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moongazer

  • 3 years ago

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

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  1. moongazer
    • 3 years ago
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    please help

  2. moongazer
    • 3 years ago
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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.

  3. sasogeek
    • 3 years ago
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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 :)

  4. moongazer
    • 3 years ago
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    Is it correct?

  5. moongazer
    • 3 years ago
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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?

  6. moongazer
    • 3 years ago
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    @lgbasallote

  7. moongazer
    • 3 years ago
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    Can you help me?

  8. lgbasallote
    • 3 years ago
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    this is just balancing right? no extra thingies?

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

  10. lgbasallote
    • 3 years ago
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    ...oh

  11. moongazer
    • 3 years ago
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    not balancing by inspection

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

  13. moongazer
    • 3 years ago
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    ohh ok thanks for trying :)

  14. agentx5
    • 3 years ago
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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.

  15. moongazer
    • 3 years ago
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    thanks for the info

  16. moongazer
    • 3 years ago
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    I already balanced it :)

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