阿摩線上測驗
登入
首頁
>
學士後西醫物理及化學
>
112年 - 112 學士後醫學系招生考試:物理及化學#114227
> 試題詳解
76. What is the pH value of a 0.010 M solution of sodium azide (NaN
3
)?
The K
a
of HN
3
= 1.0 × 10
-6
(A) 6.0
(B) 7.0
(C) 8.0
(D) 9.0
(E) 10.0
答案：
登入後查看
統計：
A(0), B(0), C(2), D(25), E(2) #3101827
詳解 (共 1 筆)
uu(已上榜)
B1 · 2023/11/26
#5974130
(共 1 字，隱藏中）
前往觀看
12
0
其他試題
72. The valve between the 3.00 L bulb, in which the gas pressure is 2.00 atm, and the 2.00 L bulb, in which the gas pressure is 2.70 atm, is opened. What is the final pressure in the two bulbs, assuming the temperature remains constant?(A) 0.71 atm (B) 1.03 atm (C) 1.62 atm (D) 2.01 atm (E) 2.28 atm
#3101823
73. The concentration of a monoprotic acid is “a” M and the degree of dissociation of the acid is “b”. What is its Ka? (A) b2 /(a-b) (B) ab2 /(1-a) (C) a2b2 /(1+a) (D) ab2 /(1-b) (E) a2b2 /(1-b)
#3101824
74. How many chiral centers are in the following structure?(A) 6 (B) 7 (C) 8 (D) 9 (E) 10
#3101825
75. Compound X and Y contain two elements A and B only. 3.0 g of compound X contains 1.4 g of A, while 27.0 g of compound Y contains 7.0 g of A. The formula of compound X is AB. Which one is the formula of compound Y? (A) AB2 (B) A4B6 (C) A2B (D) A3B4 (E) A2B5
#3101826
77. What is the net number of octahedral holes contained in the close packing of spheres unit cell like face-center cubic? (A) 8 (B) 4 (C) 12 (D) 6 (E) 3
#3101828
78. Reaction of potassium metal with excess O2(g) leads to KO2(s). Determine the oxidation state of the oxygen in KO2(s). (A) +1 (B) –2 (C) –0.5 (D) +2 (E) –1
#3101829
79. Determine the point group of the molecules shown below? (A) C3v (B) C3 (C) D3d (D) D3h (E) None of the above.
#3101830
80. Which charge of NO would have a bond order of 3? (A) +1 (B) –1 (C) 0 (D) +2 (E) –2
#3101831
81. What is the energy required to excite the electron in the hydrogen atom from the n = 1 level to the n = 2 level? (RH = 13.61 eV) (A) 3.40 eV (B) 6.81 eV (C) 10.21 eV (D) 13.61 eV (E) None of the above.
#3101832
82. Using the data shown as follows to calculate ΔG° for the reaction (1 F = 96,485 coulombs): Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s). The reduction potentials for Fe2+ and Cu2+ are as follows:Fe2+ + 2e– → Fe ε° = –0.44 VCu2+ + 2e– → Cu ε° = 0.34 V (A) –1.9 × 104 J (B) –1.5 × 105 J (C) 0 J (D) 1.9 × 104 J (E) 1.5 × 105 J
#3101833
阿摩線上測驗
登入
