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Chemistry
Stoichiometry
Precipitation Reaction
How many milliliters of a 0.200M AgNO3 solution are required to preccipitate most of the Cl^- ions from a solution containing 0.855 g of NaCl?
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How many milliliters of a 0.200M AgNO3 solution are required to preccipitate most of the Cl^- ions from a solution containing
Top answer:
NaCl + AgNO3 = AgCl + NaNO3 moles NaCl = grams/molar mass mols AgNO3 = moles NaCl M AgNO3 = mols/L
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Suppose you wanted to find out how many milliliters
of 1.0 M AgNO3 are needed to provide 169.88 g of pure AgNO3. a. What is step
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call me
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How many milliliters of .200 M AgNO3 solution are required to precipitate most of the Cl- from the solution containing .855
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See above.
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How many milliliters of 0.277 M AgNO3 solution are needed to react completely with 47.1 mL of 0.278 M NaCl solution? How many
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Use the procedure in the succeeding problem [HNO3 and Ba(OH)2] to solve this one. Also you may need
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What mass of silver nitrate (AgNO3) is required to prepare 400 g of a 3.50% solution of
AgNO3
Top answer:
3.50% AgNO3 means 3.50g AgNO3/100 g soln. You want 400 g; therefore, you will need 3.50g x (400/100)
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find the mass of silver chloride formed from 33.2ml of a 0.100M solution of silver nitrate and 200ml of 0.200M solution of
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To find the mass of silver chloride formed, we can use mass-to-mass calculations. Here's how you can
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(a) When 300.0 milliliters of a solution of 0.200 molar AgNO3 is mixed with 100.0 milliliters of a 0.0500 molar CaCl2 solution,
Top answer:
.138M
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Silver nitrate (AgNO3) reacts with sodium
chloride as indicated by the equation AgNO3 + NaCl → AgCl + NaNO3 . How many grams of
Top answer:
AgNO3 + NaCl → AgCl + NaNO3 millimoles AgNO3 = 132 mL x 0.719 = 94.9 millimols NaCl needed = 94.9
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Silver nitrate (AgNO3) reacts with sodium
chloride as indicated by the equation AgNO3 + NaCl → AgCl + NaNO3 . How many grams of
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AgNO3 + NaCl → AgCl + NaNO3 millimols AgNO3 = mL x M = 132 mL x 0.719 M = 94.9 millimols NaCl
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Silver nitrate (AgNO3) reacts with sodium
chloride as indicated by the equation AgNO3 + NaCl ! AgCl + NaNO3 . Howmany grams of
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16.43, found the answer
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