19 mLs of neutralizer to neutralize 100 mLs of acid lake water, if we needed to neutralize 50 liters of drinking water, how many grams ?
回答 (3)
2017-09-30 12:30 pm
Only the information on the titration of lake water was provided, but there is no information on that of the drinking acid. Hence, the answer is "unknown".
2017-09-30 7:15 pm
19ml of neutralizer neutralized 100ml of water. The volume of neutralizer neede to produce 50L of water would be (50,000 x 19)/100 = 9500mL of neutralizer.
2017-09-30 11:42 am
(50 L lake water) x (19 mL neutralizer / 0.100 L lake water) = 9500 mL neutralizer
The number of grams depends on the density, which was not given. However, if we take the density given in your previous question:
(9500 mL) x (3 g / 100 mL) = 285 g
The number of grams depends on the density, which was not given. However, if we take the density given in your previous question:
(9500 mL) x (3 g / 100 mL) = 285 g
收錄日期: 2021-04-18 17:50:47
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