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    (Original post by imnoteinstein)
    wb when they're different? :O
    Then, they will specify what molar enthalpy of neutralization they want you to find. eg. they will have to say : Find the molar enthalpy of neutralization of KOH.

    In this case, divide by the moles of KOH.
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    (Original post by Sandy_Vega30)
    Then, they will specify what molar enthalpy of neutralization they want you to find. eg. they will have to say : Find the molar enthalpy of neutralization of KOH.

    In this case, divide by the moles of KOH.
    thanks alott
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    (Original post by imnoteinstein)
    thanks but i think the reason is that c2h5 doesnt exist haha! nothing suggests which ones molecular but the fact that c2h5 isnt a thing :'
    OMG! I didn't even notice that! :facepalm:
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    (Original post by Sandy_Vega30)
    OMG! I didn't even notice that! :facepalm:
    sorry for so many questions but here
    https://110d0b1a7b584219c71ba4cfe896...0Chemistry.pdf
    Q22b iii, why do we multiply by 2 to find enthalpy change of reaction/ mole?
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    3 A hydrocarbon contains, by mass, 82.7% carbon and 17.3% hydrogen.The molecular formula of the hydrocarbon is A CH3
    1. B C2H6
    2. C C2H5
    3. D C4H10
      I found the empirical formula and its C, but why the answer is D??Why the empirical formula just be same as the molecular?
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    Just a quick question, what is an acceptable definition of a sub shell?
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    Which of the following enthalpy changes cannot be measured directly by experiment?
    The enthalpy change of
    1. A formation of methane.
    2. B combustion of hydrogen.
    3. C formation of carbon dioxide.
    4. D combustion of carbon monoxide.
    why A?
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    (Original post by zahragoli97)
    3 A hydrocarbon contains, by mass, 82.7% carbon and 17.3% hydrogen.The molecular formula of the hydrocarbon is A CH3
    1. B C2H6
    2. C C2H5
    3. D C4H10
      I found the empirical formula and its C, but why the answer is D??Why the empirical formula just be same as the molecular?
    The empirical formula is C (molecular mass = 29). The molecular formula may have been D because the molecular mass may have been 58.
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    (Original post by zahragoli97)
    3 A hydrocarbon contains, by mass, 82.7% carbon and 17.3% hydrogen.The molecular formula of the hydrocarbon is A CH3
    1. B C2H6
    2. C C2H5
    3. D C4H10
      I found the empirical formula and its C, but why the answer is D??Why the empirical formula just be same as the molecular?
    its D because c2h5 doesnt exist
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    1. In an experiment, 3.425 g of lead oxide was reduced to form 3.105 g of lead.The empirical formula of the lead oxide is A PbOB Pb3O2C Pb3O4D Pb4O3
    how to do this question??
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    (Original post by imnoteinstein)
    its D because c2h5 doesnt exist
    aah i see lol
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    https://110d0b1a7b584219c71ba4cfe896...0Chemistry.pdf
    answers for with explanation please i know i am bet late but please
    Q4
    q8
    q9
    q10
    q11
    q14
    q16
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    (Original post by zahragoli97)
    Which of the following enthalpy changes cannot be measured directly by experiment?
    The enthalpy change of
    1. A formation of methane.
    2. B combustion of hydrogen.
    3. C formation of carbon dioxide.
    4. D combustion of carbon monoxide.
    why A?
    its a how science works question.
    the reason is the enthalpy of combustion of all hydrocarbons can not be measured directly.
    and its A cause methane is a hydrocarbon.
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    (Original post by zahragoli97)
    3 A hydrocarbon contains, by mass, 82.7% carbon and 17.3% hydrogen.The molecular formula of the hydrocarbon is A CH3
    1. B C2H6
    2. C C2H5
    3. D C4H10
      I found the empirical formula and its C, but why the answer is D??Why the empirical formula just be same as the molecular?
    Refer to page 2. The same question was discussed earlier.
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    (Original post by katenell)
    https://110d0b1a7b584219c71ba4cfe896...0Chemistry.pdf
    answers for with explanation please i know i am bet late but please
    Q4
    q8
    q9
    q10
    q11
    q14
    q16
    Q4 B has an unpaired electron in the 2p subshell. So it's further away from the nucleus than Be's electrons (which are in the 2s subshell and are closer to the nucleus).
    Q8 88/44= 2 mol of CO2 27/18= 1.5 mol of water therefore 4 mol of CO2 and 3 mol of water. So from this you can see that there will be 4 carbons and 6 hydrogens so the answer is C, C4H6
    Q9 This is a definition you just have to learn, Avagadro constant refs to atoms in one mol of any monatomic element
    Q10 8.5/(14+3) = 0.5 mol 15/30= 0.5 mol so 0.5 mol forms 0.5 mol so its 100%
    Q11 55/1000 * 0.2 = 0.011 mol... 0.011*143.4 = 1.58g
    Q14 draw them out with displayed formula and you will see why its C2F4
    Q16 draw the hess's cycle then you will see that the calculation will be -572-(-484) = -88. But this is for 2 mol so for 1 mol its -44 KJmol-1
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    (Original post by Bliss_)
    its a how science works question.
    the reason is the enthalpy of combustion of all hydrocarbons can not be measured directly.
    and its A cause methane is a hydrocarbon.
    A is "Formation of Methane"., not combustion. Now, how would you explain this?
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    (Original post by imnoteinstein)
    sorry for so many questions but here
    https://110d0b1a7b584219c71ba4cfe896...0Chemistry.pdf
    Q22b iii, why do we multiply by 2 to find enthalpy change of reaction/ mole?
    You have to multiply by 2 because there are 2 moles of NH4CNS being used in the equation.
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    (Original post by :D :))
    Q4 B has an unpaired electron in the 2p subshell. So it's further away from the nucleus than Be's electrons (which are in the 2s subshell and are closer to the nucleus).
    Q8 88/44= 2 mol of CO2 27/18= 1.5 mol of water therefore 4 mol of CO2 and 3 mol of water. So from this you can see that there will be 4 carbons and 6 hydrogens so the answer is C, C4H6
    Q9 This is a definition you just have to learn, Avagadro constant refs to atoms in one mol of any monatomic element
    Q10 8.5/(14+3) = 0.5 mol 15/30= 0.5 mol so 0.5 mol forms 0.5 mol so its 100%
    Q11 55/1000 * 0.2 = 0.011 mol... 0.011*143.4 = 1.58g
    Q14 draw them out with displayed formula and you will see why its C2F4
    Q16 draw the hess's cycle then you will see that the calculation will be -572-(-484) = -88. But this is for 2 mol so for 1 mol its -44 KJmol-1
    thank youu but i didnot understand q10
    and i there a faster way for q 14 or is there any rule ?
    Q16 how the cycle should be drawn ?/
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    (Original post by Sandy_Vega30)
    A is "Formation of Methane"., not combustion. Now, how would you explain this?
    Yep I was thinking that too
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    (Original post by Sandy_Vega30)
    You have to multiply by 2 because there are 2 moles of NH4CNS being used in the equation.
    you are smart
    could you please answer my question
 
 
 
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