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lara [203]
3 years ago
9

45 grams of ammonia, NH3, in 0.75 L of solution.

Chemistry
1 answer:
GrogVix [38]3 years ago
4 0
2.3 moles of sodium chloride in 0.45 liters of solution . necessarile correct ... 0.025 L. (1.3 SF). 45 grams of ammonia in 0.75 L of solution. 45g x 1 mol = 2. demol ... anol x 1,50 = 3 mol NaOH x 40,019 = 120g Naolt.
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