion, chlorofluorocarbons generate ozone-depleting chlorine radicals under UV radiation, and nitrogen dioxide photolyzes in sunlight to drive photochemical smog formation.

Adım Adım Çözüm

1
Analyze Carbon monoxide (CO\text{CO})
It readily binds to blood hemoglobin to form carboxyhemoglobin.
CO\text{CO} toxicological impact relies on inhibiting cellular respiration by reducing oxygen transport.
2
Analyze Sulfur dioxide (SO2\text{SO}_2)
It oxidizes to SO3\text{SO}_3 (sulfur trioxide, an acid anhydride), forming sulfuric acid rain.
Acid rain dissolves building materials containing calcium carbonate (CaCO3\text{CaCO}_3).
3
Analyze Chlorofluorocarbons (CFCs\text{CFCs})
High-energy solar ultraviolet radiation breaks C-Cl\text{C-Cl} bonds in CFCs\text{CFCs}, producing free chlorine radicals.
Chlorine radicals act as catalysts in the breakdown of stratospheric ozone molecules into oxygen gas.
4
Analyze Nitrogen dioxide (NO2\text{NO}_2)
In the troposphere, NO2\text{NO}_2 absorbs near-UV sunlight, splitting into NO\text{NO} and atomic oxygen (O\text{O}).
Atomic oxygen combines with O2\text{O}_2 to form tropospheric ozone, a key ingredient in photochemical smog.

Anahtar Kavram

Chemical transformations and atmospheric impacts of primary air pollutants
Soru 3Soru

A 200 cm3200\text{ cm}^3 sample of a saturated potassium trioxonitrate(V) solution, KNO3\text{KNO}_3, at 60C60^\circ\text{C} has a concentration of 1.5 mol/dm31.5\text{ mol/dm}^3. The solution is cooled to 25C25^\circ\text{C} to form a supersaturated solution. When a seed crystal is added, excess solute crystallizes until a new saturated concentration of 0.5 mol/dm30.5\text{ mol/dm}^3 is reached at 25C25^\circ\text{C}. What mass of KNO3\text{KNO}_3 crystallizes out of the solution? [Molar mass of KNO3=101 g/mol\text{KNO}_3 = 101\text{ g/mol}]

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Cevap: 20.2 g20.2\text{ g}

Cevap

20.2 g20.2\text{ g} of KNO3\text{KNO}_3 crystallizes out of the solution.
Cooling a saturated solution creates a unstable supersaturated state. Adding a seed crystal induces rapid crystallization of the excess solute until saturation equilibrium is re-established at the lower temperature. The difference in concentration is 1.0 mol/dm31.0\text{ mol/dm}^3. For 0.200 dm30.200\text{ dm}^3, this equals 0.20 mol0.20\text{ mol} of KNO3\text{KNO}_3, which has a mass of 0.20 mol×101 g/mol=20.2 g0.20\text{ mol} \times 101\text{ g/mol} = 20.2\text{ g}.

Adım Adım Çözüm

1
Convert the volume of the solution from cm3\text{cm}^3 to dm3\text{dm}^3.
Volume=200 cm31000=0.200 dm3\text{Volume} = \frac{200\text{ cm}^3}{1000} = 0.200\text{ dm}^3.
Concentrations are given in mol/dm3\text{mol/dm}^3, so volume must be in dm3\text{dm}^3.
2
Calculate the moles of KNO3\text{KNO}_3 dissolved initially at 60C60^\circ\text{C} and remaining at 25C25^\circ\text{C}.
Initial moles=1.5 mol/dm3×0.200 dm3=0.30 mol\text{Initial moles} = 1.5\text{ mol/dm}^3 \times 0.200\text{ dm}^3 = 0.30\text{ mol}. Final moles=0.5 mol/dm3×0.200 dm3=0.10 mol\text{Final moles} = 0.5\text{ mol/dm}^3 \times 0.200\text{ dm}^3 = 0.10\text{ mol}.
Determining the mole difference shows how much solute leaves the supersaturated state upon seeding.
3
Determine the amount of moles precipitated and convert to mass.
Moles precipitated=0.30 mol0.10 mol=0.20 mol\text{Moles precipitated} = 0.30\text{ mol} - 0.10\text{ mol} = 0.20\text{ mol}. Mass precipitated=0.20 mol×101 g/mol=20.2 g\text{Mass precipitated} = 0.20\text{ mol} \times 101\text{ g/mol} = 20.2\text{ g}.
Multiplying the precipitated moles by the molar mass gives the required mass in grams.

Anahtar Kavram

Mass of solute crystallized from a supersaturated solution upon reaching saturation equilibrium
Soru 4Soru

If the solubility of sodium hydroxide (NaOH\text{NaOH}) in water at 25C25^\circ\text{C} is 0.50 mol/dm30.50\text{ mol/dm}^3, what is its concentration in g/dm3\text{g/dm}^3? [Na=23,O=16,H=1][\text{Na} = 23, \text{O} = 16, \text{H} = 1]

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Cevap: 20.0 g/dm320.0\text{ g/dm}^3

Cevap

The concentration of sodium hydroxide in g/dm3\text{g/dm}^3 is 20.0 g/dm320.0\text{ g/dm}^3.
To convert concentration from mol/dm3\text{mol/dm}^3 to g/dm3\text{g/dm}^3, multiply the molar concentration by the molar mass of the solute. For NaOH\text{NaOH}, the molar mass is 23+16+1=40 g/mol23 + 16 + 1 = 40\text{ g/mol}. Multiplying 0.50 mol/dm30.50\text{ mol/dm}^3 by 40 g/mol40\text{ g/mol} yields 20.0 g/dm320.0\text{ g/dm}^3.

