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順便說出原因......

因為我個ANSWER時不時都出錯......


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[ 本帖最後由 奧得 於 2014-4-22 12:58 編輯 ]

Q9 : A
- burette : rinse with titrant + water
- pipette : rinse with titrand + water
- conical flask : only rise with water

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Q10 : C
2Cu +C ---> 2Cu +CO2
- mole of Cu0 = 52.2/(63.5+16) = 0.6566 mol
- mole of C needed (if complete combustion) : = 0.6566/2 mol
- mass of C needed: 0.3283 *12 = 3.94g

( btw, i guess your answer take 2 sig fig of 0.66mol to cal , so ans will be 3.96g)

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Q11: A
- NH3 : trigonal pyramidal
- BF3:  trigonal plannar

so answer must be 1 only

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Q? :C

-if each drop if NaOH is same , temperature rise should alse be the same
- neutralization is exothermic reaction, so temp rise
- after added equal amount of NaOH (=25cm3) , REACTION IS COMPLETE
- but the temp is higher than room temp, so temp drop to 25 C


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我今年DSE，唔明再問

[ 本帖最後由 johnson 於 2014-4-22 15:19 編輯 ]

果題Q？應該係D。字數

無可能，after complete reacted , why the temp will drop sharply?
can you explain? Also, how come the time for temp drop is much faster than temp rise?

[ 本帖最後由 johnson 於 2014-4-22 16:07 編輯 ]

Q9:A
burette&pipette rinse with water to remove impurites, rinse with titrant to ensure the concentration of the titrant unchange
conical flask should rinse with water only, add titrant will affect the result as more titrant instead of 25cm^3 is used
Q10:C
Q11:A
3 may be correct as its central atom bond with 3 other atom & a lone pair electrons.
If you say more specific, 3 is incorrect as it is trigonal pyramidal.
Q?:A
as 25cm^3 of NaOH is used to neutralize HCl
NaOH+HCl→NaCl+H2O
the volume of NaOH=Volume of HCl
It is a complete neutralization, temperature rise without drop
therefore, the final reading must not be straight
Actually, it is hard to tell you the answer as temperature rise, cooler NaOH is added. Although heat is generate, as the volume(mass) of the solution change, the temperature is really hard to tell.
However, if you calculate on each 2.5 cm^3 with known value of standard enthaply change of neutralization of NaOH&HCl , you will found that volume increase won't lead to a temperature drop.
Thus, A is the answer if neglect heat lost to the surroundings.

First of all, after complete reacted, the curve is stop.
If you want to ask why the temperature is dropped which is beyond the question, I can tell you the answer.
Since the temperature of NaOH is lower than the solution, excess NaOH will act as a coolant so the temperature will be decrease.

yes , answer is A , i see the graph wrongly ( x axis is time)
as the Q said only 25cm3 is added , A should be the answer

Actually, it is hard to consider as certain amount of coolant is add to the system.
As temperature increase, the energy that the coolant can take will be greater.
It is hard to tell whether the heat that the coolant can take or the heat generate by neutralization is greater.Thus, most appropraite answer is the temperature rise will be smaller in the temperature rising curve.The following calculation is showing how complicated the curve is,
temperature rise for first 2.5cm^3=
(2.5/1000)(57000)=(27.5(volume)X1.0(density)/1000)(4200)(T-25)
T=26.23
temperature rise for 2nd 2.5cm^3=
142.5=(27.5/1000)(4200)(T-26.23)+(2.5/1000)(4200)(T-25)
T=27.25
I am not going to calculate all the answer but we can see that the temperature rise is decreasing.
1.23>1.02
Therefore, A is incorrect as well.
No solution for this question.

[ 本帖最後由 輝輝 於 2014-4-22 20:44 編輯 ]

建議Q？番學校
，同老師講：「你收皮啦！冇個答案岩啊，垃圾。」

Sor，我都睇錯volume，A無誤

A都係錯，你新手？
由於容量增加，升既溫度會越黎越少
所以應該係條曲線，唔係直線
A錯

樓上，那麼沒有一個答案是對的了


人家自修生啦……
另外輝輝，BF3不是trigonal pyramidal 而是 trigonal planer啦
一個是polar 另一個是non polar~

[ 本帖最後由 奧得 於 2014-4-22 21:58 編輯 ]

我講NH3,
份卷會唔會太易,唔問radical,唔問unfamiliar 3-D struture,唔問element+H2O/acid/alkaline

高材生QAQ
人家不夠時間做啦……