SS2: CHEMISTRY - 1ST TERM
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Periodicity and Periodic Table I | Week 15 Topics|1 Quiz
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Quantum Numbers Orbitals & Electrical Structure | Week 26 Topics|1 Quiz
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Periodicity and Periodic Table II | Week 312 Topics|1 Quiz
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Periodic Table and Atomic Properties
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Melting and Boiling Point
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Electrical and Thermal Conductivities
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Atomic Size [Radius]
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Ionic Size [Radius]
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Atomic Volume
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Ionization Energy
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Electron Affinity
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Electronegativity
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Differences between Ionization Energy and Electron Affinity
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Summary of Trends of Atomic Properties
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Theory Questions - Periodicity and Periodic Table II
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Periodic Table and Atomic Properties
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Periodicity and Periodic Properties III | Week 411 Topics|1 Quiz
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Periodicity and Periodic Properties IV | Week 55 Topics|1 Quiz
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Mass-Volume Relationship in Reaction | Week 68 Topics|1 Quiz
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Types of Reactions: Oxidation and Reduction | Week 7 & 87 Topics|1 Quiz
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Oxidation – Reduction Reaction II | Week 93 Topics|1 Quiz
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Electrode Potential and Electrochemical Cells I | Week 106 Topics|1 Quiz
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Electrode Potential and Electrochemical Cells II | Week 115 Topics|1 Quiz
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Electrolysis I | Week 128 Topics|1 Quiz
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Electrolysis II | Week 138 Topics|1 Quiz
Mass-Volume Calculations
Topic Content:
- Mass-Volume Calculations
In Mass-Volume Calculations a mass is given and a volume is to be found. This group of calculations depends upon the use of molar volume. The molar volume of any gas at standard temperature and pressure is 22.4 dm3
Example 6.4.1:
Calculate the volume of Carbon[IV]oxide gas at STP when 5 g of zinc trioxocarbonate[V] is heated strongly.
[Zn = 65, C = 12, O = 16, molar volume = 22.4 dm3 at STP]
Solution
ZnCO3[s] → ZnO[s] + CO2[g]
1 mole of ZnCO3[s] ⇒ 1 mole of CO2[g] at STP
1 × [65 + 12 + (16 × 3)] g of ZnCO3[s] ⇒ 1 × 22.4 dm3 of CO2 at STP
125 g of ZnCO3[s] ⇒ 22.4 dm3 of CO2 at STP
∴ 5 g of ZnCO3[s] ⇒ \( \frac{5}{125} \; \times \; \frac{22.4}{1}\) dm3 of CO2 at STP
= 0.896 dm3 of CO2[g] at STP
Example 6.4.2:
What volume of dry oxygen gas [measured at STP] will be produced from the decomposition of 2.45 g of potassium trioxochlorate [V] (KClO3)?
[K = 39, Cl = 35.5, O = 16 molar volume = 22.4 dm3 at STP]
Solution
2KClO3[S] → 2KCl[s] + 302[g]
From the balanced equation;
2 moles of KClO3 ⇒ 3 moles of O2 at STP
2 × [39 + 35.5 + (16 × 3)] g of KClO3 ⇒ 3 × 22.4 dm3 of O2 at STP
∴ 245 g of KClO3[s] ⇒ 67.2 dm3 of O2[g]
∴ 2.45 g of KClO3 ⇒ \( \frac{2.45}{245} \; \times \; \frac{67.2}{1}\)
= 0.67 dm3 of O2 at STP
Example 6.4.3:
Calculate the volume of Hydrogen gas measured at 30°C and 720 mmHg when 6.5 g of Zinc reacts with hydrochloric acid.
[Zn = 65, C = 12, H = 1, Cl = 35.5, molar volume = 22400 cm3 at STP]
Equation of reaction:
Zn[s] + 2HCl[aq] → ZnCl2[aq] + H2[g]
1 mole of Zn[s] ⇒ 1 mole of H2[g] at STP
65 g of Zn[s] ⇒ 22400 cm3 of H2[g] at STP
∴ 6.5 g of Zn[s] ⇒ \( \frac{6.5}{65} \; \times \; \frac{22400}{1} \) cm3 of H2[g] at STP
= 2240 cm3 of H2[g] at STP
using: \( \frac{P_1 V_1}{T_1} =\frac{P_2 V_2}{T_2} \)
P1 = 760 mmHg P2 = 720 mmHg
V1 = 2240 cm3 V2 = ?
T1 = 273 K T2 = 30°C + 273 = 303 K
V2 = \( \frac{P_1 V_1 T_2}{P_2 T_1} \)
V2 = \( \frac{760 \: \times \: 2240 \: \times \: 303}{720 \: \times \: 273} \\ \scriptsize = 2624.3 \: cm^3\)