因為我英文不怎好所以想借助大大的英文能力QQ可能有些題目很簡單我都不知道=_=
Q：
（１） Bromine is the only nonmetal that is a liquid at room temperature. Consider the isotope bromine-81,(81,35)Br . Select the combination which lists the correct atomic number, neutron number, and massnumber, respectively.
（２） Atoms of the same element with different mass numbers are called:(A) ions (B) neutrons (C) allotropes (D) chemical families (E) isotopes.
（３） An oxide ion, O2-, has:
(A) 8 protons and 10 electrons (B) 10 protons and 8 electrons (C) 8 protons and 9 electrons(D) 8 protons and 7 electrons (E) 10 protons and 7 electrons
（４） Which of the following elements are the least reactive?
(A) alkali metals (B) noble gases (C) halogens (D) alkaline earth metals (E) metalloids
（５） The compound, P4S10, is used in the manufacture of safety matches. What is its name?
(A) phosphorus sulfide (B) phosphoric sulfide (C) phosphorus decasulfide (D) tetraphosphorusdecasulfide (E) triphosphorus tetrasulfide
（６） Which is the correct formula for copper (II) phosphate?
(A) Cu2PO4 (B) Cu3(PO4)2 (C) Cu2PO3 (D) Cu(PO4)2 (E) Cu(PO3)2
（７） The size of an atomic orbital is associated with
(A) the principal quantum number (n) (B) the angular momentum quantum number (l) (C) themagnetic quantum number (ml) (D) the spin quantum number (ms) (E) the angular momentum
and magnetic quantum numbers, together.
（８）Which of the following is a correct set of quantum numbers for an electron in a 3d orbital?
(A) n = 3, l = 0, ml = -1 (B) n = 3, l = 1, ml = +3 (C) n = 3, l = 2, ml = 3 (D) n = 3, l = 3, ml = +2(E) n = 3, l = 2, ml = -2
（９）Which of these compounds is most likely to be covalent?
(A) KF (B) CaCl2 (C) SF4 (D) Al2O3 (E) CaSO4
（１０） The conjugate base of a weak acid is
(A) a strong base (B) a weak base (C) a strong acid (D) a weak acid(E) none of these
化學問題(英文版)
2009-07-22 10:36 am
回答 (3)
2009-07-22 4:11 pm
✔ 最佳答案
（１）解：Br-81的溴同位素，其原子序為35、電子數為35、質子數為35、中子數為46、質量數為81。（２）解：相同的原子，不同的質量數，稱之為同位素。(E) isotopes.
（３）解：因為氧離子獲得兩個電子，故有10個電子，但仍有8個質子。(A) 8 protons and 10 electrons.
（４）解：最低活性的元素就是鈍氣組態或惰性氣體。(B) noble gases.
（５）解：P4S10－十硫化四磷。(D) tetraphosphorusdecasulfide.
（６）解：過磷酸銅的化學式為－(B) Cu3(PO4)2。
（９）解：最接近共價鍵的化合物為－(C) SF4。
（１０） 弱酸的共軛鹼是弱鹼。(B) a weak base.
先回答這些，其餘的稍後再回答。
2009-07-22 18:51:08 補充：
（７）解：原子軌域的大小當然與主量子數有關。(A) the principal quantum number (n).
（８）解：3d軌域可以填入10個電子，所以其量子數為－n = 3, l = 2, ml = -2, -1, 0, 1, 2.
