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  • 4 years ago

"Complete the equations for these single-replacement reactions in aqueous solutions. Balance each equation. Write "No reaction" if a reactions doesn't occur. Use the activity series chart you copied into your notes." ***Are these going to all be aqueous for product and reactants??*** a.) Fe + Pb(NO3)2-> b.) Cl2 + NaI -> c.) Ca + H2O -> My answers I got, which I am pretty sure I am wrong are: a.) 2Fe (aq) +Pb(NO3)2(aq) -> 2 FeNO3 (aq?) +Pb(s?) B.) Cl2 (aq) +NaI (aq) -> NaCl2(aq) +I(aq) C.) Ca(s) + 2H2O(g)->Ca (OH)2 (aq) + H2 (g)

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  1. anonymous
    • 4 years ago
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    a. Fe (s) + Pb(NO3)2 (aq) → Fe(NO3)2 (aq) + Pb (s) For this problem, the only thing you got wrong were the states of matter and the product when Fe and NO3 react. Remember that iron has a charge of +2 or +3; in this problem I assumed that it has a charge of +2. b. Cl2 (g) + 2NaI (aq) → 2NaCl (aq) + I2 (g) Same thing, just remember when doing single displacement reactions, treat everything as an ion w/ a charge ONLY when it reacts with another element. For example, when chlorine gas (Cl2) reacts with sodium (Na), the product would be NaCl because Na+ and Cl- makes NaCl. Also, remember your diatomic elements- iodine by itself is a diatomic gas (I2). c. Ca (s) + 2H2O (l) → Ca(OH)2 (aq) + 2H2 (g) You have the right equation, but make sure to correctly balance everything. Also, the reaction occurs in aqueous solution, so H2O will likely be a liquid, not gas. Overall, most of the equations you got were correct, but just remember to balance and put the correct states of matter on the diatomic gases.

  2. anonymous
    • 4 years ago
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    For A how would it be different if it was a charge of +3? OH! Okay, thanks for number C. . . for some reason I missed that I balanced it wrong. I took H as just 2 not 4 I see where it came from now. Awesome! :) THANKS!

  3. anonymous
    • 4 years ago
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    If Fe has a charge of +3, the equation would look like 2Fe (s) + 3Pb(NO3)2 (aq) → 2Fe(NO3)3 (aq) + 3Pb (s) Yeah and no problem!

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