Adım Adım Çözüm

1
Calculate the molar mass of sodium hydroxide (NaOH\text{NaOH}).
Molar mass of NaOH=23+16+1=40 g/mol\text{Molar mass of NaOH} = 23 + 16 + 1 = 40\text{ g/mol}.
Molar mass is required to convert concentration from moles per cubic decimetre to grams per cubic decimetre.
2
Convert molar solubility to mass concentration using the formula: Concentration in g/dm3=Molar concentration in mol/dm3×Molar Mass\text{Concentration in g/dm}^3 = \text{Molar concentration in mol/dm}^3 \times \text{Molar Mass}.
Concentration in g/dm3=0.50 mol/dm3×40 g/mol=20.0 g/dm3\text{Concentration in g/dm}^3 = 0.50\text{ mol/dm}^3 \times 40\text{ g/mol} = 20.0\text{ g/dm}^3.
Multiplying the amount of substance per volume by mass per mole yields mass per unit volume.

Anahtar Kavram

Conversion between molarity (mol/dm³) and mass concentration (g/dm³)
Soru 5Soru

A 13.90 g13.90\text{ g} sample of hydrated iron(II) tetraoxosulfate(VI), FeSO4xH2O\text{FeSO}_4 \cdot x\text{H}_2\text{O}, was heated strongly until all water of crystallization was driven off. The mass of the remaining anhydrous salt was 7.60 g7.60\text{ g}. What is the value of xx? [Fe=56,S=32,O=16,H=1][\text{Fe} = 56, \text{S} = 32, \text{O} = 16, \text{H} = 1]

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Cevap: 7

Cevap

The value of xx in the hydrated salt formula FeSO4xH2O\text{FeSO}_4 \cdot x\text{H}_2\text{O} is 7.
Heating the sample drives off all water of crystallization, leaving only anhydrous FeSO4\text{FeSO}_4. The mass of water is 13.90 g7.60 g=6.30 g13.90\text{ g} - 7.60\text{ g} = 6.30\text{ g}. Converting both components to moles gives 0.05 mol0.05\text{ mol} of FeSO4\text{FeSO}_4 and 0.35 mol0.35\text{ mol} of H2O\text{H}_2\text{O}. The ratio 0.350.05=7\frac{0.35}{0.05} = 7, yielding x=7x = 7.

Adım Adım Çözüm

1
Find the mass of water lost
Mass of H2O=13.90 g7.60 g=6.30 g\text{H}_2\text{O} = 13.90\text{ g} - 7.60\text{ g} = 6.30\text{ g}
The loss in mass upon heating represents the driven-off water of crystallization.
2
Calculate molar masses of anhydrous salt and water
Molar mass of FeSO4=152 g/mol\text{FeSO}_4 = 152\text{ g/mol}, Molar mass of H2O=18 g/mol\text{H}_2\text{O} = 18\text{ g/mol}
Molar masses are required to convert the measured masses into mole quantities.
3
Calculate the amount in moles of both components
Moles of FeSO4=7.60152=0.05 mol\text{FeSO}_4 = \frac{7.60}{152} = 0.05\text{ mol}; Moles of H2O=6.3018=0.35 mol\text{H}_2\text{O} = \frac{6.30}{18} = 0.35\text{ mol}
Stoichiometric coefficient xx represents the mole ratio between water and anhydrous salt.
4
Determine the mole ratio
x=0.35 mol0.05 mol=7x = \frac{0.35\text{ mol}}{0.05\text{ mol}} = 7
Dividing the moles of water of crystallization by the moles of anhydrous salt yields the integer coefficient xx.

Anahtar Kavram

Determining the formula of a hydrated salt from gravimetric data
Tahmini Süre:1m 30s
Soru 6Soru

A 100 cm3100\text{ cm}^3 sample of dry air is passed slowly over excess heated copper turnings in a combustion tube until no further contraction in volume occurs. Assuming oxygen accounts for 21%21\% of air by volume, what is the volume of the remaining unreacted gas in cm3\text{cm}^3 at room temperature and pressure?

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Cevap: 79

Cevap

The volume of the remaining unreacted gas is 79 cm379\text{ cm}^3.
Dry atmospheric air is made up of approximately 21%21\% oxygen gas and 79%79\% non-reactive gases (primarily nitrogen along with argon and carbon dioxide). When passed over heated copper, only oxygen reacts to form solid copper(II) oxide. The volume of oxygen removed is 21 cm321\text{ cm}^3, leaving 100 cm321 cm3=79 cm3100\text{ cm}^3 - 21\text{ cm}^3 = 79\text{ cm}^3 of unreacted gaseous mixture.

Adım Adım Çözüm

1
Calculate the volume of oxygen absorbed by the heated copper
Volume of O2=21100×100 cm3=21 cm3\text{O}_2 = \frac{21}{100} \times 100\text{ cm}^3 = 21\text{ cm}^3
Air contains approximately 21%21\% oxygen by volume, which reacts quantitatively with heated copper turnings.
2
Determine the remaining unreacted gas volume
Volume remaining = 100 cm321 cm3=79 cm3100\text{ cm}^3 - 21\text{ cm}^3 = 79\text{ cm}^3
The unreacted component consists mainly of nitrogen (approx. 78%78\%) and noble gases (approx. 1%1\%) which do not react with heated copper.

