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    (Original post by imnoteinstein)
    i now realise how dumb the question was - thankyou!
    No Problem.
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    Can anyone explain to me how to do this question??
    1. A 27.0 g sample of an unknown hydrocarbon, CxHy, was burned completely in excessoxygen to form 88.0 g of carbon dioxide and 27.0 g of water. [Molar masses / g mol–1: CO2 = 44; H2O = 18]
      Which of the following is a possible formula of the unknown hydrocarbon?
    A CH4
    B C2H6
    C C4H6
    D C6H6
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    Hello guys! I would really appreciate if someone could give Chemrevise Unit 1 (pdf) notes! Cannot open the site for some reason :c
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    (Original post by Bliss_)
    Attachment 535087
    hey
    yeah i'm doing it on the same date as you are
    lets just hope its easy
    on a completely random note can anyone tell me why the answer is C and not A
    don't all ionic cpds have little covalent character?
    Not all ionic compounds have a little covalent character. The fluorine anion in the above example is so electronegative and so small that it holds onto its electron cloud tightly. So, it is very very difficult to polarize.
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    (Original post by IAROX15)
    Not all ionic compounds have a little covalent character. The fluorine anion in the above example is so electronegative and so small that it holds onto its electron cloud tightly. So, it is very very difficult to polarize.
    oh, i see. thanks man that helped clear up my concept.
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    (Original post by zahragoli97)
    Can anyone explain to me how to do this question??
    1. A 27.0 g sample of an unknown hydrocarbon, CxHy, was burned completely in excessoxygen to form 88.0 g of carbon dioxide and 27.0 g of water. [Molar masses / g mol–1: CO2 = 44; H2O = 18]
      Which of the following is a possible formula of the unknown hydrocarbon?
    A CH4
    B C2H6
    C C4H6
    D C6H6
    Find the moles of CO2 and H20 which is 2 and 1.5 respectively. 1 mole of CxHy gives 2 moles of CO2 therefore to balance the equation the hydrocarbon must have 2 carbons. Therefore the obvious choice from the options is B.
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    (Original post by Sandy_Vega30)
    Find the moles of CO2 and H20 which is 2 and 1.5 respectively. 1 mole of CxHy gives 2 moles of CO2 therefore to balance the equation the hydrocarbon must have 2 carbons. Therefore the obvious choice from the options is B.
    (Original post by zahragoli97)
    Can anyone explain to me how to do this question??
    1. A 27.0 g sample of an unknown hydrocarbon, CxHy, was burned completely in excessoxygen to form 88.0 g of carbon dioxide and 27.0 g of water. [Molar masses / g mol–1: CO2 = 44; H2O = 18]
      Which of the following is a possible formula of the unknown hydrocarbon?
    A CH4
    B C2H6
    C C4H6
    D C6H6
    Hey guys, I think the answer is C because if you use Sandy's method you would have to multiply the mole ratio by two to get whole numbers because the number of Moles of H20 is 1.5 hence the moles of carbon will be 4 mol. The only option is C then because there are 4 carbon atoms.. I'm not sure so pls correct me if I am wrong.:excited:
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    (Original post by wenogk)
    Hey guys, I think the answer is C because if you use Sandy's method you would have to multiply the mole ratio by two to get whole numbers because the number of Moles of H20 is 1.5 hence the moles of carbon will be 4 mol. The only option is C then because there are 4 carbon atoms.. I'm not sure so pls correct me if I am wrong.:excited:
    I agree with u i got the same answer and also Thanku @Sandy your explanations to previous questions are superb :congrats:
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    (Original post by wenogk)
    Hey guys, I think the answer is C because if you use Sandy's method you would have to multiply the mole ratio by two to get whole numbers because the number of Moles of H20 is 1.5 hence the moles of carbon will be 4 mol. The only option is C then because there are 4 carbon atoms.. I'm not sure so pls correct me if I am wrong.:excited:
    This is the equation showing moles of CO2 and H2O:
    CxHy + 5.5 O2 ------> 2CO2 + 1.5H2O

    This is the equation after multiplying by 2:

