MC + CONVENTIONAL QUESTION

小霞的fans · 發表於 22/3/2009 01:58 PM

1. IF THE AVOGADRO CONSTANT IS L mol^-1 , how many molecules are their in 22g of carbon dioxide gas?

relative atomic mass : C=12.0 ,O=16.0

A 1/2 L
B.1L
C.2L
D.3L

我完全唔識點做 .. .

7.which of the following compounds contains the largest number of positive ions?

relative atomic masses: Ca=40.0 , C=12.0 , O=16.0 , H=1.0

A.1 MOLE OF NH4CL
B.1 MOLE OF NA2CO3
C.100G OF CaCO3
D.18G OF H20

我又係唔明點做 .. 佢N 同NA 既ATOMIC MASS都冇比..我會撞A的 ..

15.CONSIDER THE FOLLOWING EQUATION
X2(g)+3Y2(g) 變左 2XY3(g)

if 4 moles of X2 are allowed to reacr with 6moles of Y2 , what is the number of moles of XY3 formed?

A.2
B.4
C.8
D.10

題目都睇唔明 . 又係唔識做 == .. 教教ME PLS ..

conventional question ..

a sample of impure magnesium 3.00g was dissolved in excess dilute hydrochloric acid to give a solution
containing MG^2+ ions. an excess of sodium hydroxide solution was then added to percipitate all Mg^2+ ions as Mg(OH)2 (s).
the percipitate was filtered off, washed , dried and finally heated to convert all into 4.00g of MgO (s)

1. writh a balanced chemical wquation for the conversion of Mg(OH)2 TO MgO

佢個Mg(OH)2 都唔知搞過d咩變左MgO = =

識左呢題我就識下面果D架啦><唔該哂

fish · 發表於 22/3/2009 03:35 PM

原帖由 小霞的fans 於 22/3/2009 13:58 發表

 登錄/註冊後可看大圖

1. IF THE AVOGADRO CONSTANT IS L mol^-1 , how many molecules are their in 22g of carbon dioxide gas?

relative atomic mass : C=12.0 ,O=16.0

A 1/2 L
B.1L
C.2L
D.3L

不懂計數的話可留意unit, 別人要求你答Litre,又比了 L/ mol 你,你當然要找到mol才可解決題目

Number of mole = mass / molar mass
                           = 22g / (44g/mol)
                           = 0.5 mol

Amount of molecules = Avo. const. * mole
                                       = (1L/mol) * 0.5mol
                                    = 0.5L

Answer: A

7.which of the following compounds contains the largest number of positive ions?
relative atomic masses: Ca=40.0 , C=12.0 , O=16.0 , H=1.0

A.1 MOLE OF NH4CL
B.1 MOLE OF NA2CO3
C.100G OF CaCO3
D.18G OF H20

我又係唔明點做 .. 佢N 同NA 既ATOMIC MASS都冇比..我會撞A的 ..

會考大多不會每題都給你molar mass的,他們喜歡叫你看Periodic table的
再者,不給molar mass的話可能代表無需要

number of molecules, ions 都是跟avogadro's constant 的,unit是 mol-1
所以對問number of molecules, ions 的題目時很多時候都會需要轉unit去mole
還有,拜託請不要給我看到"CL=chlorine" 等嘔心物......佔分的

先由Formula看
1mole of NH4Cl.....有1mol NH4+
1mole of Na2CO3.....有2 mol Na+

100g of CaCO3...計數:
mole = mass/ molar mass
   = 100g/ (100g/mol)
   = 1 mol
Formula中只有一個Ca,所以是1mol of Ca2+

18g of H2O...不是ionic compound根本不用計,就算是auto-ionization/ self-ionization都不會多

Na+有最多mole,當然有是多粒的 positive ions了

Answer: B

15.CONSIDER THE FOLLOWING EQUATION
X2(g)+3Y2(g) 變左 2XY3(g)

if 4 moles of X2 are allowed to reacr with 6moles of Y2 , what is the number of moles of XY3 formed?

