WJEC A-Level Chemistry 1410QS
Formulae and Equations
Basic Ideas About Atoms
Chemical Calculations
Bonding
Solid Structures
The Periodic Table
Simple Equilibria and Acid-Base Reactions
Thermochemistry
Rates of Reaction
The Wider Impact of Chemistry
Organic Compounds
Hydrocarbons
Halogenoalkanes
Alcohols and Carboxylic Acids
Instrumental Analysis
Redox and Standard Electrode Potential
Redox Reactions
Chemistry of the P-Block
Chemistry of the D-Block Transition Metals
Chemical Kinetics
Enthalpy Changes for Solids and Solutions
Entropy and Feasibility of Reactions
Equilibrium Constants
Acid-Base Equilibria
Stereoisomerism
Aromaticity
Alcohols and Phenols
Aldehydes and Ketones
Carboxylic Acids and Their Derivatives
Amines
Amino Acids, Peptides and Proteins
Organic Synthesis and Analysis
Practical Skills and Techniques
Planning, Analysis and Evaluation
Core Experiments and Their Chemistry
Mathematical Skills in Chemistry
Mathematical Skills in Chemistry
Symbols, rearranging equations and logarithms: rearranging an equation
- 3 things to know
- All units
Overview
Fourteen recall questions on symbols, rearranging equations and logarithms: rearranging an equation.
What you need to know
Answer each one before you reveal it.
Rearrange c = n/V to find the volume of 0.200 mol dm⁻³ solution that contains 0.0500 mol of solute.
A first-order reaction has a rate of 4.50 × 10⁻⁴ mol dm⁻³ s⁻¹ when [A] = 0.150 mol dm⁻³. Calculate the rate constant.
Questions from this subtopic
Each one with the model answer underneath.
- 1
Why is the equilibrium sign drawn with two half-arrows of equal length in most equations, and what would unequal arrows suggest?
Model answer
Two equal half-arrows show that both the forward and reverse processes occur and that neither is being singled out as dominant. Drawing one half-arrow longer is sometimes used informally to suggest that the equilibrium position lies to that side, but the position of equilibrium is properly conveyed by the value of the equilibrium constant, not by the drawing.
- 2
Rearrange n = m/M to make M the subject, and use it to find the molar mass of a substance of which 3.20 g is 0.100 mol.
Model answer
Multiplying both sides by M gives nM = m, and dividing by n gives M = m/n. Substituting, M = 3.20 ÷ 0.100 = 32.0 g mol⁻¹.
- 3
Make m the subject of n = m/M and calculate the mass of 0.250 mol of sodium chloride. (M = 58.5 g mol⁻¹)
Model answer
Multiplying through by M gives m = nM. So m = 0.250 × 58.5 = 14.6 g to three significant figures.
- 4
Rearrange c = n/V to find the volume of 0.200 mol dm⁻³ solution that contains 0.0500 mol of solute.
Model answer
V = n/c = 0.0500 ÷ 0.200 = 0.250 dm³, which is 250 cm³.
- 5
Make T the subject of pV = nRT and calculate the temperature of 0.500 mol of gas occupying 1.20 × 10⁻² m³ at 1.01 × 10⁵ Pa. (R = 8.31 J K⁻¹ mol⁻¹)
Model answer
Dividing both sides by nR gives T = pV/(nR). T = (1.01 × 10⁵ × 1.20 × 10⁻²) ÷ (0.500 × 8.31) = 1212 ÷ 4.155 = 292 K.
- 6
Show how pV = nRT can be rearranged to give molar mass, and calculate M for 0.320 g of a gas occupying 250 cm³ at 1.00 × 10⁵ Pa and 298 K.
Model answer
Since n = m/M, substituting gives pV = mRT/M and therefore M = mRT/(pV). Converting the volume, 250 cm³ = 2.50 × 10⁻⁴ m³. M = (0.320 × 8.31 × 298) ÷ (1.00 × 10⁵ × 2.50 × 10⁻⁴) = 792.4 ÷ 25.0 = 31.7 g mol⁻¹.
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