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aleksandrvk [35]
3 years ago
10

how many grams of solid silver nitrate would you need to prepare 200.0 mL of a 0.150 M AgNO3 solution

Chemistry
1 answer:
aleksklad [387]3 years ago
4 0
0.150 M AgNO3 = x mol / 0.200 Liters
x mol = 0.03 mol AgNO3
0.03 mol AgNO3 (169.9g AgNO3 / 1 mol AgNO3) We are converting moles to grams here with stoichiometry.

Final answer = 5.097 grams, but if you want it in terms of sig figs then it is 5.09 grams.
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Answer:

B) -4.1 units

Explanation:

According to this question, a state property X has a value 89.6 units. It undergoes the certain changes as follows:

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