From the definition of molar concentration, 5.00 mL of 0.100M silver nitrate solution contains 0.100 X 5.00/1000 or 5.00 X 10-4 gram formula masses of silver nitrate and 5.00 mL of 0.200 X 5.00/1000 or 1.00 X 10-3 gram formula masses of sodium bromide. Because all of the ions involve in this question are monovalent ones, that means that the limiting reagent for the precipitation reaction is silver, and the very low solubility product constant for silver bromide means that the amount of unreacted silver can be disregarded in calculating the answer within the number of significant digits justified by the precision of specifying the amounts of materials reacted. Therefore, 5.00 x 10-4 gram formula masses of silver bromide are formed. The gram formula mass of silver bromide is 187.77, and the mass formed is therefore 0.939 grams to the justified number of significant digits.
One common test for bromide ions is the silver nitrate test, where adding silver nitrate to a solution containing bromide ions produces a cream-colored precipitate of silver bromide. Another test is the starch-iodide test, which involves adding starch and iodine solution to the sample, causing a blue color to form in the presence of bromide ions.
When silver nitrate and sodium bromide are mixed together in a laboratory setting, a white precipitate of silver bromide is formed. This reaction is a double displacement reaction where the silver ions from silver nitrate combine with the bromide ions from sodium bromide to form an insoluble compound. The formation of the precipitate indicates that a chemical reaction has occurred.
When aqueous bromide and aqueous silver nitrate are mixed, a white precipitate of silver bromide is formed due to a double displacement reaction. The balanced chemical equation for this reaction is: AgNO3(aq) + KBr(aq) → AgBr(s) + KNO3(aq)
When ethyl bromide, an alkyl halide, reacts with alcoholic silver nitrate (AgNO3), silver bromide (AgBr) and ethanol are produced. This reaction is a substitution reaction where the bromine in ethyl bromide is replaced by the nitrate ion from silver nitrate.
When silver nitrate and lithium bromide are mixed, a white precipitate of silver bromide forms in the solution due to a double displacement reaction. The balanced chemical equation for the reaction is: AgNO3 + LiBr -> AgBr + LiNO3 The silver bromide is insoluble in water and will separate out as a solid.
One common test for bromide ions is the silver nitrate test, where adding silver nitrate to a solution containing bromide ions produces a cream-colored precipitate of silver bromide. Another test is the starch-iodide test, which involves adding starch and iodine solution to the sample, causing a blue color to form in the presence of bromide ions.
When silver nitrate is added to a solution containing bromide ions, a white precipitate of silver bromide will form. This is due to a chemical reaction between the silver nitrate and bromide ions, resulting in the insoluble silver bromide precipitate.
The reaction between sodium bromide and silver nitrate forms silver bromide and sodium nitrate. The product is a white precipitate of silver bromide, while sodium nitrate remains dissolved in the solution as a spectator ion.
Silver bromide and sodium nitrate will react to form silver nitrate and sodium bromide as the products. The precipitate formed will be silver bromide, which is insoluble in water and will appear as a white solid in the reaction mixture.
When chloride and bromide ions are mixed with silver nitrate, a white precipitate of silver chloride and a pale yellow precipitate of silver bromide are formed. This reaction is a qualitative test to distinguish between chloride, bromide, and nitrate ions.
When bromide ions are added to silver nitrate, a white precipitate of silver bromide is formed. This is because silver ions react with bromide ions to form the insoluble silver bromide precipitate. This reaction is commonly used in qualitative analysis to detect the presence of bromide ions.
Impossale to tell. Added: If the salt solution contains chloride ion (eg. from table salt, NaCl) then silver chloride will precipitate. Both sodium and nitrate ions stay unchanged in solution.Ag+aq + Cl-aq --> AgCls
When aqueous bromide and aqueous silver nitrate mix, a white precipitate of silver bromide is formed. This is a chemical reaction that involves the exchange of ions: Ag⁺ + Br⁻ → AgBr↓.
Silver nitrate is commonly used to test for the presence of halide ions, such as chloride, bromide, and iodide ions. When silver nitrate is added to a solution containing these ions, characteristic precipitates are formed: white for chloride ions, cream for bromide ions, and yellow for iodide ions.
When silver nitrate and sodium bromide are mixed together in a laboratory setting, a white precipitate of silver bromide is formed. This reaction is a double displacement reaction where the silver ions from silver nitrate combine with the bromide ions from sodium bromide to form an insoluble compound. The formation of the precipitate indicates that a chemical reaction has occurred.
When aqueous bromide and aqueous silver nitrate are mixed, a white precipitate of silver bromide is formed due to a double displacement reaction. The balanced chemical equation for this reaction is: AgNO3(aq) + KBr(aq) → AgBr(s) + KNO3(aq)
When ethyl bromide, an alkyl halide, reacts with alcoholic silver nitrate (AgNO3), silver bromide (AgBr) and ethanol are produced. This reaction is a substitution reaction where the bromine in ethyl bromide is replaced by the nitrate ion from silver nitrate.