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anonymous

  • one year ago

Ok this question is driving me up the wall. Write ionic and net ionic equations for the following reactions: AgNO3 + Na2SO4 → As far as I am understanding from the book for the ionic equation I should remove the soluble chemicals from the insoluble chemicals and combined the insoluble chemicals. (Ag+)+(NO3-)+ (Na+)+(SO4(2-)) However for the net ionic equation if I follow the book I must remove all solubles from the equation which is everything except. (Ag+) So would my net ionic equation would just be??? Please explain! AgNO3 + Na2SO4 → (Ag+)

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  1. Photon336
    • one year ago
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    I think it would be this: Let me explain @whitewolfman426 \[ 2Ag ^{+} + SO4^{2-} -> Ag2SO4\]

  2. Photon336
    • one year ago
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    you omit the common ions

  3. Photon336
    • one year ago
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    let's work this out \[2AgNO3(aq) + Na2SO4 (aq) = Ag2SO4(s) + 2NaNO3(aq) \] the ionic equation would be this \[2Ag ^{+} + 2NO_{3}^{-} + 2Na ^{+} + SO_{4}^{2-} = Ag_{2}SO_{4}(s) + 2Na ^{+} + 2NO3^{-}\] the net ionic equation is \[2Ag ^{+}+ SO_{4}^{2-} = Ag_{2}SO_{4}(s)\]

  4. Photon336
    • one year ago
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    if you notice something NO3 and Na+ are on both sides of our reaction and you notice something right? they don't participate in our reaction and are called spectator ions. they're just there so we can ignore them

  5. Photon336
    • one year ago
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    @WhiteWolfman426 @zzbd @diana12475 make sense?

  6. Photon336
    • one year ago
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    @WhiteWolfman426 that's called a double replacement reaction familiar with that?

  7. anonymous
    • one year ago
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    @Photon336 through this myself. Why did you in the original equation switch 2SO4 and NO3?

  8. anonymous
    • one year ago
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    @Photon336 nope

  9. anonymous
    • one year ago
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    @Photon336 quick explanation if you will.

  10. Photon336
    • one year ago
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    |dw:1444012856587:dw|

  11. Photon336
    • one year ago
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    @WhiteWolfman426 take a look at the reaction above. that's basically what happens in a double replacement reaction.

  12. anonymous
    • one year ago
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    @Photon336 would I have to do this with every equation?

  13. Photon336
    • one year ago
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    I guess it depends on the reaction, but you would have to look to see what's going on first @WhiteWolfman426

  14. Photon336
    • one year ago
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    like for solubility there are rules that you have to memorize. I think all nitrates are soluble

  15. anonymous
    • one year ago
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    @Photon336 we have a chart in the book we are supposed to follow but this still doesn't make sense. I can theoretically say oh wait this doesn't work let me switch everything around.

  16. anonymous
    • one year ago
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    @Photon336 Also, how do you determine which ones you are going to keep together. According to the chart in my book 3/4 of the problem is soluble.

  17. anonymous
    • one year ago
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    @Photon336 Which gets eliminated in the net ionic formula

  18. anonymous
    • one year ago
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    For the net ionic equation, You have to separate each compound into the ions.

  19. anonymous
    • one year ago
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    ONLY if they are soluble

  20. Photon336
    • one year ago
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    yeah I did for the first part @diana12475 @WhiteWolfman426 was asking why it was a double replacement reaction

  21. anonymous
    • one year ago
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    if they are not, then you leave the compounds as is. you then go on to eliminate the soluble ions. It is double replacement only if you can separate the compounds into ions

  22. Photon336
    • one year ago
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    well said @diana12475

  23. anonymous
    • one year ago
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    @diana12475 @Photon336 Again according to the book NO3, Na,2SO4 are all soluble except Ag

  24. anonymous
    • one year ago
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    @diana12475 @Photon336 So necessarily the net ionic equation would just have Ag in it

  25. Photon336
    • one year ago
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    Na+ and NO3- are soluble

  26. Photon336
    • one year ago
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    that equation suggests that you form a precipitate, Ag2SO4 which is insoluble

  27. Photon336
    • one year ago
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    silver will almost always be insoluble regardless of what it's in * dont remember exactly @WhiteWolfman426 @diana12475 @pencile thoughts?

  28. anonymous
    • one year ago
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    @Photon336 @diana12475 Attached is the chart I have in my book

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  29. Photon336
    • one year ago
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    ok @WhiteWolfman426 so right away

  30. Photon336
    • one year ago
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    all those compounds listed would be SOLUBLE but if they are paired with Hg2+ or Ag+ they are INSOLUBLE.