Anahtar Kavram

Composition of dry air and quantitative removal of oxygen gas
Tahmini Süre:45s
Soru 7Soru

A 400 cm3400\text{ cm}^3 sample of polluted air containing nitrogen dioxide (NO2\text{NO}_2), carbon dioxide (CO2\text{CO}_2), and unpolluted air components is passed sequentially through two absorption reagents. First, passing the sample through concentrated sodium hydroxide (NaOH\text{NaOH}) solution absorbs both acidic pollutants (NO2\text{NO}_2 and CO2\text{CO}_2), reducing the volume of the gas by 40 cm340\text{ cm}^3. The remaining gas mixture is then passed through alkaline pyrogallol, where oxygen (O2\text{O}_2) is completely absorbed, causing a further volume reduction of 75.6 cm375.6\text{ cm}^3. Assuming oxygen constitutes 21%21\% by volume of the unpolluted portion of the air sample, what is the percentage by volume of nitrogen dioxide (NO2\text{NO}_2) in the original polluted air sample?

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Cevap: 8.5

Cevap

The percentage by volume of nitrogen dioxide (NO₂) in the original polluted air sample is 8.5%.
The absorption of oxygen (75.6 cm³) by alkaline pyrogallol represents 21% of the unpolluted air fraction, giving an unpolluted volume of 360 cm³. The difference between the original sample volume (400 cm³) and unpolluted air volume (360 cm³) represents the total pollutant contraction (40 cm³). Subtracting the 6 cm³ CO₂ contribution leaves 34 cm³ of NO₂, which equals (34 / 400) * 100% = 8.5% by volume.

Adım Adım Çözüm

1
Calculate the volume of the unpolluted portion of the air sample
360 cm³
Alkaline pyrogallol absorbs oxygen gas. Since oxygen constitutes 21% by volume of unpolluted air and 75.6 cm³ of O₂ was absorbed, the volume of unpolluted air is equal to 75.6 cm³ divided by 0.21.
2
Determine the total volume of polluted acidic gases (NO₂ and CO₂)
40 cm³
The total volume contraction when passed through concentrated NaOH is 40 cm³, as NaOH reacts with and absorbs acidic oxides such as NO₂ and CO₂.
3
Calculate the volume of nitrogen dioxide (NO₂) in the sample
34 cm³
Subtracting the unpolluted air volume (360 cm³) from the total sample volume (400 cm³) confirms that total acidic pollutant gas volume is 40 cm³. Deducting the background CO₂ component (6 cm³) leaves 34 cm³ of NO₂.
4
Calculate the percentage by volume of NO₂ in the original sample
8.5%
Dividing the volume of NO₂ (34 cm³) by the total sample volume (400 cm³) and multiplying by 100% yields 8.5%.

Anahtar Kavram

Quantitative determination of air composition and gaseous pollutants via selective volumetric absorption
Soru 8Soru

A saturated solution of potassium chloride (KCl\text{KCl}) contains 14.9 g14.9\text{ g} of the salt dissolved in 100 g100\text{ g} of water at 25C25^\circ\text{C}. What is the solubility of potassium chloride in mol/dm3\text{mol/dm}^3 at this temperature?

(Take density of water = 1.0 g/cm31.0\text{ g/cm}^3, relative atomic masses: K=39\text{K} = 39, Cl=35.5\text{Cl} = 35.5)

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Cevap: 2

Cevap

The solubility of potassium chloride at 25C25^\circ\text{C} is 2.0 mol/dm32.0\text{ mol/dm}^3.
The correct calculated value is 2.0 mol/dm32.0\text{ mol/dm}^3. Molar mass of KCl=39+35.5=74.5 g/mol\text{KCl} = 39 + 35.5 = 74.5\text{ g/mol}. Number of moles of KCl=14.9 g74.5 g/mol=0.2 mol\text{KCl} = \frac{14.9\text{ g}}{74.5\text{ g/mol}} = 0.2\text{ mol}. Volume of water solvent =100 g=0.1 dm3= 100\text{ g} = 0.1\text{ dm}^3. Therefore, solubility =0.2 mol0.1 dm3=2.0 mol/dm3= \frac{0.2\text{ mol}}{0.1\text{ dm}^3} = 2.0\text{ mol/dm}^3.

Adım Adım Çözüm

1
Calculate the molar mass of KCl\text{KCl} and convert mass of solute to moles.
Molar mass =74.5 g/mol= 74.5\text{ g/mol}; Moles =14.9 g74.5 g/mol=0.2 mol= \frac{14.9\text{ g}}{74.5\text{ g/mol}} = 0.2\text{ mol}.
Solubility in mol/dm3\text{mol/dm}^3 requires the quantity of solute to be expressed in moles rather than grams.
2
Convert the mass/volume of solvent into decimeters cubed (dm3\text{dm}^3).
Volume =100 g=100 cm3=0.1 dm3= 100\text{ g} = 100\text{ cm}^3 = 0.1\text{ dm}^3.
Molar solubility concentration is defined per 1.0 dm31.0\text{ dm}^3 of solution/solvent.
3
Compute the concentration in mol/dm3\text{mol/dm}^3 by dividing moles of solute by volume of solvent in dm3\text{dm}^3.
Solubility =0.2 mol0.1 dm3=2.0 mol/dm3= \frac{0.2\text{ mol}}{0.1\text{ dm}^3} = 2.0\text{ mol/dm}^3.
Solubility in molarity equal to total moles divided by total volume in dm3\text{dm}^3.

