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eli123

  • 2 years ago

can someone please help me find these values?

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  1. eli123
    • 2 years ago
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  2. eli123
    • 2 years ago
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    @phi

  3. eli123
    • 2 years ago
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    @robtobey

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

  5. aaronq
    • 2 years ago
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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.

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

  7. aaronq
    • 2 years ago
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    the other way around

  8. eli123
    • 2 years ago
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    oh ok and for h2? @aaronq

  9. aaronq
    • 2 years ago
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    you have to find the volume first

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

  11. aaronq
    • 2 years ago
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    unless you made other gases in your experiment

  12. aaronq
    • 2 years ago
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    yes

  13. eli123
    • 2 years ago
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    no i didnt

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

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

  16. aaronq
    • 2 years ago
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    otherwise you have to know the density of the gas

  17. eli123
    • 2 years ago
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    therefore, how would the equation look?

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

  19. aaronq
    • 2 years ago
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    make sure you have the same units

  20. eli123
    • 2 years ago
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    oooh

  21. eli123
    • 2 years ago
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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?

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

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

  24. aaronq
    • 2 years ago
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    a ratio is just a division volume of gas/moles of gas

  25. eli123
    • 2 years ago
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    so volume of h2/moles of h2?

  26. eli123
    • 2 years ago
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    @aaronq

  27. aaronq
    • 2 years ago
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    yeah

  28. eli123
    • 2 years ago
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    @aaronq i get a huge number

  29. eli123
    • 2 years ago
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    37.00ml/0.002

  30. aaronq
    • 2 years ago
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    yep, thats what the want

  31. eli123
    • 2 years ago
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    18500?

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

  33. aaronq
    • 2 years ago
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    mL/mol

  34. eli123
    • 2 years ago
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    so its right?

  35. eli123
    • 2 years ago
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    @aaronq

  36. aaronq
    • 2 years ago
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    to what they're asking you, yes

  37. eli123
    • 2 years ago
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    but its such a big number? @aaronq

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

  39. eli123
    • 2 years ago
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    18.5 L DAMM

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

  41. eli123
    • 2 years ago
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    ARE YOU POSITIVE?

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

  43. eli123
    • 2 years ago
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    how would that be?

  44. aaronq
    • 2 years ago
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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

  45. aaronq
    • 2 years ago
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    did they give you the concentration of the acid?

  46. eli123
    • 2 years ago
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    no

  47. eli123
    • 2 years ago
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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

  48. eli123
    • 2 years ago
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    basically the magnesium was reacting with the hcl

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

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

  51. eli123
    • 2 years ago
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    limiting reactant?

  52. aaronq
    • 2 years ago
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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

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

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

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

  56. eli123
    • 2 years ago
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    sorry!!!

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

  58. eli123
    • 2 years ago
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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

  59. aaronq
    • 2 years ago
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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

  60. eli123
    • 2 years ago
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    oh my god

  61. eli123
    • 2 years ago
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    moles of mg that reacted?

  62. eli123
    • 2 years ago
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    coefficient is 2?

  63. aaronq
    • 2 years ago
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    yeah because thats your limiting reactant..

  64. aaronq
    • 2 years ago
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    for Mg is 2 for H2 is 1

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

  66. aaronq
    • 2 years ago
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    so it's coefficient is 1 for both

  67. aaronq
    • 2 years ago
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    sorry, i'm like lost today idk why

  68. eli123
    • 2 years ago
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    i dont get it

  69. aaronq
    • 2 years ago
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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

  70. eli123
    • 2 years ago
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    thats the ratio?

  71. aaronq
    • 2 years ago
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    yep the ratio is 1:1

  72. eli123
    • 2 years ago
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    but what value would i put?

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

  74. eli123
    • 2 years ago
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    0.001439?

  75. aaronq
    • 2 years ago
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    yeah, even round up to 0.0015

  76. aaronq
    • 2 years ago
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    or down to 0.0014

  77. eli123
    • 2 years ago
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    ok wait where did the 0.001439 come from>?

  78. aaronq
    • 2 years ago
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    its the number of moles of magnesium

  79. eli123
    • 2 years ago
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    ooooh yes thats my 0.001

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

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

  82. aaronq
    • 2 years ago
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    which ratio?

  83. eli123
    • 2 years ago
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    ratio of volume displaced to moles h2 gas prodced

  84. aaronq
    • 2 years ago
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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

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

  86. aaronq
    • 2 years ago
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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

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

  88. eli123
    • 2 years ago
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    and thats the ratio?

  89. eli123
    • 2 years ago
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    wow i feel stupid

  90. aaronq
    • 2 years ago
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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

  91. eli123
    • 2 years ago
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    oh my i get the big number again hahaha

  92. aaronq
    • 2 years ago
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    haha don't feel stupid

  93. aaronq
    • 2 years ago
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    your friend in the other thread used 0.00144 moles

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

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

  96. eli123
    • 2 years ago
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    exactly

  97. eli123
    • 2 years ago
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    but look when i divide 0.035/ 24.31

  98. aaronq
    • 2 years ago
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    0.0014397367338544

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

  100. aaronq
    • 2 years ago
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    zeros dont count though

  101. aaronq
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    well zeros on the left side

  102. eli123
    • 2 years ago
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    oh..

  103. eli123
    • 2 years ago
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    still..

  104. eli123
    • 2 years ago
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    so now its 3 sig figures?

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

  106. eli123
    • 2 years ago
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    so only 14 are significant

  107. aaronq
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    yes

  108. eli123
    • 2 years ago
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    what about 0.016/24.31

  109. eli123
    • 2 years ago
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    0.0066?

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