

Originally posted by Shikangninjagold:
Bro. At least answer my question and don't go so toxic.
Okay. I am sorry for being rude.
Let me try to answer your question.
Originally posted by Shikangninjagold:So I am a student from Hwa Chong Institution. I don't really like the school and I want to transfer to National Junior College. How do I do that? When can I do that? How will the process be like?
Thx
Regards,
Boy
Disclaimer: I am not from Hwa Chong Institution or National Junior College. I am from a neighborhood secondary school.
Hwa Chong cutoff point is 260. National Junior College cutoff point is 258. If you can qualify for Hwa Chong Institution, you can qualify for National Junior College, so you can give it a try.
Based on my experience, secondary one students can transfer school at the end of secondary one, and enrol into the new school at secondary two.
I never transfer school before, so I don't know the process. You may want to ask your school general office, or write an email to the Ministry of Education to ask them about it.
Edited by gekpohboy 23 Apr `17, 10:22PM



It's not just drug dealers he's after now.
Duterte was heard telling workers: “If you lose your job, I’ll give you one. Kill all the drug addicts.” He then said: “Help me kill addicts” and “Let’s kill addicts everyday.”
https://sg.news.yahoo.com/philippinepresidentrodrigotellsjobless091703370.html



Originally posted by gekpohboy:
First and foremost, fix up your English.
Your English like that, how come can go to those topnotch school like Hwa Chong Institution and National Junior College?
You may want to consider transferring to a neighbourhood school instead.
A 188 school... or better still, Normal Academic... it's better for you.
Bro. At least answer my question and don't go so toxic.



Cameo 23/4
SBS6575U 5 (BBDEP SP)
SG5404Z 10 (BNDEP 20)
SBS3295Y 21 (AMDEP SP)
SBS6624L 22 (AMDEP 21)
SBS8589S 45 (BNDEP 127)
SBS8550C 59 (AMDEP 265)
SBS3015P 65 (BNDEP 19)
SG5376X 65 (BNDEP SP)
SBS7459R 117 (HGDEP 80)
SG5405X 293 (BNDEP 20)SMB3605X 67 (KJDEP 190)
SMB1449T 167 (AMDEP 852)
SMB1544B 169 (AMDEP 167)
SG1034J 171 (AMDEP SP)
SMB175P 803 (AMDEP 855)
TIB1131G 811 (AMDEP SP)
SG1050L 852 (AMDEP SP)
SMB3C 853C (AMDEP 805)
SG5762P 854 (AMDEP SP)
SG5782G 854 (AMDEP 963)
SMB1326P 855 (AMDEP 963)
SG5773H 857 (AMDEP 963)
SMB1531M 860 (AMDEP 169)
SMB1631H 860 (AMDEP SP)
SG5752T 965 (AMDEP SP)
SMB3126L 969 (AMDEP 858)
SMB3166X 969 (AMDEP 859)
SMB3525T 969 (AMDEP SP)
SMB3585U 969 (AMDEP 857)
SMB1602S 980 (AMDEP 851)SBS3372H 66 (BUDEP SP)
SG5010B 66 (BUDEP SP)
SG5022S 66 (BUDEP SP)SBS6483B 3 (LYDEP SP)
SG1018G 3 (LYDEP SP)
SBS3441S 85 (LYDEP SP)
SG5109B 85 (LYDEP SP)



Rides 23/4
Taxi Bedok North Ave 4 to Chai Chee St
 Layover @Bethesda Cathedral 
SBS8589S 45 Chai Chee St to Jln Eunos
SBS6575U 5 Jln Eunos to Newton Rd
 Lunch @McDonald's United Square 
SG1182P 851 Thomson Rd to Ang Mo Kio Ave 6
SMB1338E 852 Ang Mo Kio Ave 6 to Lentor Ave
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SG1050L 852 Yishun Temp Int to Lentor Ave
SMB3166X 969 Lentor Ave to Tampines Int
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 Dinner @Fengshan FC 



Hi Carychidestar, do you want to consider joining my BedokFunland JC tuition? Personalized guidance will help you a lot more effectively than can be done through an online forum. You can PM me if you're interested.
1st qn : qn is flawed becoz the counteranion is not specified (so let's assume it's uninegative), and the molarity of sulphite solution not given (so let's assume it's also 0.1M). 2 mol of M3+ reacts with 1 mol of SO3 2, which when oxidized to SO4 2, releases 2 mol of e. Hence 2 mol of M3+ accepts 2 mol of e, ie. final OS of metal = +2.
2nd qn : each mol of FeC2O4 releases 3 mol of e upon oxidation (as each Fe2+ is oxidized to Fe3+, and each C2O4 2 is oxidized to 2 CO2). Each mol of MnO4 accepts 5 mol of e upon reduction in acidic pH. Hence each mol of FeC2O4 will react with 3/5 mol of MnO4.
Originally posted by Carychidestar:Hi ultima,how do you solve these two questions?
'50cm3 of a 0.1 mol dm3 of a solution of a metallic salt reacted exactly with 25cm3 of sulphite solution.The half equation for the oxidation of sulphite ion is
SO3^2 + H2O > SO4^2 + 2H^+ +2e
If the original exidation number of the metal in the salt was 3, calculate the new oxidation number of the metal'
I only managed to find the no.moles of the salt and I'm stucked.
'Halfequations
2CO2 + 2e > C2O4^2
FE^3+ + e > FE^2+
MNO4^ + 8H^+ + 5e > MN^2+ + 4H2O
Calculate the no.moles of MNO4^ ion that will react with 1 mol of iron(II) ethanedioate;FeC2O4 in acidic solution'
This one,I dont even know where to start since there are no numbers provided.
Edited by UltimaOnline 23 Apr `17, 8:58PM