補充題：
（１）解：SCl2的分子形狀為彎曲形和水分子一樣，而不是直線形，其電子點式如下－
‥ ‥ ‥
：Cl：Ｓ： Ｈ：Ｏ：
‥ ‥ ‥
：Cl： Ｈ
‥
2009-07-22 19:10:53 補充：
（２）解：首先，假設此化合物之化學式為CxHyOz，並寫出其燃燒的化學反應方程式如下－
CxHyOz＋O2 → 2CO2＋3H2O
由題目中可得各化合物之莫耳數如下：
CxHy：620/62＝10 mol
CO2：880/44＝20 mol
H2O：542/18＝30 mol
由上述結果可推知，x＝2，y＝6，且其分子量為62，可得z＝2。
此化合物之分子式為C2H6O2，實驗式為CH3O。
2009-07-22 19:11:24 補充：
（３）解：因其化學反應方程式如下－
2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)
所以其莫耳數比－Al：HCl＝2：6＝1：3，故其莫耳數算式如下－
Al：27/27＝1 mol
HCl：V*6＝3 mol V＝0.5 升
（a）需鹽酸體積為0.5升，即500毫升。
2009-07-22 19:11:52 補充：
因為200 ml 6 M的鹽酸會用完，所以其莫耳數比－HCl：AlCl3＝3：1，故其莫耳數算式如下－
HCl：0.2*6＝1.2 mol
AlCl3：W/(27+35.5*3)＝0.4 mol W＝53.4 克
（b）產生AlCl3的質量為53.4克。
2009-07-22 19:20:22 補充：
（４）解：因為化學反應方程式的反應焓＝生成物的生成焓總和－反應物的生成焓總和，所以反應焓的算式如下－
ΔH°rxn = 2ΔH°f[CO2(g)] +ΔH°f[H2O(l)] - ΔH°f[C2H2(g)] - (5/2)ΔH°f[O2(g)]
-1299 = 2*(-394) + (-286) - ΔH°f[C2H2(g)] - (5/2)*0
故C2H2的生成焓為－ΔH°f[C2H2(g)] = +225 kJ/mol。
(5)
(a) 1s2 2s2 2p6 3s2 3p6 3d2 4s2
(b) 3d2 4s2
2009-07-22 19:23:34 補充：
（５）解：Ti-22的基態電子組態為1s2 2s2 2p6 3s2 3p6 3d2 4s2。而最外層的電子為4s2及3d2。因為3d2的能階較4s2高，所以3d2會比4s2先與其他的元素反應。
2014-10-30 5:12 pm
2009-07-22 6:31 pm
(1)
Atomic number (原子序) = 35
Neutron number (中子數) = 81 - 35 = 46
Mass number (質量數) = 81
(2) (E) (同位素)
(3) (A)
O atom (原子) has 8 protons (質子) and 8 electrons (電子).
O2- ion (離子) has 2 more electrons than O atom, i.e. 10 electrons.
(4) (B) (貴氣體，即惰性氣體)
(5) (D) (四磷化十硫).
(6) (B).
Copper(II) ion (銅(II) 離子) is Cu2+(aq), while phosphate ion (磷酸離子) is PO43-(aq).
(7) (A) (主量子數).
(8) (E)
For the five 3d orbitals :
n = 3
l = 2
ml = -2, -1, 0, 1, 2
(9) (C)
Two non-metals tends to form covalent compounds.
(10) (B)
The weaker the acid, the stronger is its conjugate base (共軛鹼).
For example, CH3COOH is a weak acid, its conjugate base CH3COO- is a weak base.
(1)
Cl : S : Cl
(Add four more dots to S, and add 6 more dots to each Cl).
(2)
Molar mass of C, H, O, CO2 and H2O are 12, 1, 16, 44 and 18 respectively.
In 620 g of the compound:
Mass of C = Mass of C in CO2 = 880 x (12/44) = 240 g
Mass of H = Mass of H in H2O = 542 x (1x2/18) = 60.2 g
Mass of O = 620 - (240 + 60.2) = 319.8 g
Mole ratio C : H : O = 240/12 : 60.2/1 : 319.8/16 = 1 : 3 : 1
Empirical formula = CH3O
3.
(a)
Molar mass of Al = 27 g/mol
No. of moles of Al = 27/27 = 1 mol
Mole ratio Al : HCl = 2 : 6 = 1 : 3
Minimum no. of moles of HCl = 1 x 3 = 3 mol
Minimum volume of HCl = 3/6 = 0.5 L = 500 mL
(b)
200 mL < 500 mL. Hence, HCl is the limiting reactant (completely reacted).
