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moongazer Group Title

Balance by redox CrBr3 + NaOH + Cl2 ----> NaCrO4 + NaBrO + NaCl + H2O please help :)

  • 2 years ago
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  1. moongazer Group Title
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    please help

    • 2 years ago
  2. moongazer Group Title
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    Cr(+3)Br3(-1) + Na(+1)O(-2)H(+1) + Cl2(0) ----> Na(+1)Cr(+7)O4(-2) + Na(+1)Br(+1)O(-2) + Na(+1)Cl(-1) + H2(+1)O(-2) The oxidation number is enclosed by parenthesis.

    • 2 years ago
  3. sasogeek Group Title
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    i'm not good at chem, but i think you should find their oxidation states, then make sure that sum of oxidation states on left = that on right :)

    • 2 years ago
  4. moongazer Group Title
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    Is it correct?

    • 2 years ago
  5. moongazer Group Title
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    I already did that but I can't balance it correctly. Did I made a mistake in identifying their oxidation number?

    • 2 years ago
  6. moongazer Group Title
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    @lgbasallote

    • 2 years ago
  7. moongazer Group Title
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    Can you help me?

    • 2 years ago
  8. lgbasallote Group Title
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    this is just balancing right? no extra thingies?

    • 2 years ago
  9. moongazer Group Title
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    balancing by state change method. we will identify the oxidizing numbers then balance

    • 2 years ago
  10. lgbasallote Group Title
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    ...oh

    • 2 years ago
  11. moongazer Group Title
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    not balancing by inspection

    • 2 years ago
  12. lgbasallote Group Title
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    well im not that good in chemistry to know all those. not my specialty sorry

    • 2 years ago
  13. moongazer Group Title
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    ohh ok thanks for trying :)

    • 2 years ago
  14. agentx5 Group Title
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    Ok this is something I can help with after class, but first... 1. @sasogeek is correct, net electron gain & loss must equal out on both sides 2. @lgbasallote is correct, you need to balance the equation, and I would do this first (easier). Not sure how you were introduced to this in class, but you should balance moles of atoms first (most of the time)... 3. Chromium is your wildcard (meaning you should do it LAST for charges), but I can tell you from experience with this element that +7 would be HIGHLY unstable. Unless you're dealing with intermediary reactions, you won't have this charge on the valence electrons. Stable charges are +1, +2, +3, +4, +5, and +6. The +3 cation charge is the most common because of the natural D-orbital configuration for this element (most common = most stable) 4. Memorize: The substance which loses electrons is oxidized. (L.E.O.) The substance which gains electrons is reduced. (G.E.R.) Can you identify what is being reduced and what is being oxidized? 5. When writing charges, you want to pay attention to treating polyatomic anions as whole groups. So OH\(^-\) , SO\(_4 \ ^{2-}\) , etc. The reason why you should get into the practice of this is because of larger organize polyatomic anions and cations which have resonance structures (i.e.: benzene rings, nucleotides, etc.) I'll be back to check to see if you can figure out the balancing portion first.

    • 2 years ago
  15. moongazer Group Title
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    thanks for the info

    • 2 years ago
  16. moongazer Group Title
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    I already balanced it :)

    • 2 years ago
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