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12/30 = percentage of blue marbles = 40% 40% of 20 =8 12-8 =4 John has 4 more blue marbles

Answered by welchje | 2024-06-10

The required, John has 4 more **blue marbles **than Jane.
What is the Ratio?
The **ratio **can be defined as the **comparison **of the **fraction **of one **quantity **towards others. e.g.- water in milk.
Here, Let's start by finding the ratio of red to blue marbles for John:
Red marbles: 18
Blue marbles: 12
Red to the blue ratio: 18/12 = 3/2
Since the ratio of red to blue marbles is the same for Jane and John, we can use this ratio to find the number of blue marbles that Jane has:
Blue-to-red ratio : 2/3
Total number of Jane's marbles: 20
Blue marbles: (2/5) * 20 = 8
Red marbles: 20 - 8 = 12
Now we can find how many more blue marbles John has than Jane:
John's blue marbles: 12
Jane's blue marbles: 8
Difference : 12 - 8 = 4
Therefore, John has 4 more **blue marbles **than Jane.
Learn more about **Ratio **here:
brainly.com/question/13419413
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Answered by priyankapandeyVT | 2024-06-17

John has 4 more blue marbles than Jane. This is determined by calculating the red to blue marble ratio for both individuals and finding the difference in their blue marbles. John's count of blue marbles is 12, while Jane's is 8.
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Answered by priyankapandeyVT | 2024-10-02

1. Reaksi a: Cu + H⁺ + NO₃⁻ → Cu²⁺ + NO₃⁻ + H₂OSetengah Reaksi Oksidasi:Cu → Cu²⁺ + 2e⁻Setengah Reaksi Reduksi:NO₃⁻ + 4H⁺ + 3e⁻ → NO₂ + 2H₂OGabungkan:Cu + 2NO₃⁻ + 8H⁺ → Cu²⁺ + 2NO₂ + 4H₂O2. Reaksi b: Zn + H⁺ + NO₃⁻ → Zn²⁺ + NH₄⁺ + H₂OSetengah Reaksi Oksidasi:Zn → Zn²⁺ + 2e⁻Setengah Reaksi Reduksi:NO₃⁻ + 10H⁺ + 8e⁻ → NH₄⁺ + 3H₂OGabungkan:Zn + 2NO₃⁻ + 10H⁺ → Zn²⁺ + NH₄⁺ + 3H₂O3. Reaksi c: H₂O₂ + MnO₄⁻ + H⁺ → Mn²⁺ + H₂O₂ + O₂Setengah Reaksi Oksidasi:H₂O₂ → O₂ + 2e⁻Setengah Reaksi Reduksi:MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂OGabungkan:3H₂O₂ + 2MnO₄⁻ + 8H⁺ → 2Mn²⁺ + 3O₂ + 4H₂O4. Reaksi d: SO₃²⁻ + Cl₂ + H⁺ → HSO₄⁻ + H₂OSetengah Reaksi Oksidasi:SO₃²⁻ → HSO₄⁻ + 2e⁻Setengah Reaksi Reduksi:Cl₂ + 2e⁻ → 2Cl⁻Gabungkan:SO₃²⁻ + Cl₂ + 2H⁺ → HSO₄⁻ + 2Cl⁻ + H₂O5. Reaksi e: P₄ + H⁺ + NO₃⁻ → H₃PO₄ + CNO⁻ + H₂OSetengah Reaksi Oksidasi:P₄ → 4P³⁺ + 12e⁻Setengah Reaksi Reduksi:NO₃⁻ + 8H⁺ + 6e⁻ → CNO⁻ + 3H₂OGabungkan:P₄ + 2NO₃⁻ + 16H⁺ → 4H₃PO₄ + 2CNO⁻ + 6H₂O6. Reaksi f: CN⁻ + MnO₄⁻ + OH⁻ → MnO₂ + CNO⁻ + H₂OSetengah Reaksi Oksidasi:CN⁻ → CNO⁻ + e⁻Setengah Reaksi Reduksi:MnO₄⁻ + 2OH⁻ + 3e⁻ → MnO₂ + H₂OGabungkan:2CN⁻ + MnO₄⁻ + 4OH⁻ → MnO₂ + 2CNO⁻ + 2H₂O7. Reaksi g: [Fe(CN)₆]³⁻ + N₂H₄ → [Fe(CN)₆]⁴⁻ + N₂ + H₂OSetengah Reaksi Oksidasi:N₂H₄ → N₂ + 4e⁻Setengah Reaksi Reduksi:[Fe(CN)₆]³⁻ + e⁻ → [Fe(CN)₆]⁴⁻Gabungkan:[Fe(CN)₆]³⁻ + N₂H₄ → [Fe(CN)₆]⁴⁻ + N₂ + H₂O8. Reaksi h: Fe(OH)₂ + O₂ + OH⁻ → Fe(OH)₃ + H₂OSetengah Reaksi Oksidasi:Fe(OH)₂ → Fe(OH)₃ + e⁻Setengah Reaksi Reduksi:O₂ + 2e⁻ + 2OH⁻ → 2OH⁻Gabungkan:2Fe(OH)₂ + O₂ + 2OH⁻ → 2Fe(OH)₃ + H₂O9. Reaksi i: C₆H₅OH + MnO₄⁻ + H₂O → C₂H₃O₂ + MnO₂ + H₂OSetengah Reaksi Oksidasi:C₆H₅OH → C₂H₃O₂ + 3e⁻Setengah Reaksi Reduksi:MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + H₂OGabungkan:C₆H₅OH + MnO₄⁻ + 4H⁺ → C₂H₃O₂ + MnO₂ + H₂O10. Reaksi j: Cl₂ + H₂O → H⁺ + Cl⁻ + HOCISetengah Reaksi Oksidasi:Cl₂ + 2e⁻ → 2Cl⁻Setengah Reaksi Reduksi:H₂O + 2e⁻ → HOCI + 2H⁻Gabungkan:Cl₂ + H₂O + 2e⁻ → HOCI + 2Cl⁻11. Reaksi k: SO₃²⁻ + H₂O → SO₄²⁻ + H₂OSetengah Reaksi Oksidasi:SO₃²⁻ → SO₄²⁻ + 2e⁻Setengah Reaksi Reduksi:H₂O → H₂O (tidak ada perubahan)12. Reaksi l: MnO₄⁻ + H₂O → Mn²⁺ + OH⁻Setengah Reaksi Oksidasi:MnO₄⁻ → Mn²⁺ + 5e⁻Setengah Reaksi Reduksi:H₂O → OH⁻ (tidak ada perubahan)

Answered by pupusdwisuryoanggono | 2025-07-22