PractiseQuestion bank
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12 questions
SA-AC-TM-0121 markL1 FoundationCalculatorConfigurations
The highest oxidation state of manganese is
Salam Academy Original Practice · Proposed mark scheme
SA-AC-TM-0011 markL2 StandardCalculatorConfigurations
The electron configuration of Fe³⁺ is
Salam Academy Original Practice · Proposed mark scheme
SA-AC-TM-0021 markL2 StandardCalculatorConfigurations
The ground-state configuration of chromium is
Salam Academy Original Practice · Proposed mark scheme
SA-AC-TM-0031 markL2 StandardCalculatorConfigurations
Scandium is not a transition element because
Salam Academy Original Practice · Proposed mark scheme
SA-AC-TM-0061 markL2 StandardCalculatorRedox titrations
In a KMnO₄ titration, the end point is
Salam Academy Original Practice · Proposed mark scheme
SA-AC-TM-0071 markL2 StandardCalculatorProperties
Transition elements show variable oxidation states because
Salam Academy Original Practice · Proposed mark scheme
SA-AC-TM-0081 markL2 StandardCalculatorProperties
Iron in the Haber process is a
Salam Academy Original Practice · Proposed mark scheme
SA-AC-TM-0101 markL2 StandardCalculatorProperties
Fe²⁺ catalyses S2O82−+2I−→2SO42−+I2 because Salam Academy Original Practice · Proposed mark scheme
SA-AC-TM-0111 markL2 StandardCalculatorRedox titrations
KMnO₄ titrations are acidified with dilute sulfuric acid rather than hydrochloric acid because
Salam Academy Original Practice · Proposed mark scheme
SA-AC-TM-0043 marksL3 ExamCalculatorRedox titrations
25.0 cm³ of Fe²⁺(aq) reacts with 20.0 cm³ of 0.0200 mol dm⁻³ KMnO₄. Calculate [Fe²⁺] (mol dm⁻³).
Salam Academy Original Practice · Proposed mark scheme
SA-AC-TM-0053 marksL3 ExamCalculatorRedox titrations
A solution of Fe²⁺ needs 15.0 cm³ of 0.0100 mol dm⁻³ K₂Cr₂O₇. How many moles of Fe²⁺ (×10⁻⁴) were present?
Salam Academy Original Practice · Proposed mark scheme
SA-AC-TM-0091 markL3 ExamCalculatorProperties
The reaction between MnO₄⁻ and C₂O₄²⁻ speeds up after starting because
Salam Academy Original Practice · Proposed mark scheme