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eli123

can someone please help me find these values?

  • one year ago
  • one year ago

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  1. eli123
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  2. eli123
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    @phi

    • one year ago
  3. eli123
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    @robtobey

    • one year ago
  4. NoelGreco
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    For someone to comment, he's have to know more about the lab.

    • one year ago
  5. aaronq
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    the moles you can simply find by dividing the masses by their molar mass... for the volume of H2 formed, you need to write the equation of the reaction in your exp and work with that.

    • one year ago
  6. eli123
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    @aaronq so for magnesium it would be 24.31/0.035?

    • one year ago
  7. aaronq
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    the other way around

    • one year ago
  8. eli123
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    oh ok and for h2? @aaronq

    • one year ago
  9. aaronq
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    you have to find the volume first

    • one year ago
  10. eli123
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    is the volume of hydrogen gas produced the same as teh volume of liquid displaced? @aaronq

    • one year ago
  11. aaronq
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    unless you made other gases in your experiment

    • one year ago
  12. aaronq
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    yes

    • one year ago
  13. eli123
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    no i didnt

    • one year ago
  14. eli123
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    so the volume divided by the mass of h2 will give me the moles of h2?

    • one year ago
  15. aaronq
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    well for ideal gases you can just use the fact that 1 mole = 22.414 L at STP

    • one year ago
  16. aaronq
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    otherwise you have to know the density of the gas

    • one year ago
  17. eli123
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    therefore, how would the equation look?

    • one year ago
  18. aaronq
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    so for ideal gases, moles = volume of gas/(22.414 L/mol)

    • one year ago
  19. aaronq
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    make sure you have the same units

    • one year ago
  20. eli123
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    oooh

    • one year ago
  21. eli123
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    @aaronq for the moles of magnesium igot 0.001 for all the trials, but my friend wrote it in dimensional analysis . whats betteR?

    • one year ago
  22. aaronq
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    it's the same thing, since it's a gas, the volume is proportional to the amount of moles

    • one year ago
  23. eli123
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    ok and how would i find the ratio of volume displaced in mL to moles h2 gas produced? @aaronq

    • one year ago
  24. aaronq
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    a ratio is just a division volume of gas/moles of gas

    • one year ago
  25. eli123
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    so volume of h2/moles of h2?

    • one year ago
  26. eli123
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    @aaronq

    • one year ago
  27. aaronq
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    yeah

    • one year ago
  28. eli123
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    @aaronq i get a huge number

    • one year ago
  29. eli123
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    37.00ml/0.002

    • one year ago
  30. aaronq
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    yep, thats what the want

    • one year ago
  31. eli123
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    18500?

    • one year ago
  32. aaronq
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    they actually want the volume displaced/molesof H2 which is the same thing

    • one year ago
  33. aaronq
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    mL/mol

    • one year ago
  34. eli123
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    so its right?

    • one year ago
  35. eli123
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    @aaronq

    • one year ago
  36. aaronq
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    to what they're asking you, yes

    • one year ago
  37. eli123
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    but its such a big number? @aaronq

    • one year ago
  38. aaronq
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    yep thats not that big of a number, remember it's in mL ..

    • one year ago
  39. eli123
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    18.5 L DAMM

    • one year ago
  40. aaronq
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    lol that's not a lot, it really should be around 22.414 L

    • one year ago
  41. eli123
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    ARE YOU POSITIVE?

    • one year ago
  42. aaronq
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    well unless they want you to calculate using Mg (s) + HCl -> H2 + MgCl (aq)

    • one year ago
  43. eli123
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    how would that be?

    • one year ago
  44. aaronq
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    this would be the theoretical way you balance the equation and use diminutional analysis, but really your answer is gonna be like off by very little, also, because of rounding

    • one year ago
  45. aaronq
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    did they give you the concentration of the acid?

    • one year ago
  46. eli123
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    no

    • one year ago
  47. eli123
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    all we had to do was place a piece of magnesium inside a buret that contained hcl and this was then converted and all the hcl migrated down to the bottom of the buret

    • one year ago
  48. eli123
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    basically the magnesium was reacting with the hcl

    • one year ago
  49. eli123
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    and after that we had to find the values in that table

    • one year ago
  50. aaronq
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    hm well they really wouldn't need to tell you, do you know what a limiting reagent is?

    • one year ago
  51. eli123
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    limiting reactant?

    • one year ago
  52. aaronq
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    yep, it's the same thing. you could use the number of moles of that and translate it to moles of H2 produced 2Mg (s) + 2HCl -> H2 + 2MgCl (aq) 0.001439 moles of Mg/2 = x moles of H2/1 moles of H2 = 0.0028794734677088 thats the theoretical yield

    • one year ago
  53. eli123
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    well that makes more sense because thats what we are currently learning in class haha

    • one year ago
  54. aaronq
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    oh wait sorry i multiplied, it should be moles of H2 = 0.0007198683669272

    • one year ago
  55. aaronq
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    haha oh true, well maybe let me know next time? haha

    • one year ago
  56. eli123
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    sorry!!!