    2CxHy + 11O2 ----> 4CO2 + 3H2O

    Now, for every 2 moles of hydrocarbon, there are 4 moles of CO2 produced. Try balancing out the Carbons. The only way to produce 4 carbons on both sides is if x were 2. Hence the hydrocarbon is C2H6. I still think the answer is B. zahragoli97 Could you please confirm?
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    (Original post by Saad69)
    I agree with u i got the same answer and also Thanku @Sandy your explanations to previous questions are superb :congrats:
    Thank you!
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    (Original post by Sandy_Vega30)
    This is the equation showing moles of CO2 and H2O:
    CxHy + 5.5 O2 ------> 2CO2 + 1.5H2O

    This is the equation after multiplying by 2:

    2CxHy + 11O2 ----> 4CO2 + 3H2O

    Now, for every 2 moles of hydrocarbon, there are 4 moles of CO2 produced. Try balancing out the Carbons. The only way to produce 4 carbons on both sides is if x were 2. Hence the hydrocarbon is C2H6. I still think the answer is B. zahragoli97 Could you please confirm?
    (Original post by Saad69)
    I agree with u i got the same answer and also Thanku @Sandy your explanations to previous questions are superb :congrats:
    The answer is C it is the 8th question of the IAL May 2014 paper
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    (Original post by wenogk)
    The answer is C it is the 8th question of the IAL May 2014 paper
    Oh! Okay then. My bad. Sorry.
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    (Original post by imnoteinstein)
    https://110d0b1a7b584219c71ba4cfe896...0Chemistry.pdf
    Q9, answers A but why is the temp change the same? if e=mct then shouldnt the heat evolved be more bc the "mass" (volume) has increased?
    Definitely you have a valid point. However i take it to be like this. Each of the neutralization reactions can be represented by the same ionic equation. Enthalpy of neutralization is the enthalpy change when one water is formed. IF both have the same concentration and same volume then the number of moles is the same hence temperature change remains the same.
    I.E 50cm^3 2 moldm-3 concentration for both means that the moles of each is 0.1 mol.
    20cm^3 2moldm-3 concentration for both means that the moles of each is 0.04 mol.

    In both cases mol ratio is 1 is to 1 hence delta T is the answer.

    Hope that helped
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    (Original post by zahragoli97)
    Which of the following quantities, used in the calculation of the lattice energy oflithium oxide, Li2O, has a negative value?
    1. A The enthalpy change of atomization of lithium.
    2. B The first ionization energy of lithium.
    3. C The first electron affinity of oxygen.
    4. D The second electron affinity of oxygen
    the answer is C can anyone explain to me why?
    The answer is C because when the oxygen atom gains the electron for the first time, energy is given out hence it is negative. When the second electron needs to be gained by the oxygen, work has to be done due to increased electron to electron repulsion hence must put in energy. When you put in energy, it means an endothermic process. Hence a positive value. Hope that helps.
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    If anyone has some good revision notes now would be the best time to share them
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    Can anyone solve and explain part B? June 2013 paper

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    could anyone explain this to me and the way i should write it in the exam ?
    in a period and when the question asks
    1)why the trend of IE i increasing across the period ?
    2)why between to element it decreases ?
    3)why melting / and boiling point increases across period?
    4 )why between two element it decreases ?
    5) why it increases after this element ?
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    (Original post by Sandy_Vega30)
    No Problem.
    could you please answer my question
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    could anyone answer my question
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    (Original post by katenell)
    could anyone explain this to me and the way i should write it in the exam ?
    in a period and when the question asks
    1)why the trend of IE i increasing across the period ?
    2)why between to element it decreases ?
    3)why melting / and boiling point increases across period?
    4 )why between two element it decreases ?
    5) why it increases after this element ?
    1 IE increases across the group due to reduction in radius size hence stronger nuclear attraction.
    2. I don't understand this. Could you be a bit more specific.
    3. the same answer as 1. AN increase in the nuclear attraction due to reduction in the size of radius. This means that the attraction is much harder to break and hence more heat energy required.
    4 and 5. I don't understand the question. What decreases and increases?
 
 
 
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