No. of moles of HCl reacted = 6 x (200/1000) = 1.2 mol
Mole ratio HCl : AlCl3 = 6 : 2 = 3 : 1
No. of moles of AlCl3 formed = 1.2/3 = 0.4 mol
Molar mass of AlCl3 = (27 + 35.5x2) = 98 g/mol
Mass of HCl formed = 0.4 x 98 = 39.2 g
(4)
ΔH°rxn = 2ΔH°f[CO2(g)] +ΔH°f[H2O(l)] - ΔH°f[C2H2(g)] - (5/2)ΔH°f[O2(g)]
-1299 = 2(-394) + (-286) -ΔH°f[C2H2(g)] - (5/2)(0)
ΔH°f[C2H2(g)] = +225 kJ/mol
(5)
(a) 1s2 2s2 2p6 3s2 3p6 3d2 4s2
(b) 3d2 4s2
Atomic number (原子序) = 35
Neutron number (中子數) = 81 - 35 = 46
Mass number (質量數) = 81
(2) (E) (同位素)
(3) (A)
O atom (原子) has 8 protons (質子) and 8 electrons (電子).
O2- ion (離子) has 2 more electrons than O atom, i.e. 10 electrons.
(4) (B) (貴氣體，即惰性氣體)
(5) (D) (四磷化十硫).
(6) (B).
Copper(II) ion (銅(II) 離子) is Cu2+(aq), while phosphate ion (磷酸離子) is PO43-(aq).
(7) (A) (主量子數).
(8) (E)
For the five 3d orbitals :
n = 3
l = 2
ml = -2, -1, 0, 1, 2
(9) (C)
Two non-metals tends to form covalent compounds.
(10) (B)
The weaker the acid, the stronger is its conjugate base (共軛鹼).
For example, CH3COOH is a weak acid, its conjugate base CH3COO- is a weak base.
(1)
Cl : S : Cl
(Add four more dots to S, and add 6 more dots to each Cl).
(2)
Molar mass of C, H, O, CO2 and H2O are 12, 1, 16, 44 and 18 respectively.
In 620 g of the compound:
Mass of C = Mass of C in CO2 = 880 x (12/44) = 240 g
Mass of H = Mass of H in H2O = 542 x (1x2/18) = 60.2 g
Mass of O = 620 - (240 + 60.2) = 319.8 g
Mole ratio C : H : O = 240/12 : 60.2/1 : 319.8/16 = 1 : 3 : 1
Empirical formula = CH3O
3.
(a)
Molar mass of Al = 27 g/mol
No. of moles of Al = 27/27 = 1 mol
Mole ratio Al : HCl = 2 : 6 = 1 : 3
Minimum no. of moles of HCl = 1 x 3 = 3 mol
Minimum volume of HCl = 3/6 = 0.5 L = 500 mL
(b)
200 mL < 500 mL. Hence, HCl is the limiting reactant (completely reacted).
No. of moles of HCl reacted = 6 x (200/1000) = 1.2 mol
Mole ratio HCl : AlCl3 = 6 : 2 = 3 : 1
No. of moles of AlCl3 formed = 1.2/3 = 0.4 mol
Molar mass of AlCl3 = (27 + 35.5x2) = 98 g/mol
Mass of HCl formed = 0.4 x 98 = 39.2 g
(4)
ΔH°rxn = 2ΔH°f[CO2(g)] +ΔH°f[H2O(l)] - ΔH°f[C2H2(g)] - (5/2)ΔH°f[O2(g)]
-1299 = 2(-394) + (-286) -ΔH°f[C2H2(g)] - (5/2)(0)
ΔH°f[C2H2(g)] = +225 kJ/mol
(5)
(a) 1s2 2s2 2p6 3s2 3p6 3d2 4s2
(b) 3d2 4s2
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