  31. anonymous
    • one year ago
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    @Photon336 but isn't the point of having them in the ionic equation is to separate the soluble and insoluble?

  32. Photon336
    • one year ago
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    @WhiteWolfman426 the point is to write out all the ions first, and then determine which ones are going to be soluble and which are going to be in soluble because remember the ions may break apart in solution and then react with each other. so let me show you

  33. Photon336
    • one year ago
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    @Whitewolfman426 look at this. say we take NaNO3 and we put it in water right? first thing we do is we look at our rules. this thing is going to interact with the water and it's not going to precipitate, how do we know? based on our solubility rules. |dw:1444014327758:dw| now say if we had Ag2SO4 THE other substance in our reaction. @WhiteWolfman426 what happens here is that a reaction is still going to take place, the ions are going to react, but if the arent soluble they will just collect at the bottom. |dw:1444014493292:dw|

  34. anonymous
    • one year ago
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    @Photon336 Ok I can see that since they both are opposite charges they will stay connected to one another in water. So if we want to come out with that answer we would have to do double replacement for it if everything turns out to break up into individual ionic forms.

  35. Photon336
    • one year ago
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    @WhiteWolfman426 I think most of these reactions are double replacement. but I think from our discussion we can say that three steps are involved here: STEP #1 WORK out double replacement reaction as I stated STEP #2 CHECK each resulting compound whether its soluble based on the rules in your chart. STEP #3 Whatever compound isn't soluble you write out the equation that lead to that compound being formed.

  36. anonymous
    • one year ago
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    @Photon336 So going back to double replacement. How would you know if a double replacement is necessary for the problem you are working on?

  37. Photon336
    • one year ago
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    that's a good question

  38. anonymous
    • one year ago
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    @Photon336 Sorry for asking all these questions. This class I am taking online so it's pretty much sink or swing and the book is as useful as a paperweight.

  39. Photon336
    • one year ago
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    no i mean you're asking the right questions man don't worry

  40. Photon336
    • one year ago
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    okay so @WhiteWolfman426 so like generally for all precipitation reactions are double replacement

  41. Photon336
    • one year ago
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    Let's go back to this figure. if you say managed to identify the compound that collected at the bottom of the flask. and you found that YOU had AD, and you knew that you started from AB + CD, what type of reaction would it be? like look at this and tell me how you would know. @whitewolfman426 AB + CD --> AD + ? |dw:1444015377101:dw|

  42. anonymous
    • one year ago
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    @Photon336 obviously it would be BC because that is left of the soluble ions that can attract to each other since they are opposite polarities.

  43. Photon336
    • one year ago
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    nicely done, but @WhiteWolfman426 in your experiment you would know your starting materials what they are. and then if you see something collecting at the bottom and you identify and figure out what that is, you'll know some kind of precipitation reaction happened. and for precipitation there is only one mechanism for that DOUBLE Replacement .

  44. anonymous
    • one year ago
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    @Photon336 Ok let me get this straight in my head. I take the one substance AB I put it in the water it dissolves. Then I put the next substance CD into the same water only part of it dissolves. They then would reconfigure themselves to form AD and BC?

  45. Photon336
    • one year ago
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    essentially that's what happens

  46. Photon336
    • one year ago
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    but usually one substance will collect at the bottom and won't dissolve.

  47. anonymous
    • one year ago
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    @Photon336 ok that sort of makes seance but we have an equation here were 3/4 of it is soluble by nature only leaving one to collect at the bottom. Am I to believe the one that is insoluble will naturally or just be like I am going to pair with this one now.

  48. Photon336
    • one year ago
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    wait where did you get 3/4ths?

  49. anonymous
    • one year ago
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    if you look at the original problem and my chart from my book everything is soluble in water except for Ag so if we happened to put those combinations in water the only thing that would be left is Ag and since everything is already dissolved in the water there won't be anything left to be collected in the bottom except for Ag.

  50. Photon336
    • one year ago
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    remember @WhiteWolfman426 Ag will react with the other compound too and they won't break apart because they aren't soluble. so it will be AG2SO4.

  51. Photon336
    • one year ago
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    think about it before the reaction when you put both substances AB and CD into water, yes the break up to get A + B + C + D but A will react with D and B will react with C and if BC is soluble it will just be ions, floating around. and if AD isn't soluble it would collect at the bottom.

  52. anonymous
    • one year ago
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    @Photon336 Then double displacement won't work in this problem because it doesn't make seance for separating it then adding it to another ion.