A.2
B.4
C.8
D.10

題目都睇唔明 . 又係唔識做 == .. 教教ME PLS ..

3.
考你懂不懂excess/ limited reactant的一條問題
From the equation, 1mol of X2will react with 3 moles of Y2and give 2 moles of XY3
所以
From the equation, ratio of X2:Y2= 1:3

現在我給你的材料有 4mole X2及 6 mole Y2, 我想知有多少XY3出來
Ratio of reactant we have:
ratio of X2: Y2is 2:3

跟上面的1:3比起來,我們有太多X2了(由另一個角度看就是我們沒有足夠的Y2)
所以X2是excess reagent而 Y2是limiting reagent

由於Y2是limiting agent,它會決定有多少product XY3出來
由上面的equation看:
ratio of Y2: XY3= 3:2

我們現在有6moles Y2,是3的兩倍,所以會有比equation 2倍XY3,即是4 moles,出了來

要列式的話:
Let n (mole) be the amount of XY3produced
ratio of Y2: XY3= 3/2 = 6/n

交叉雙乘, n=4

conventional question ..

a sample of impure magnesium 3.00g was dissolved in excess dilute hydrochloric acid to give a solution
containing MG^2+ ions. an excess of sodium hydroxide solution was then added to percipitate all Mg^2+ ions as Mg(OH)2 (s).
the precipitate was filtered off, washed , dried and finally heated to convert all into 4.00g of MgO (s)

1. write a balanced chemical equation for the conversion of Mg(OH)2 TO MgO

佢個Mg(OH)2 都唔知搞過d咩變左MgO = =

識左呢題我就識下面果D架啦><唔該哂

the precipitate was filtered off, washed , dried and finally heated to convert all into 4.00g of MgO (s)

你乖乖地讀一次這句好了= =

成段文中只有這位置講了Mg(OH)2--> MgO, 答案很大機會在這兒

句中只有2個reaction的機會,一是wash,另一個是heat
人家wash沒說用甚麼洗,你當然可無視wash, 而現實中的"wash"多是為cleaning,不是為reaction
只剩下一個choice - heat
heat
Mg(OH)2(s)--> MgO(s) +H2O(g)

[ 本帖最後由 fish 於 22/3/2009 03:49 PM 編輯 ]

小霞的fans · 發表於 22/3/2009 04:29 PM

原帖由 fish 於 21/3/2009 07:35 PM 發表

 登錄/註冊後可看大圖

不懂計數的話可留意unit, 別人要求你答Litre,又比了 L/ mol 你,你當然要找到mol才可解決題目

Number of mole = mass / molar mass
= 22g / (44g/mol)
...

你解釋得好清楚呀>W< 唔該哂 . .
我想問呢1mole of Na2CO3 .. 係唔係姐係2 mole 既Na react with 1 mole of CO3 就form到1mole of Na2CO3 ?

Na後面個2竟係代表2 mole ? 我唔係好明呢度 ..

fish · 發表於 22/3/2009 10:05 PM

原帖由 小霞的fans 於 22/3/2009 16:29 發表
Na後面個2竟係代表2 mole ? 我唔係好明呢度 .

那個subscript(下方的數字)是指 1mol Na2CO3含有多少Na+ 或CO32-
所以題目中的1mole of Na2CO3中就有 2 mole of Na+, 如果人家說是 2mole of Na2CO3,那就是4 moles Na+, 2mole CO32-

我想問呢1mole of Na2CO3 .. 係唔係姐係2 mole 既Na react with 1 mole of CO3 就form到 1mole of Na2CO3 ?