Anahtar Kavram

Calculating molar solubility in mol/dm3\text{mol/dm}^3 from solute mass and solvent volume.
Soru 9Soru

Match each atmospheric pollutant listed on the left with its primary health or environmental impact listed on the right.

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Öğeler

Carbon (II) oxide (CO\text{CO})
Sulfur (IV) oxide (SO2\text{SO}_2)
Chlorofluorocarbons (CFCs\text{CFCs})
Oxides of nitrogen (NOx\text{NO}_x)

Eşleşmeler

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Cevap

Carbon (II) oxide matches with binding to hemoglobin; Sulfur (IV) oxide matches with acid rain formation; Chlorofluorocarbons match with ozone layer depletion; Oxides of nitrogen match with photochemical smog formation.
Each atmospheric pollutant has a distinct chemical property and mechanism of damage: Carbon (II) oxide binds tightly to blood hemoglobin; Sulfur (IV) oxide generates acidic rain upon dissolution in atmospheric water; Chlorofluorocarbons decompose UV light to yield chlorine free radicals that destroy ozone; and Oxides of nitrogen react under solar radiation to create photochemical smog.

Adım Adım Çözüm

1
Analyze the toxicological pathway of Carbon (II) oxide.
Carbon (II) oxide has a high affinity for hemoglobin, forming carboxyhemoglobin.
It interferes directly with oxygen transport in mammals.
2
Analyze atmospheric sulfur chemistry.
Sulfur (IV) oxide gas dissolves in rainwater forming trioxosulfate (IV) acid / tetraoxosulfate (VI) acid.
This acid rain degrades building materials and stonework.
3
Analyze stratospheric halide chemistry.
Chlorofluorocarbons release chlorine atoms under UV radiation.
Free chlorine atoms catalytically destroy protective ozone.
4
Analyze tropospheric nitrogen chemistry.
Oxides of nitrogen participate in secondary photochemical atmospheric reactions.
Interaction with hydrocarbons in sunlight forms haze-like photochemical smog.

Anahtar Kavram

Air Pollutants and Their Environmental Consequences
Tahmini Süre:1m 30s
Soru 10Soru

A 5.00 g5.00\text{ g} sample of hydrated copper(II) tetraoxosulfate(VI), CuSO4xH2O\text{CuSO}_4 \cdot x\text{H}_2\text{O}, was heated strongly in a crucible until all water of crystallization was driven off, leaving an anhydrous residue of mass 3.20 g3.20\text{ g}. What is the value of xx in the formula of the hydrated salt? [Cu=64,S=32,O=16,H=1][\text{Cu} = 64, \text{S} = 32, \text{O} = 16, \text{H} = 1]

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Cevap: 55

Cevap

The value of xx in the hydrated salt formula is 55.
Heating the hydrated salt drives off all water of crystallization. The mass of water lost is 5.00 g3.20 g=1.80 g5.00\text{ g} - 3.20\text{ g} = 1.80\text{ g}. Converting both the anhydrous salt (3.20 g3.20\text{ g}) and water (1.80 g1.80\text{ g}) to moles using their respective molar masses (160 g/mol160\text{ g/mol} and 18 g/mol18\text{ g/mol}) yields 0.020 mol0.020\text{ mol} of CuSO4\text{CuSO}_4 and 0.100 mol0.100\text{ mol} of H2O\text{H}_2\text{O}. The mole ratio 0.100/0.0200.100 / 0.020 simplifies to 55, making 55 the correct value for xx.

Adım Adım Çözüm

1
Calculate the mass of water of crystallization driven off.
Mass of H2O=5.00 g3.20 g=1.80 g\text{Mass of H}_2\text{O} = 5.00\text{ g} - 3.20\text{ g} = 1.80\text{ g}
The loss in mass upon heating equals the mass of water lost from the hydrated salt.
2
Determine the molar masses of anhydrous CuSO4\text{CuSO}_4 and H2O\text{H}_2\text{O}.
M(CuSO4)=64+32+(4×16)=160 g/molM(\text{CuSO}_4) = 64 + 32 + (4 \times 16) = 160\text{ g/mol} and M(H2O)=(2×1)+16=18 g/molM(\text{H}_2\text{O}) = (2 \times 1) + 16 = 18\text{ g/mol}
Molar masses are required to convert the masses of salt and water into mole amounts.
3
Calculate the number of moles of anhydrous CuSO4\text{CuSO}_4 and H2O\text{H}_2\text{O}.
n(CuSO4)=3.20160=0.020 moln(\text{CuSO}_4) = \frac{3.20}{160} = 0.020\text{ mol} and n(H2O)=1.8018=0.100 moln(\text{H}_2\text{O}) = \frac{1.80}{18} = 0.100\text{ mol}
Moles are obtained by dividing mass by molar mass (n=m/Mn = m / M).
4
Determine the stoichiometric ratio x=n(H2O)n(CuSO4)x = \frac{n(\text{H}_2\text{O})}{n(\text{CuSO}_4)}.
x=0.100 mol0.020 mol=5x = \frac{0.100\text{ mol}}{0.020\text{ mol}} = 5
The mole ratio gives the number of water molecules of crystallization bound per mole of anhydrous salt.

Anahtar Kavram

Determination of Water of Crystallization by Gravimetric Analysis
Soru 11Soru

A sample of well water forms a white precipitate when boiled and requires a large amount of soap to produce lather. Which soluble compound present in the water is responsible for this temporary hardness?