  53. anonymous
    • one year ago
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    In this problem we have two positives ions and two negative ions which when broken up can and most likely would react to their original pairing since they are closest together when they first got into the water. Just putting them in water doesn't guarantee that a double displacement can happen.

  54. Photon336
    • one year ago
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    okay, we may not agree on how it happens BUT how else would you explain why we found Ag(2)SO4 to be at the bottom and insoluble.

  55. anonymous
    • one year ago
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    @Photon336 Honestly IDFK! This double displacement thing sounds right on paper but in practice isn't guaranteed and IDFK chemistry well enough or even understand it well enough to be even attempting a class online that I though was going to be easy then turned into a giant Cluster F^$#*

  56. Photon336
    • one year ago
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    @WhiteWolfman426 from a point of view like I see where you're coming from as to why it wouldn't happen but try to say how did one of those compound precipitated.

  57. Photon336
    • one year ago
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    it's hard if you cant visualize this

  58. Photon336
    • one year ago
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    but the easiest way is to look at your table

  59. anonymous
    • one year ago
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    OK @Photon336 let me try it and see if it comes out like it supposed to. Double Displacement BaCl2 + ZnSO4 = BaSO4+ ZnCl2

  60. Photon336
    • one year ago
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    we're working through this until you get it if you have time take a look at this @whitewolfman426 look at your chart as you work out the problems too |dw:1444016990398:dw|

  61. anonymous
    • one year ago
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    \[Ba^{2}+SO _{4}^{-2}+ Zn^{+}+Cl ^{-}\]

  62. anonymous
    • one year ago
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    would be ionic equation

  63. Rushwr
    • one year ago
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    I think the net ionic equation and the ionic equation are the same here cuz no solid is formed !

  64. Photon336
    • one year ago
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    @WhiteWolfman426 now see the two products? refer to your chart in your textbook

  65. anonymous
    • one year ago
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    net ionic equation \[BaSO _{4}+ZnCl\]

  66. Photon336
    • one year ago
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    are either of those soluble/ insoluble based on the rules?

  67. anonymous
    • one year ago
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    Ba isn't soluble. It doesn't specify zinc but I am assuming since zinc is a metal it isn't soluble.

  68. Photon336
    • one year ago
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    @rushwr do you agree @whitewolfman426 try this out STEP #1 write the double replacement reaction. STEP #2 identify which product is insoluble STEP #2 Write the equation to form product that's insoluble

  69. Photon336
    • one year ago
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    then write out the net ionic equation for this: Na2S (aq) + ZnCl2 (aq) >> 2NaCl (aq) + ZnS (s)

  70. anonymous
    • one year ago
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    Ba not Na!

  71. Photon336
    • one year ago
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    @WhiteWolfman426 so follow what I rote above and try the problem that I gave you

  72. Rushwr
    • one year ago
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    Here barium ion reacts with sulfate ion to make insoluble barium sulfate right?

  73. anonymous
    • one year ago
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    @Photon336 2NaCl (aq) + ZnS (s) = \[ZnS_(s)\]

  74. Rushwr
    • one year ago
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    @WhiteWolfman426 Ur equation was for BaSO4 and ZnCl right? Let's get started with tht okai ?

  75. anonymous
    • one year ago
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    @Rushwr @Photon336 gave me practice question.

  76. anonymous
    • one year ago
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    @Rushwr previous problem BaCl2 + ZnSO4 =

  77. anonymous
    • one year ago
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    double displacement BaCl2 + ZnSO4 = BaSO4+ ZnCl2

  78. Photon336
    • one year ago
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    @WhiteWolfman426 identify which one is insoluble

  79. Photon336
    • one year ago
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    the one that's insoluble write out the equation that would lead you to that compound

  80. anonymous
    • one year ago
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    @Photon336 Ba and Zn isn't soluble which are both positive ions so they won't connect to make one insoluble unit. BaCl2 + ZnSO4 = BaSO4+ ZnCl2 ionic equation Ba2+(SO_4^-2)+(Zn^+)+(Cl^-)

  81. anonymous
    • one year ago
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    @Photon336 net ionic equation would be BaSO4+ZnCl Since neither Ba and Zn are soluble anything linked to an insoluble doesn't break down in accordance to this conversation.

  82. Photon336
    • one year ago
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    @Rushwr what do you think?

  83. Rushwr
    • one year ago
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    @WhiteWolfman426 Look at the steps PHOTON336 gave u STEP #1 write the double replacement reaction. STEP #2 identify which product is insoluble STEP #2 Write the equation to form product that's insoluble

  84. anonymous
    • one year ago
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    @Photon336 @Rushwr I did didn't I?!