Chemical formula只代表那compound含有多少該元素(mole ratio),不一定代表如何制做
所以formula 代表了 1 mole Na2CO3含有2 mole Na+ 及 1 mole CO32-

制Na2CO3的方法CE應沒教過.....是個多步驟的reaction來的

看得明+能理解就可...我還怕中英夾雜對部份人來說是難明

小霞的fans · 發表於 23/3/2009 04:16 PM

唔該fish . .我明白點計數啦^^

有d reaction 既問題我唔明 . .

zinc sulphate can be perpared by the recation between
1.zinc nitrate and magnesium sulphate
2.zinc oxide and dilute sulphuric acid
3.zinc and dilute sulphuric acid

A.1 and 2 only
B.1 and 3 only
C.2 and 3 only
D.1,2,3
唔係全部都會FORM到架咩 ? 同埋呢zinc nitrate and magnesium sulphate 係solid定aqueous solution呀 ? 佢冇提到呀 ..
雖然呢2樣野都係soluble in water .. 咁佢地未落水之前唔係solid黎 ? 我唔係好明..

如是者 ..有1題類似我都唔識做 ..

Lead(ll) chloride can be prepared by the reaction between
1.lead and hydrochloric acid
2.lead(ll) oxide and hydrocloric acid
3.lead(ll) nitrate and sodium chloride

A.1 only
B.3 only
C.1 and 2 only
D.1 and 3 only

唔明既理由同上題一樣 . .唔係全部都form lead(ll) chloride的 ?
2果兩樣野係solid 定 aqueous solution ?
其實點先叫 prepare到 ?

同埋呢 .. 用咩test可以分辨到 dilute hydrochloric acid 同埋 dilute nitric acid ?
本書好似冇的 .. .

[ 本帖最後由小霞的fans 於 22/3/2009 08:18 PM 編輯 ]

fish · 發表於 24/3/2009 01:20 AM

原帖由 小霞的fans 於 23/3/2009 16:16 發表

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zinc sulphate can be perpared by the recation between
1.zinc nitrate and magnesium sulphate
2.zinc oxide and dilute sulphuric acid
3.zinc and dilute sulphuric acid

A.1 and 2 only
B.1 and 3 only
C.2 and 3 only
D.1,2,3
唔係全部都會FORM到架咩 ? 同埋呢zinc nitrate and magnesium sulphate 係solid定aqueous solution呀 ? 佢冇提到呀 ..
雖然呢2樣野都係soluble in water .. 咁佢地未落水之前唔係solid黎 ? 我唔係好明..

看第一個reaction
Assume the reaction goes completely
Zn(NO3)2(aq) + MgSO4(aq) --> ZnSO4(aq) + Mg(NO3)2(aq)
兩個product都是water soluble, 你不能直接分開它們,所以不可以用來prepare ZnSO4

你知另外兩個就即是明了
如果要求的metal可以再被oxidize (尤是Fe 2+)的話就要小心一下 nitric acid/ concentrated sulphuric acid/ HOT dilute sulphuric acid了
題目若不說明是solid/ aqueous/ liquid/ gas 的話就想想那compound的normal condition(你總不能在lab入面找到liquid/ gaseous Zn(NO3) or MgSO4吧-w-)

Lead(ll) chloride can be prepared by the reaction between
1.lead and hydrochloric acid
2.lead(ll) oxide and hydrocloric acid
3.lead(ll) nitrate and sodium chloride

A.1 only
B.3 only
C.1 and 2 only
D.1 and 3 only

PbCl2這條我沒十足的信心,但我會選B 3 only.
原因為Pb + HCl要夠熱才可react
考慮一下reactivity series, K> Na> Ca> Mg> Al> (coke)> Zn> Fe> Sn> Pb> H+> Cu> Ag> Hg> Au> Pt
要Pb跟 H+ react的話, 速度慢得很,所以要加熱
人家沒說hot HCl....我會assume room temperature

PbO + HCl mixture亦要夠熱才可react (印象中書中是沒有的,但學校應會做類似的實驗)
原因不敢肯定, CE亦沒有說過…..可無視-->我覺得應該是intermediate的問題

PbNO3+ NaCl….明顯的一個precipitation reaction

唔明既理由同上題一樣 . .唔係全部都form lead(ll) chloride的 ?
2果兩樣野係solid 定 aqueous solution ?
其實點先叫 prepare到 ?

