7. You have 75.0 mL of 0.10 M HA. After adding 30.0 mL of 0.10 M NaOH, the pH is 5.50. What is the Ka value of HA ? (A) 2.1 x 10-6 (B) 3.8 x 10-6 (C) 4.7 x 10-2 (D) 7.4 x 10-8 (E) 6.3 x 10-4

7. You have 75.0 mL of 0.10 M HA. After adding 30.0 mL of 0.10 M NaOH, the pH is 5.50. What is the Ka value of HA ?
(A) 2.1 x 10^-6
(B) 3.8 x 10^-6
(C) 4.7 x 10^-2
(D) 7.4 x 10^-8
(E) 6.3 x 10^-4

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統計： A(26), B(6), C(6), D(3), E(0) #2344885

詳解 (共 1 筆)

呂欣怡

B1 · 2021/07/02

#4859153

(共 1 字，隱藏中）

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11. Assuming that the combustion of hydrogen gas provides three times as much energy per gram as gasoline, calculate the volume of liquid H2 (density = 0.0710 g/mL) required to furnish the energy contained in 80.0 L (about 20 gal) of gasoline (density = 0.740 g/mL). Calculate also the volume that this hydrogen would occupy as a gas at 1.00 atm and 25°C. At 1.00 atm and 25°C, the hydrogen gas needed to replace 20 gal of gasoline occupies a volume of L. (A) 277 (B) 19700 (C) 59200 (D) 239000 (E) 977000

7. You have 75.0 mL of 0.10 M HA. After adding 30.0 mL of 0.10 M NaOH, the pH is 5.50. What is the Ka value of HA ?
(A) 2.1 x 10^-6
(B) 3.8 x 10^-6
(C) 4.7 x 10^-2
(D) 7.4 x 10^-8
(E) 6.3 x 10^-4

詳解 (共 1 筆)

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