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  • #1280

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    A-Level Chemistry Tuition Singapore/H2 Chemistry Tuition/JC Chemistry Tutor

    Please post your A-Level Chemistry/H2/H1/JC questions to Mr Ong and Alvin.

    From A-Level Chemistry Tuition Singapore – Tutor

    #1282

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    A-Level Chemistry Tuition Singapore/H2 Chemistry Tuition/JC Chemistry Tutor

    Please be inform that we have a class specially dedicated to A-Level H1 Chemistry Tuition Singapore to help the stucents to ACE H1 Chemistry

    Sunday : 12 pm to 2 pm

    From A-Level Chemistry Tuition Singapore – Tutor

    #1290

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    Hi Wood Yan

    SAJC 2009 Prelim P3 Q5c
    PJC 2008 Prelin P3 Q3d

    Please go to the resource to download the solutions

    From A-level Chemistry Tuition Tutor Singapore

    #1360

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    Hi A-Level/JC Chemistry Tuition Students – Singapore

    For the same molecular formula, the cis isomer has a higher boiling point than the trans isomer.

    Reasons:

    _ The cis isomer is polar while the trans isomer is non-polar.

    _ The polar cis isomer has permanent dipole- permanent dipole forces of
    attraction between its molecules while the non-polar trans isomer has weak induced dipole-induced dipole forces of attraction between its molecules.

    _ More energy is required to overcome the stronger permanent dipolepermanent dipole forces between the molecules of cis isomer.

    _ Hence, the cis isomer has a higher boiling point than the trans isomer.

    Why is the melting point of the cis isomers lower?

    You might have thought that the same argument would lead to a higher melting point for cis isomers as well, but there is another important factor operating.

    In order for the intermolecular forces to work well, the molecules must be able to pack together efficiently in the solid.

    Trans isomers pack better than cis isomers. The “U” shape of the cis isomer doesn’t pack as well as the straighter shape of the trans isomer.
    The poorer packing in the cis isomers means that the intermolecular forces aren’t as effective as they should be and so less energy is needed to melt the molecule – a lower melting point.

    #1364

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    A-Level Chemistry Tuition Singapore/H2 Chemistry Tuition/JC Chemistry Tutor

    Hi JC/A-Level Chemistry Tuition Students Singapore

    Good News!, there is a lesson on Sunday 25 March 8.30am to 10 am on Ionic Equilibrium.

    Come and join us. It is FOC

    From A-Level Chemistry Tuition Singapore Tutor

    #1443

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    A-Level Chemistry Tuition Singapore/H2 Chemistry Tuition/JC Chemistry Tutor

    Hi JC/A-Level Chemistry Tuition Students Singapore

    Good News!, there is a lesson on Sunday 6 March 1 pm to 2.30 pm on Ksp

    Come and join us. It is FOC

    From A-Level Chemistry Tuition Singapore Tutor

    #1452

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    Which is a stronger acid, phenol or 2,4,6-trichlorophenol? why?

    2,4,6-trichlorophenol is more acdic because the Cl groups are electron with-drawing and will make the O more electron deficient, so the deprotonation of the O will be more favorable

    From A-Level Chemistry Tuition Singapore Tutor

    #1453

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    Which one is more soluble, 2-nitrophenol or phenol? Why?

    2-nitrophenol is less soluble due to the intra molecular hydrogen bonding.

    From A-Level Chemistry Tuition Singapore Tutor

    #1532

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    Hi JC1 Students

    This week we will be doing some quick check on the understanding of AMS, Redox, Chemical Bonding and Gases.

    Cheers!

    Alvin and Mr Ong – A-level/JC/H2 Chemistry Tutors

    #1533

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    Hi JC2 Students

    This week we will complete Nitrogen Compounds. Continue to work hard on Organic Chemistry. Remember the mind mapping.

    Will be starting revision on the Physical Chemistry.

    Cheers!