要prepare到就要用簡易的physical method separate到....最好出兩個不同physical state的東東
簡易的physical method如filtration, evaporation, etc.
如上面的ZnSO4(aq) + Mg(NO3)2(aq)...你不能用physical,method分開兩個aqueous solution吧
但
PbNO3(aq) + 2NaCl(aq) --> PbCl2(s) + 2NaNO3(aq)
一個solid一個aqueous, filter一下再用少少凍水沖沖就很美好了 (essay 題的話filtration + rinse with little cold water 可佔分)

同埋呢 .. 用咩test可以分辨到 dilute hydrochloric acid 同埋 dilute nitric acid ?
本書好似冇的 .. .

HCl VS HNO3
想想它們的 properties就可發現有不少test可用的了
All nitrates are soluble
All halides are soluble except Pb(II), Ag
所以隨便倒些silver nitrate/ lead nitrate入兩acid中就可以了,有precipitate的是HCl

HNO3is an oxidizing acid but not HCl
直接放Copper metal 入兩acid中, 有reaction的是HNO3

長問題的話要describe一下reaction, 即reaction 時的變化, product的外表, etc.

偶好像對inorganic chem生疏了..orz

[ 本帖最後由 fish 於 24/3/2009 01:47 AM 編輯 ]

比超仔 · 發表於 24/3/2009 06:23 PM

原帖由 fish 於 24/3/2009 01:20 AM 發表

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PbCl2這條我沒十足的信心,但我會選B 3 only.
原因為Pb + HCl要夠熱才可react
考慮一下reactivity series, K> Na> Ca> Mg> Al> (coke)> Zn> Fe> Sn> Pb> H+> Cu> Ag> Hg> Au> Pt
要Pb跟 H+ react的話, 速度慢得很,所以要加熱
人家沒說hot HCl....我會assume room temperature

PbO + HCl mixture亦要夠熱才可react (印象中書中是沒有的,但學校應會做類似的實驗)
原因不敢肯定, CE亦沒有說過…..可無視-->我覺得應該是intermediate的問題

PbNO3+ NaCl….明顯的一個precipitation reaction

關於這題答案沒錯
但不對的原因主要是因為
insoluble coat formed on the surface of the solid(Pb or PbO) stops further reaction

P.S. PbCl2 is an insoluble solid

fish · 發表於 24/3/2009 11:53 PM

原帖由 比超仔 於 24/3/2009 18:23 發表

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關於這題答案沒錯
但不對的原因主要是因為
insoluble coat formed on the surface of the solid(Pb or PbO) stops further reaction

P.S. PbCl2 is an insoluble solid

對啊~忘了insoluble coating這一點啊~謝謝=]

小霞的fans · 發表於 26/3/2009 07:15 PM

原帖由 fish 於 23/3/2009 05:20 AM 發表

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看第一個reaction
Assume the reaction goes completely
Zn(NO3)2 (aq) + MgSO4 (aq) --> ZnSO4 (aq) + Mg(NO3)2 (aq)
兩個product都是water soluble, 你不能直接分開它們,所以不可以用來prepare ZnSO4

你 ...

咁ZnO+2HCl
咁會出ZnCl2(aq)+H2O(l) ..
咁點樣分開呢2個product ??

fish · 發表於 26/3/2009 09:56 PM

原帖由 小霞的fans 於 26/3/2009 19:15 發表

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咁ZnO+2HCl
咁會出ZnCl2(aq)+H2O(l) ..
咁點樣分開呢2個product ??

跟別人曬鹽一樣...evaporation

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MC + CONVENTIONAL QUESTION

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