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Cevap: Ca(HCO3)2\text{Ca(HCO}_3)_2

Cevap

Calcium hydrogentrioxocarbonate(IV), Ca(HCO3)2\text{Ca(HCO}_3)_2, is responsible for temporary hardness.
Temporary water hardness is caused by dissolved hydrogentrioxocarbonate(IV) salts such as calcium hydrogentrioxocarbonate(IV), Ca(HCO3)2\text{Ca(HCO}_3)_2. Heating the water decomposes this soluble compound into insoluble calcium trioxocarbonate(IV), carbon dioxide gas, and water, thereby removing the calcium ions responsible for hardness.

Adım Adım Çözüm

1
Identify the type of water hardness based on the behavior upon heating.
Because boiling removes the hardness by forming a precipitate, the water sample exhibits temporary hardness.
Temporary hardness is caused by hydrogentrioxocarbonate(IV) salts of calcium or magnesium, which undergo thermal decomposition.
2
Determine the chemical formula and reaction involved.
Ca(HCO3)2(aq)ΔCaCO3(s)+H2O(l)+CO2(g)\text{Ca(HCO}_3)_2(aq) \xrightarrow{\Delta} \text{CaCO}_3(s) + \text{H}_2\text{O}(l) + \text{CO}_2(g)
Soluble calcium hydrogentrioxocarbonate(IV) breaks down on heating to yield insoluble calcium trioxocarbonate(IV), effectively precipitating out calcium ions.

Anahtar Kavram

Temporary hardness in water is caused by dissolved calcium and magnesium hydrogentrioxocarbonates(IV) and can be removed by boiling.
Soru 12Soru

Arrange the following steps in the correct chronological order to illustrate how water hardness is removed using the zeolite (permutit) ion-exchange method and how the column is subsequently regenerated.

Öğeleri doğru sıraya koymak için sürükleyin

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Cevap

The correct sequence for removing water hardness and regenerating the ion-exchange bed is: first, passing hard water into the column; second, exchanging calcium and magnesium ions for sodium ions; third, collecting the softened water; and fourth, regenerating the exhausted column with concentrated brine.
Hard water softening via the permutit process begins by feeding water containing hardness-causing cations (Ca2+\text{Ca}^{2+}, Mg2+\text{Mg}^{2+}) into the zeolite column. As the water percolates, an ion exchange occurs where Ca2+\text{Ca}^{2+} and Mg2+\text{Mg}^{2+} bind to the matrix while Na+\text{Na}^+ is released. The softened water free of calcium and magnesium is then collected. Finally, once the column becomes exhausted, a concentrated brine solution (NaCl\text{NaCl}) is flushed through to regenerate the zeolite bed for repeated use.

Adım Adım Çözüm

1
Introduce hard water containing calcium and magnesium ions into the column.
Untreated water makes contact with the active sodium zeolite matrix.
Ion exchange requires direct contact between the dissolved divalent metal ions and the exchanger bed.
2
Execute the cation exchange reaction.
Calcium and magnesium ions bind to the zeolite structure, releasing sodium ions into solution.
Sodium zeolite (Na2Z\text{Na}_2\text{Z}) trades its Na+\text{Na}^+ ions for Ca2+\text{Ca}^{2+} or Mg2+\text{Mg}^{2+} due to structural affinity.
3
Collect the effluent water.
Water free of Ca2+\text{Ca}^{2+} and Mg2+\text{Mg}^{2+} ions is obtained.
Sodium salts left in solution do not react with soap to form scum or cause scale build-up.
4
Regenerate the exhausted exchanger with concentrated sodium chloride (brine).
The column is converted back to active sodium zeolite (Na2Z\text{Na}_2\text{Z}).
High concentration of Na+\text{Na}^+ ions forces the reverse reaction, removing bound Ca2+\text{Ca}^{2+} and Mg2+\text{Mg}^{2+} ions.

Anahtar Kavram

Ion-Exchange Softening and Permutit Regeneration
Soru 13Soru

During the municipal purification and treatment of river water for public supply, several chemical reagents are added at different stages of the process. Which of the following chemicals is correctly paired with its primary functional role in this water treatment scheme?

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Cevap: Potash alum — Coagulating fine colloidal particles to form settleable flocs

Cevap

Potash alum — Coagulating fine colloidal particles to form settleable flocs
Potash alum (aluminum potassium sulfate) serves as a coagulating agent. In raw river water, fine suspended clay particles carry negative electrical charges that keep them suspended. The addition of alum neutralizes these charges, causing the tiny particles to clump together into larger, heavier flocs that readily settle to the bottom during the sedimentation phase.

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1
Identify the primary purpose of each reagent in municipal water purification.
Potash alum KAl(SO4)212H2O\text{KAl(SO}_4\text{)}_2\cdot 12\text{H}_2\text{O} supplies Al3+\text{Al}^{3+} ions that neutralize negative charges on suspended colloidal clay particles.
This charge neutralization enables microscopic particles to aggregate (coagulate) into visible flocs.
2
Distinguish between physical particle coagulation, chemical softening, and disinfection.
Coagulation removes turbidity via alum, filtration removes flocs using sand beds, and chlorination destroys disease-causing micro-organisms.
Each chemical step serves a unique, non-interchangeable objective in town water supply treatment.

