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What mass of silver chloride can be produced from 1.88 L of a 0.139 M solution of silver nitrate?
Since the four named compounds are the only reactants and products, this question can be answered from the law of conservation of mass: The amount of silver nitrate must be 14.35 + 8.5 - 5.85 or 17.0 grams.
What is the percent of silver in silver chloride?
Silver chloride is composed of one silver atom and one chloride atom. The molar mass of silver chloride is 143.32 g/mol and the molar mass of silver is 107.87 g/mol. Therefore, the percentage of silver in silver chloride is (107.87/143.32) x 100 = 75.3%.
How many grams of silver chloride can be produced from the reaction of 20 g of silver nitrate with an excess amount of sodium chloride AgNO3 (aq) NaCl (aq) AgCl (s) NaNO3 (aq)?
20 grams of silver nitrate would produce an equivalent amount of silver chloride if all the silver nitrate is converted. The molar ratio of AgNO3 to AgCl is 1:1, so 20 grams of AgNO3 would produce 20 grams of AgCl.
In the reaction of silver nitrate with sodium chloride how many grams of silver chloride will be produced from 100 g of silver nitrate when it is mixed with an excess of sodium chloride?
When silver nitrate reacts with sodium chloride, silver chloride is formed according to the equation: AgNO3 + NaCl -> AgCl + NaNO3. The molar ratio of silver nitrate to silver chloride is 1:1. Therefore, 100 g of silver nitrate will produce 143.32 g of silver chloride.
Silver chloride often used in silver plating contains 75.27 percent Ag Calculate the mass of silver chloride required to plate 285mg of pure silver?
To calculate the mass of silver chloride needed to plate 285mg of pure silver, you can start by determining the mass of silver in the silver chloride. Since silver chloride contains 75.27% silver, the mass of silver in the silver chloride is 0.7527 * mass of silver chloride. Once you have the mass of silver in the silver chloride, you can set up a ratio to find the mass of silver chloride needed to plate 285mg of pure silver.
How many grams of silver chloride are equal to a mol of silver chloride?
Silver chloride - AgClAg (107.89 grams) + Cl (35.45 grams) = 143.34 grams
How many grams of chromium (III) chloride are required to produce 75.0 g of silver chloride?
The formula of chromium chloride is CrCl3 and the formula for silver chloride is AgCl. The relevant formula unit masses are 158.36 for chromium (III) chloride and 143.32 for silver chloride. The gram atomic masses of chlorine, chromium, and silver are 35.453, 51.996, and 107.866 respectively. Therefore, the mass fraction of chloride in chromium (III) chloride is [3(35.453)/158.36] or 0.671628 and the mass fraction of chloride in silver chloride is 35.453/143.32 or 0.24737. Therefore, to form 75.0 g of silver chloride, (0.24737)(75.0) or 18.55 g of chloride is needed, and this amount of chloride is contained in 18.55/0.6716 or 27.6 g of chromium (III) chloride, to the justified number of significant digits.
How many grams of silver chloride will precipitate by silver nitrate 5.85g of sodium chloride?
The answer is 14,35 g AgCl.
What mass of silver chloride can be produced from 1.88 L of a 0.139 M solution of silver nitrate?
Since the four named compounds are the only reactants and products, this question can be answered from the law of conservation of mass: The amount of silver nitrate must be 14.35 + 8.5 - 5.85 or 17.0 grams.
How many grams of silver chloride can be produced if you start with 4.62 grams of barium chloride 2AgNO3 plus BaCl2 and mdash and gt 2AgCl plus Ba(NO3)?
6,36 g of silver chloride are obtained.
How many grams of silver chloride are formed when 10.0 g of silver nitrate reacts with 15.0 g of barium chloride?
To find the limiting reactant, we need to determine how many grams of silver chloride can be produced from each reactant and compare the results. Calculate the amount of silver chloride that can be produced from 10.0 g of silver nitrate. Calculate the amount of silver chloride that can be produced from 15.0 g of barium chloride. The reactant that produces the lesser amount of silver chloride will be the limiting reactant.
What is the percent of silver in silver chloride?
Silver chloride is composed of one silver atom and one chloride atom. The molar mass of silver chloride is 143.32 g/mol and the molar mass of silver is 107.87 g/mol. Therefore, the percentage of silver in silver chloride is (107.87/143.32) x 100 = 75.3%.
How many grams of silver nitrate are needed to react with 50 grams of sodium chloride?
You need 145,337 g silver nitrate.
When silver nitrate reacts with barium chloride silver chloride and barium nitrate are formed How many grams of silver chloride are formed when 10.8 g of silver nitrate reacts with 15 g of barium?
9.11 g
How many grams of chromium III chloride are required to produce 75.0 g of silver chloride?
The formula of chromium chloride is CrCl3 and the formula for silver chloride is AgCl. The relevant formula unit masses are 158.36 for chromium (III) chloride and 143.32 for silver chloride. The gram atomic masses of chlorine, chromium, and silver are 35.453, 51.996, and 107.866 respectively. Therefore, the mass fraction of chloride in chromium (III) chloride is [3(35.453)/158.36] or 0.671628 and the mass fraction of chloride in silver chloride is 35.453/143.32 or 0.24737. Therefore, to form 75.0 g of silver chloride, (0.24737)(75.0) or 18.55 g of chloride is needed, and this amount of chloride is contained in 18.55/0.6716 or 27.6 g of chromium (III) chloride, to the justified number of significant digits.
What mass of silver chloride will be produced when 113 grams of silver nitrate is reacted?
The answer is 95,34 g.
How many grams calcium are in 100 grams calcium chloride?
The formula of anhydrous calcium chloride is CaCl2, and its gram formula mass is 110.99. The gram atomic mass of calcium is 40.08. Therefore, the grams of calcium in 100 grams of calcium chloride is 100(40.08/110.99) or 36.11 grams, to the justified number of significant digits.