  85. Rushwr
    • one year ago
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    According to that we have to write the balance double displacement reaction. The double displacement reaction for this reaction is , BaCl2 + ZnSO4 = BaSO4 + ZnCl2 Now look at the products side. There u have BaSO4 and ZnCl2 We know BaSO4 is insoluble.

  86. Rushwr
    • one year ago
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    Did u get up to this point ?

  87. anonymous
    • one year ago
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    @Rushwr and ZnCl2 isn't soluble either. In fact I looked it up on the internet and it was said that it isn't either. So the net ionic equation would be BaSO4+ZnCl since again according to this conversation when you have an insoluble ion the rest of it doesn't dissolve.

  88. Rushwr
    • one year ago
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    @WhiteWolfman426 But BaSO4 is insoluble right ? Out of ZnCl2 and BaSO4 , BaSO4 is insoluble.!!!!!!!!!!!!!!

  89. anonymous
    • one year ago
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    @Rushwr they both aren't soluble!!!!!!!!!!!!!

  90. Rushwr
    • one year ago
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    ZnCl2 is soluble ! @WhiteWolfman426

  91. anonymous
    • one year ago
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    @Rushwr NVM just looked it up yes it's soluble.

  92. Photon336
    • one year ago
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    Don't do the solubility rules from memory

  93. Photon336
    • one year ago
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    i do that and almost always forget it

  94. anonymous
    • one year ago
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    @Photon336 @Rushwr Look my book covers just about s&*$# and the examples are s@^&* so how am I supposed to know of the top of my f&*(* head that ZnCl is soluble without having to look it up on the internet. When I looked up Zn alone it said it was insoluble. When the f^^&* does that change the f&*(&ing rule all of a sudden!!!!!!

  95. anonymous
    • one year ago
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    @Photon336 @Rushwr Sorry didn't mean to get mean there just been fighting chemistry and barely passing this online class for the last to chapters and my teacher isn't very helpful so I am very on edge!

  96. Rushwr
    • one year ago
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    I'll tell u some solubility rules later after we finish these question okai !

  97. Rushwr
    • one year ago
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    Now we know BaSO4 is insoluble. The net ionic equation consists of the ions needed to make that insoluble compound. Therefore are net ionic equation would be Ba2+ + SO4^2 --> BaSO4(s) Our ionic equation would be Ba2+ + 2Cl- + Zn2+ + SO4^2- --> BaSO4(s) + Zn2+ + 2Cl-

  98. anonymous
    • one year ago
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    @Rushwr @Photon336 Thank you for your time an patience I got to go before I break something. I am tired and I have been working on this chemistry all day so yeah I am done for tonight. Thank you again for your patience.

  99. Rushwr
    • one year ago
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    Did u get tht part ? The last comment I posted ?

  100. Rushwr
    • one year ago
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    Salts containing Group I elements (Li+, Na+, K+, Cs+, Rb+) are soluble . There are few exceptions to this rule. Salts containing the ammonium ion (NH4+) are also soluble. Salts containing nitrate ion (NO3-) are generally soluble. Salts containing Cl -, Br -, or I - are generally soluble. Important exceptions to this rule are halide salts of Ag+, Pb2+, and (Hg2)2+. Thus, AgCl, PbBr2, and Hg2Cl2 are insoluble. Most silver salts are insoluble. AgNO3 and Ag(C2H3O2) are common soluble salts of silver; virtually all others are insoluble. Most sulfate salts are soluble. Important exceptions to this rule include CaSO4, BaSO4, PbSO4, Ag2SO4 and SrSO4 . Most hydroxide salts are only slightly soluble. Hydroxide salts of Group I elements are soluble. Hydroxide salts of Group II elements (Ca, Sr, and Ba) are slightly soluble. Hydroxide salts of transition metals and Al3+ are insoluble. Thus, Fe(OH)3, Al(OH)3, Co(OH)2 are not soluble. Most sulfides of transition metals are highly insoluble, including CdS, FeS, ZnS, and Ag2S. Arsenic, antimony, bismuth, and lead sulfides are also insoluble. Carbonates are frequently insoluble. Group II carbonates (CaCO3, SrCO3, and BaCO​3) are insoluble, as are FeCO3 and PbCO3. Chromates are frequently insoluble. Examples include PbCrO4 and BaCrO4. Phosphates such as Ca3(PO4)2 and Ag3PO4 are frequently insoluble. Fluorides such as BaF2, MgF2, and PbF2 are frequently insoluble.

  101. Rushwr
    • one year ago
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    Those are some solubility rules for u ! Just read them and get an idea ! :)

  102. anonymous
    • one year ago
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    @Rushwr Thank you!!!

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