    Alvin and Mr Ong – A-level/JC/H2 Chemistry Tutors

    #1536

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    Hi Yan Yan

    Solution for Atomic Structure
    Qns 22 NYJC 09/4

    Q26 PJC 09/H1/4
    For Cl the 4s orbital is filled before the 3d orbitals
    For Cr and any transition metals the 3d orbitals lie at the lower energy tha 4s orbitals
    Ans B

    Q28 SAJC 09/4
    Element X is in Group III while element Y is in Group VI. Hence the formula formed between X and Y is X2Y3.
    Ans B

    Q42 Nov 09/4
    1st & 2nd IE of Al = 577 + 1820 = 2397 kJ mol-1
    1st & 2nd IE of:
    (A) Co = 957 + 1640 = 2397 kJ mol-1
    (B) Cr = 653 + 1590 = 2243 kJ mol-1
    (C) Cu = 745 + 1960 = 2705 kJ mol-1
    (D) Ni = 736 + 1750 = 2486 kJ mol-1
    Ans A

    Q44 DHS 09/3
    The greatest jump is between the 6th and 7th I.E. Hence element X must be in group VI. Therefore, either O or Te is the answer. But with reference to the Data Booklet, the I.E. values do not correspond to that of oxygen. Hence, element X is Te.
    Ans D

    Q48 MI 09/3

    Q50 ACJC 09/5
    A: It is a Group II element as the big increase from its 2nd to 3rd IE suggests that the 3rd electron is removed from an inner shell.
    Since it is in Period 2 Group II, the element is Be.
    B: BeO is an amphoteric oxide (diagonal relationship with Al2O3)
    C: Incorrect statement as BeCl2 is a simple covalent molecule.
    D: Be, being a metal, can conduct electricity both in solid and molten states due to the presence of mobile delocalised electrons.
    Ans C

    Q54 SRJC 09/H1/4
    E: Nitrogen; F: Oxygen
    1st IE of N: N(1s22s22p3) → N+ (1s22s22p2)
    1st IE of O: O(1s22s22p4) → O+ (1s22s22p3)
    Conclusion: The 1st IE of O is lower than the 1st IE of N as less energy is required to remove the 2p electron from the paired electrons in 2p orbital (inter-electronic repulsion). E > F.
    2nd IE of N: N+ (1s22s22p2) → N2+ (1s22s22p1)
    2nd IE of O: O+(1s22s22p3) → O2+ (1s22s22p2)
    Conclusion: The 2nd IE of O is higher than the 2nd IE of N as both involves the removal of 2p electron from singly filled p-orbital. However, the removal of the 2p electron from O+ will experience a greater electrostatic forces of attraction from the nucleus than that of N+ as O+ has greater nuclear charge (+8) than N+ (+7). E < F
    Ans D

    Q64 NJC09/32

    #1587

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    Hi All A-Level/JC/H2/H1 Chemistry Students

    Starting 28 May, we are conducting Chemistry Revision Programme.
    Please refer to the Schedule for more detials.

    We can also customise the schedule to suit you

    From
    A-Level/JC/H2/H1 Chemistry Tutors

    #1797

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    A-Level Chemistry Tuition Singapore/H2 Chemistry Tuition/JC Chemistry Tutor

    Hi All A-Level/JC/H2/H1 Chemistry Students

    We had completed first week organic revision on organic reactions and mechanism. Physical Chemistry on Atoms Molecules Stoichiometry, Atomic Structure, Gases and Chemical Energetics also been covered in the first week.

    Please contact the tutors if you have more questions.

    Good News, if you have more questions on Chemical Energetics please attend 13th June Monday 8am to 10am. No charge in this session

    From
    A-Level/JC/H2/H1 Chemistry Tutors

    #1821

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    A-Level Chemistry Tuition Singapore/H2 Chemistry Tuition/JC Chemistry Tutor

    Hi All A-Level/JC/H2/H1 Chemistry Students

    We had completed second week organic revision on Amino Acids and Proteins. Next week we will be revising Chemical Bonding on Monday, Reactions Kinetics on Tuesday, Chemical Eqm on Wednesday, Ionic Eqm on Thursday, Electrochemistry on Friday

    Please contact the tutors if you have more questions.

    From Alvin and Mr Ong

    #1873

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    A-Level Chemistry Tuition Singapore/H2 Chemistry Tuition/JC Chemistry Tutor

    Hi All A-Level/JC/H2/H1 Chemistry Students

    Saytzeff’s rule

    The more substituted alkene is the more stable alkene because
    1) Alkyl groups are electron donating and hence they intensify the electron cloud of the double bond, strengthening it.

    2) Alkyl groups are sterically large, hence they would be the most stable if they are further away from each other.

    However it must be noted that there are exceptions to Saytzeff’s rule, which is Hofmann’s rule. Hofmann states that the least substituted alkene is the most stable alkene. But for the purpose of your syllabus, stick to what Saytzeff says.

    From Alvin

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