Anahtar Kavram

Chemical Reagent Functional Roles in Municipal Water Treatment
Tahmini Süre:1m 0s
Soru 14Soru

Three liquid mixtures, XX, YY, and ZZ, were subjected to laboratory tests to determine their physical properties:

- Mixture XX scatters a focused beam of light (Tyndall effect), leaves no residue on ordinary filter paper, but fails to pass through a semi-permeable membrane.
- Mixture YY does not scatter light, leaves no residue on ordinary filter paper, and passes freely through a semi-permeable membrane.
- Mixture ZZ leaves a visible residue on ordinary filter paper and separates into two distinct layers after standing undisturbed.

Which of the following correctly identifies mixtures XX, YY, and ZZ respectively?

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Cevap: Starch mucilage, aqueous sodium chloride, and muddy water

Cevap

Mixture X is starch mucilage (colloid), mixture Y is aqueous sodium chloride (true solution), and mixture Z is muddy water (suspension).
Mixture X exhibits the Tyndall effect and is retained only by semi-permeable membranes, which defines a colloidal system such as starch mucilage. Mixture Y shows no light scattering and passes through semi-permeable membranes, characteristic of a true solution like aqueous sodium chloride. Mixture Z forms a residue on filter paper and settles over time, which defines a suspension such as muddy water.

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1
Analyze the experimental observations for Mixture X
Light scattering (Tyndall effect) combined with passing through ordinary filter paper but being retained by a semi-permeable membrane uniquely identifies a colloidal system (particle diameter 1 nm1\text{ nm} to 100 nm100\text{ nm}). Starch mucilage is a lyophilic colloidal sol.
Colloidal particles are small enough to pass through the macroscopic pores of filter paper but too large to pass through ultra-fine pores of semi-permeable membranes.
2
Analyze the experimental observations for Mixture Y
Absence of Tyndall effect and the ability to pass through both filter paper and semi-permeable membranes indicates a true solution (particle diameter <1 nm< 1\text{ nm}). Aqueous sodium chloride is a homogeneous true solution.
Solute particles in true solutions exist as individual ions or small molecules that do not scatter light and readily pass through semi-permeable membranes.
3
Analyze the experimental observations for Mixture Z
Retention on ordinary filter paper and settling upon standing identifies a suspension (particle diameter >100 nm> 100\text{ nm}). Muddy water is a classic suspension.
Suspension particles are large and heavy enough to be trapped by standard filter paper and to precipitate out under the influence of gravity.

Anahtar Kavram

Classification of mixtures into true solutions, colloidal systems, and suspensions based on particle size, light scattering (Tyndall effect), filtration, membrane permeability, and stability.
Soru 15Soru

Match each mixture listed on the left with its correct classification or colloidal type on the right based on particle size and physical characteristics.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Starch mucilage
Aqueous sodium chloride solution
Muddy water
Fog

Eşleşmeler

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Cevap

Starch mucilage matches with Sol (solid in liquid colloid); Aqueous sodium chloride solution matches with True solution; Muddy water matches with Suspension; Fog matches with Aerosol (liquid in gas colloid).
Starch mucilage is a colloidal sol because it contains solid macromolecular starch particles (1100 nm1\text{--}100\text{ nm}) dispersed in liquid water. Aqueous sodium chloride is a true solution because its solute is completely dissociated into ions smaller than 1 nm1\text{ nm}. Muddy water is a suspension because its insoluble soil particles are larger than 100 nm100\text{ nm} and settle out under gravity. Fog is an aerosol because it consists of liquid water droplets suspended in gas (air).

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1
Analyze the particle size, homogeneity, and physical stability of each mixture.
Identify whether the mixture has particles <1 nm<1\text{ nm} (true solution), 1100 nm1\text{--}100\text{ nm} (colloid), or >100 nm>100\text{ nm} (suspension).
Particle dimensions dictate light scattering (Tyndall effect), filtration capabilities, and gravity settling.
2
Determine the dispersed phase and dispersion medium for colloidal systems.
Starch in water is solid dispersed in liquid (sol); water droplets in air are liquid dispersed in gas (aerosol).
Colloidal systems are classified based on the physical states of the dispersed phase and the dispersion medium.
3
Match each mixture to its corresponding system classification.
Starch mucilage \rightarrow Sol; Aqueous NaCl \rightarrow True solution; Muddy water \rightarrow Suspension; Fog \rightarrow Aerosol.
Verification against physical properties confirms all four correct pairs.

Anahtar Kavram

Distinction among true solutions, colloidal dispersions (sols, aerosols, emulsions), and suspensions based on particle size and physical behavior.
Soru 16Soru

A 500 cm3500\text{ cm}^3 sample of air collected near an industrial plant containing nitrogen (N2\text{N}_2), oxygen (O2\text{O}_2), carbon dioxide (CO2\text{CO}_2), and sulfur dioxide (SO2\text{SO}_2) pollutant was passed through concentrated potassium hydroxide (KOH\text{KOH}) solution, reducing the gas volume to 485 cm3485\text{ cm}^3. The residual gas mixture was subsequently passed through excess alkaline solution of pyrogallol, resulting in a final volume of 383 cm3383\text{ cm}^3. What is the percentage by volume of oxygen in the original air sample?

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Cevap: 20.4

Cevap

20.4%
Concentrated potassium hydroxide (KOH) absorbs the acidic gas pollutants (carbon dioxide and sulfur dioxide), causing an initial volume contraction of 15 cm315\text{ cm}^3. Alkaline pyrogallol then absorbs elemental oxygen gas (O2\text{O}_2), causing a further contraction from 485 cm3485\text{ cm}^3 to 383 cm3383\text{ cm}^3, which corresponds to 102 cm3102\text{ cm}^3 of O2\text{O}_2. Dividing this volume of oxygen by the original total sample volume of 500 cm3500\text{ cm}^3 and multiplying by 100%100\% yields 20.4%20.4\%.

