# 超急!!! 原文化學計算題!!!!

上星期老師交代的作業,剩下兩題怎麼算都算不出來,算到後面都不知道在算甚麼@@

有請各位大大幫忙了!! 萬分感謝!!!

1.How many milliliter of 3.00 M H2SO4 are required to react with 4.35g of solid containing 23.2 wt% Ba(NO3)2 if the reaction is Ba2+ +SO42- ---> BaSO4(s)?

2.It is recommended that drinking water contain 1.6 ppm fluorede(F-) for prevention tooth decay. Consider a reservoir with a diameter of 4.50X10X10 m and a depth of 10.0 m. How many grams of F- should be added to give 1.6 ppm?Fluoride is provided H2SiF6. How many grams of H2SiF6 contain this much F-?

### 1 Answer

- 小白Lv 79 years agoFavorite Answer
第２題的直徑是4.5 x 10 x 10 = 450 m??

2011-09-28 21:41:24 補充：

1. 設需要 V mL 的 H2SO4 水溶液

鋇離子莫耳數 = 4.35 x 23.2% / 261.37 = 3.861 x 10^-3 mol

完全反應則鋇離子與硫酸根莫耳數相等，即

3.861 x 10^-3 = V/1000 x 3.00 ==> V = 1.287 mL

2. 水庫的水體積 = 450^2 x 3.14 / 4 x 10 = 1.59 x 10^6 m3

因水的密度為1 t/m3，故有 1.59 x 10^6 噸重

含氟離子 1.6 ppm，即 1.6 x 1.59 = 2.544 噸 = 2.544 x 10^6 g

換算成 H2SiF6 = 2.544 x 10^6 x 144 / (19 x 6) = 3.21 x 10^6 g

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