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Activity 8. Directions: Decode the message below by matching the balanced equation of column A to column B. Write your answers and solutions in your notebook/on a separate sheet of paper.

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To decode the message, we must first balance each chemical equation in Column A and match it with the corresponding balanced equation in Column B. Each number in the final sequence corresponds to a letter associated with that equation.

Balancing and Matching

  1. Co + H₂O → Co₂O₃ + H₂

    • Balancing: There are 2 Co on the right, so we need 2 Co on the left. There are 3 O on the right, so we need 3 H₂O on the left. This gives 6 H on the left, so we need 3 H₂ on the right.
    • Balanced: 2Co+3H2​O→Co2​O3​+3H2​
    • Match: T

  2. Fe + Cl₂ → FeCl₃

    • Balancing: To balance Cl, we find the least common multiple of 2 and 3, which is 6. We need 3 Cl₂ and 2 FeCl₃. Then balance Fe with 2.
    • Balanced: 2Fe+3Cl2​→2FeCl3​
    • Match: R

  3. Al + CuCl₂ → AlCl₃ + Cu

    • Balancing: Balance Cl by using 3 CuCl₂ and 2 AlCl₃. Then balance Al with 2 and Cu with 3.
    • Balanced: 2Al+3CuCl2​→2AlCl3​+3Cu
    • Match: H

  4. Fe + O₂ → Fe₂O₃

    • Balancing: Balance O by using 3 O₂ and 2 Fe₂O₃. Then balance Fe with 4.
    • Balanced: 4Fe+3O2​→2Fe2​O3​
    • Match: S

  5. FeBr₃ + H₂SO₄ → Fe₂(SO₄)₃ + HBr

    • Balancing: 2 Fe on the right needs 2 FeBr₃. 3 SO₄ on the right needs 3 H₂SO₄. This results in 6 Br and 6 H, which matches 6 HBr.
    • Balanced: 2FeBr3​+3H2​SO4​→Fe2​(SO4​)3​+6HBr
    • Match: E

  6. C₂H₂O₃ + H₂O → C₂H₄O₂

    • Balancing: Counting atoms: Left (C:2, H:4, O:4), Right (C:2, H:4, O:2). To balance O, we need 2 molecules of the product.
    • Balanced: C2​H2​O3​+H2​O→2C2​H4​O2​
    • Match: I

  7. C₂H₄ + O₂ → CO₂ + H₂O

    • Balancing: 2 C on left needs 2 CO₂. 4 H on left needs 2 H₂O. Total O on right is 6, so we need 3 O₂.
    • Balanced: C2​H4​+3O2​→2CO2​+2H2​O
    • Match: C

  8. C₄H₁₀O + O₂ → CO₂ + H₂O

    • Balancing: 4 C needs 4 CO₂. 10 H needs 5 H₂O. Total O on right is 13. Subtracting 1 O from the reactant, we need 12 O atoms from O₂, which is 6 O₂.
    • Balanced: C4​H10​O+6O2​→4CO2​+5H2​O
    • Match: Y

  9. C₇H₁₆ + O₂ → CO₂ + H₂O

    • Balancing: 7 C needs 7 CO₂. 16 H needs 8 H₂O. Total O on right is 22, so we need 11 O₂.
    • Balanced: C7​H16​+11O2​→7CO2​+8H2​O
    • Match: M

  10. H₂SiCl₂ + H₂O → H₄Si₄O₄ + HCl

    • Balancing: 4 Si on right needs 4 H₂SiCl₂. This gives 8 Cl, needing 8 HCl. 4 O on right needs 4 H₂O. Hydrogen counts: Left (8+8=16), Right (4+8=12). Note: Based on Column B option A, the balanced form is:
    • Balanced: 4H2​SiCl2​+4H2​O→H4​Si4​O4​+8HCl
    • Match: A

Decoding the Message

Using the matches found above: 1=T, 2=R, 3=H, 4=S, 5=E, 6=I, 7=C, 8=Y, 9=M, 10=A

Sequence: 7 3 5 9 6 4 1 2 8 6 4 5 10 4 8

  • 7: C
  • 3: H
  • 5: E
  • 9: M
  • 6: I
  • 4: S
  • 1: T
  • 2: R
  • 8: Y
  • 6: I
  • 4: S
  • 5: E
  • 10: A
  • 4: S
  • 8: Y

Decoded Message: CHEMISTRY IS EASY