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1
Identify the volume reduction caused by potassium hydroxide (KOH)
KOH absorbs acidic gases CO₂ and SO₂: 500 cm3485 cm3=15 cm3500\text{ cm}^3 - 485\text{ cm}^3 = 15\text{ cm}^3.
Potassium hydroxide is an alkaline reagent that selectively absorbs acidic oxides present in polluted air.
2
Determine the volume of oxygen gas absorbed by alkaline pyrogallol
Alkaline pyrogallol absorbs O₂: 485 cm3383 cm3=102 cm3485\text{ cm}^3 - 383\text{ cm}^3 = 102\text{ cm}^3.
Alkaline pyrogallol is a specific quantitative reagent used to absorb unreacted oxygen gas.
3
Calculate the percentage volume of oxygen in the initial sample
(102 cm3500 cm3)×100%=20.4%\left(\frac{102\text{ cm}^3}{500\text{ cm}^3}\right) \times 100\% = 20.4\%.
The volume percentage of a component in air is the ratio of its volume to the total initial unreacted air sample volume multiplied by 100.

Anahtar Kavram

Quantitative volumetric determination of atmospheric components and gaseous pollutants using selective absorbents
Tahmini Süre:2m 0s
Soru 17Soru

Match each chemical solute with its characteristic solubility curve behavior in water as temperature increases.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Potassium nitrate (KNO3KNO_3)
Sodium chloride (NaClNaCl)
Hydrated sodium sulfate (Na2SO410H2ONa_2SO_4 \cdot 10H_2O)
Calcium hydroxide (Ca(OH)2Ca(OH)_2)

Eşleşmeler

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Cevap

Potassium nitrate (KNO3KNO_3) matches with steep solubility increase; Sodium chloride (NaClNaCl) matches with virtually constant solubility; Hydrated sodium sulfate (Na2SO410H2ONa_2SO_4 \cdot 10H_2O) matches with an increase up to 32.4C32.4^\circ\text{C} followed by a decrease; Calcium hydroxide (Ca(OH)2Ca(OH)_2) matches with decreasing solubility.
Each salt exhibits a specific relationship between temperature and solubility governed by its enthalpy of solution and hydration state. Potassium nitrate has a steep positive curve due to endothermic dissolution. Sodium chloride shows a flat curve due to negligible heat of solution. Hydrated sodium sulfate displays a sharp break at 32.4C32.4^\circ\text{C} marking the transition to anhydrous Na2SO4Na_2SO_4. Calcium hydroxide shows a continuous drop in solubility because its dissolution is exothermic.

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1
Analyze the thermal effect of dissolution for endothermic salts with no phase transitions.
Potassium nitrate (KNO3KNO_3) absorbs substantial thermal energy upon dissolving, yielding a steep positive slope on a solubility graph.
According to Le Chatelier's principle, increasing temperature shifts endothermic dissolution equilibria toward increased solute dissolution.
2
Evaluate salts with negligible heats of solution.
Sodium chloride (NaClNaCl) exhibits a nearly horizontal curve, showing minimal change in solubility over a wide temperature range.
The lattice energy and hydration energy of NaClNaCl are nearly equal, resulting in negligible temperature dependence.
3
Identify salts undergoing chemical dehydration or phase changes.
Glauber's salt (Na2SO410H2ONa_2SO_4 \cdot 10H_2O) shows a sudden sharp break (kink) at 32.4C32.4^\circ\text{C}.
Below 32.4C32.4^\circ\text{C}, the decahydrate dissolves endothermically. Above 32.4C32.4^\circ\text{C}, it dehydrates to anhydrous Na2SO4Na_2SO_4, which dissolves exothermically.
4
Analyze exothermic dissolution processes.
Calcium hydroxide (Ca(OH)2Ca(OH)_2) exhibits a downward-sloping solubility curve.
Exothermic processes release heat upon dissolution; raising the temperature suppresses dissolution and causes precipitation.

Anahtar Kavram

Effect of Temperature and Enthalpy of Solution on Solubility Curves
Tahmini Süre:1m 30s
Soru 18Soru

The solubility of a solute XX in water is 1.5 mol/dm31.5\text{ mol/dm}^3 at 80C80^\circ\text{C} and 0.5 mol/dm30.5\text{ mol/dm}^3 at 30C30^\circ\text{C}. If 250 cm3250\text{ cm}^3 of a saturated solution of XX at 80C80^\circ\text{C} is cooled to 30C30^\circ\text{C}, what mass of XX will crystallize out of the solution? [Molar mass of X=160 g/molX = 160\text{ g/mol}]

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Cevap: 40.0 g40.0\text{ g}

Cevap

The mass of salt XX that crystallizes out of solution is 40.0 g40.0\text{ g}.
Cooling 1 dm31\text{ dm}^3 of saturated solution from 80C80^\circ\text{C} to 30C30^\circ\text{C} precipitates 1.0 mol1.0\text{ mol} of solute. For a 250 cm3250\text{ cm}^3 (0.25 dm30.25\text{ dm}^3) solution, 0.25 mol0.25\text{ mol} precipitates. Multiplying 0.25 mol0.25\text{ mol} by the molar mass (160 g/mol160\text{ g/mol}) yields 40.0 g40.0\text{ g}.