    • one year ago
  57. eli123
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    so in other words, to find lets say trial #1

    • one year ago
  58. eli123
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    where the volume of hydrgen gas produced was 37.00mL and the moles of h2 gas produced were 0.002 mol

    • one year ago
  59. aaronq
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    you would convert the mass of Mg to moles then using the balanced equation, build a relationship between H2 and Mg using their respective coefficients.. this just means: moles of Mg/coefficient=moles of H2/coefficient

    • one year ago
  60. eli123
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    oh my god

    • one year ago
  61. eli123
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    moles of mg that reacted?

    • one year ago
  62. eli123
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    coefficient is 2?

    • one year ago
  63. aaronq
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    yeah because thats your limiting reactant..

    • one year ago
  64. aaronq
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    for Mg is 2 for H2 is 1

    • one year ago
  65. aaronq
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    wait no i balanced it wrong because it's Mg + 2HCl -> MgCl2 + H2

    • one year ago
  66. aaronq
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    so it's coefficient is 1 for both

    • one year ago
  67. aaronq
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    sorry, i'm like lost today idk why

    • one year ago
  68. eli123
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    i dont get it

    • one year ago
  69. aaronq
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    Mg (s) + 2HCl -> H2 + MgCl2 0.001439 moles of Mg/1 = x moles of H2/1 moles of H2 = 0.001439 thats the theoretical yield which is close to what you would get if you just divided the volume of gas produced by 22.414 L

    • one year ago
  70. eli123
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    thats the ratio?

    • one year ago
  71. aaronq
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    yep the ratio is 1:1

    • one year ago
  72. eli123
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    but what value would i put?

    • one year ago
  73. aaronq
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    i would go with the theoretical since thats what you're doing in class

    • one year ago
  74. eli123
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    0.001439?

    • one year ago
  75. aaronq
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    yeah, even round up to 0.0015

    • one year ago
  76. aaronq
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    or down to 0.0014

    • one year ago
  77. eli123
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    ok wait where did the 0.001439 come from>?

    • one year ago
  78. aaronq
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    its the number of moles of magnesium

    • one year ago
  79. eli123
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    ooooh yes thats my 0.001

    • one year ago
  80. aaronq
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    haha yep, i mean round to whatever you think it's best, or whatever they're teaching you at school

    • one year ago
  81. eli123
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    but for the ratio isnt moles of h2 divided by moles of magnesium? so 0.002h2/0.001mg

    • one year ago
  82. aaronq
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    which ratio?

    • one year ago
  83. eli123
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    ratio of volume displaced to moles h2 gas prodced

    • one year ago
  84. aaronq
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    well the amount of moles of H2 produced is equal to the moles of Mg that reacted according to the stoichiometric ratios of the equation... so to answer your question it would be 37 mL/0.001 moles.. which is what you had before, if i remember correctly

    • one year ago
  85. eli123
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    37 being the volume of hydrogen gas produced and 0.001 moles of magnesium reacted?

    • one year ago
  86. aaronq
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    well, technically yes, but the question is asking for the volume displaced divided by the moles of H2 produced..which are the same values

    • one year ago
  87. eli123
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    oh so volume of liquid displaced divided by the moles of h2

    • one year ago
  88. eli123
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    and thats the ratio?

    • one year ago
  89. eli123
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    wow i feel stupid

    • one year ago
  90. aaronq
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    that they're asking you for in the question yes. They're basically seeing how close that value is to the ideal gas constant volume of 1 mole =22.414 L, since 37mL/0.0014 mol = 26428.5714285714285714 mL/mol 26.428 L/mol so it's close if you use more than 0.001, which i think you should do lol

    • one year ago
  91. eli123
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    oh my i get the big number again hahaha

    • one year ago
  92. aaronq
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    haha don't feel stupid

    • one year ago
  93. aaronq
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    your friend in the other thread used 0.00144 moles

    • one year ago
  94. eli123
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    0.0014 is in the correct significant figures? i keep using 0.001

    • one year ago
  95. aaronq
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    hm idk though if you don't use that your error is pretty large

    • one year ago
  96. eli123
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    exactly

    • one year ago
  97. eli123
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    but look when i divide 0.035/ 24.31

    • one year ago
  98. aaronq
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    0.0014397367338544

    • one year ago
  99. eli123
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    and the right sig figures are four so you round it to 0.001?

    • one year ago
  100. aaronq
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    zeros dont count though

    • one year ago
  101. aaronq
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    well zeros on the left side

    • one year ago
  102. eli123
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    oh..

    • one year ago
  103. eli123
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    still..

    • one year ago
  104. eli123
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    so now its 3 sig figures?

    • one year ago
  105. aaronq
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    i would use 2 sig digits b/c 0.035 .. and have 0.0014

    • one year ago
  106. eli123
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    so only 14 are significant

    • one year ago
  107. aaronq
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    yes

    • one year ago
  108. eli123
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    what about 0.016/24.31

    • one year ago
  109. eli123
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    0.0066?

    • one year ago
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