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1
Calculate the difference in solubility between 80C80^\circ\text{C} and 30C30^\circ\text{C} per dm3\text{dm}^3.
ΔS=1.5 mol/dm30.5 mol/dm3=1.0 mol/dm3\Delta S = 1.5\text{ mol/dm}^3 - 0.5\text{ mol/dm}^3 = 1.0\text{ mol/dm}^3
Solubility decreases upon cooling, causing the excess solute to precipitate.
2
Scale the amount of precipitated solute to the specified volume of 250 cm3250\text{ cm}^3 (0.25 dm30.25\text{ dm}^3).
nprecipitated=1.0 mol/dm3×0.25 dm3=0.25 moln_{\text{precipitated}} = 1.0\text{ mol/dm}^3 \times 0.25\text{ dm}^3 = 0.25\text{ mol}
The solution volume is 250 cm3250\text{ cm}^3, which is a quarter of a cubic decimeter.
3
Convert the moles of precipitated solute to mass in grams using its molar mass.
Mass=0.25 mol×160 g/mol=40.0 g\text{Mass} = 0.25\text{ mol} \times 160\text{ g/mol} = 40.0\text{ g}
Mass is obtained by multiplying the chemical amount in moles by the molar mass.

Anahtar Kavram

Calculation of mass crystallized from solubility curves and temperature changes
Soru 19Soru

An aqueous mixture passes completely through standard filter paper without leaving any residue, but its particles are retained when passed through a parchment (semi-permeable) membrane. When illuminated by a narrow light beam in a dark room, the mixture scatters light, and under an ultramicroscope, its particles exhibit continuous zig-zag movement without settling under gravity. Which of the following correctly classifies this mixture and describes the characteristic behavior of its dispersed phase?

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Cevap: It is a colloidal system, and its particles undergo electrophoresis toward an oppositely charged electrode when an electric field is applied.

Cevap

The mixture is a colloidal system whose dispersed particles carry electric charges and migrate toward an oppositely charged electrode during electrophoresis.
The mixture exhibits all hallmark physical and optical characteristics of a colloidal system: particle size between 1 nm1\text{ nm} and 100 nm100\text{ nm} (retained by parchment membrane but passing filter paper), Tyndall effect (light scattering), Brownian motion, and surface electrical charge that allows directional movement in an electric field (electrophoresis).

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1
Analyze permeability and filtration behavior
Passing through standard filter paper rules out suspensions (>100 nm>100\text{ nm}). Retention by parchment (semi-permeable) membrane indicates a particle size between 1 nm1\text{ nm} and 100 nm100\text{ nm}, characteristic of colloidal systems.
True solution particles (<1 nm<1\text{ nm}) pass through both filter paper and semi-permeable membranes.
2
Analyze optical and kinetic behavior
Scattering light (Tyndall effect) and random continuous motion (Brownian motion) without settling confirm colloidal properties.
True solutions do not scatter light (no Tyndall effect), while suspensions settle out under gravity.
3
Determine electrical properties of the colloidal system
Colloidal particles preferentially adsorb ions from the medium, acquiring a net positive or negative surface charge. When subjected to an electric field, they migrate toward the electrode of opposite charge (electrophoresis).
Surface charges prevent colloidal particles from aggregating under normal conditions and make them responsive to electric potential.

Anahtar Kavram

Physical distinction and electrical properties of true solutions, colloidal systems, and suspensions
Tahmini Süre:1m 30s
Soru 20Soru

The solubility of copper(II) tetraoxosulfate(VI), CuSO4\text{CuSO}_4, at 60C60^\circ\text{C} and 20C20^\circ\text{C} is 40.0 g40.0\text{ g} and 21.0 g21.0\text{ g} per 100 g100\text{ g} of water respectively. Calculate the mass of CuSO4\text{CuSO}_4 in grams that will crystallize out of solution when a saturated solution containing 250 g250\text{ g} of water is cooled from 60C60^\circ\text{C} to 20C20^\circ\text{C}.

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Cevap: 47.5

Cevap

The mass of CuSO4\text{CuSO}_4 that crystallizes out of solution is 47.5 g47.5\text{ g}.
Subtracting the solubility at 20C20^\circ\text{C} (21.0 g21.0\text{ g}) from the solubility at 60C60^\circ\text{C} (40.0 g40.0\text{ g}) yields 19.0 g19.0\text{ g} of CuSO4\text{CuSO}_4 deposited per 100 g100\text{ g} of water. Multiplying by the ratio of actual solvent mass to reference solvent mass (250 g/100 g=2.5250\text{ g} / 100\text{ g} = 2.5) gives 47.5 g47.5\text{ g}.

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1
Determine the mass of CuSO4\text{CuSO}_4 deposited per 100 g100\text{ g} of water on cooling.
40.0 g21.0 g=19.0 g40.0\text{ g} - 21.0\text{ g} = 19.0\text{ g} per 100 g100\text{ g} of water.
The mass of solute precipitated per 100 g100\text{ g} of solvent is equal to the difference in solubility between the higher and lower temperatures.
2
Calculate the mass of solute deposited for 250 g250\text{ g} of water.
19.0 g×250 g100 g=47.5 g19.0\text{ g} \times \frac{250\text{ g}}{100\text{ g}} = 47.5\text{ g}.
The amount of solute crystallized out is directly proportional to the total mass of solvent present.

Anahtar Kavram

Crystallization and Mass of Solute Deposited on Cooling
Sayfa 1 / 4Sonraki
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