See it's an easy one..!!
AgNO3 + HCl -> AgCl + HNO3
100 mL of 0.068 M AgNO3 contains AgNO3 = 0.068 mol
So,
25 mL of 0.068 M AgNO3 contains AgNO3 = (0.068 * 25) / 1000 = 0.0017 mol
From the equation, we can see
1 mol of AgNO3 gives 1 mol of AgCl
0.0017 mol of AgNO3 gives 0.0017 mol of AgCl
Amount of AgCl can be found this way.!
No. of moles = given mass/ molecular mass
molecular mass of AgCl = 107+35.5 = 143.5 g
Therefore,
Given mass = No. of moles*molecular mass = 0.0017*143.5 = 0.244g
Note : In your question, you have written 0.068 "m" .. (small) m represents for Molality and (capital) M represents for Molarity..!
Hope I helped.. :)
NaI­(aq)­­ + AgNO3(aq) à NaNO3(aq) + AgI(s)
Since the Ksp of AgBr is less than the Ksp of AgNO3, you can predict that the AgBr will precipitate out of solution and leave NO3- in the solution
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the reaction is as follows-AgNO3 + KCl ----->AgCl +KNO3here the silver nitrate(AgNO3) reacts with potassium chloride(KCl) to form potassium nitrate(KNO3) and insoluble AgCl.
A red colouration is obtained. This is a test for ketones
Yes, it is correct.
To find the mass of AgCl formed, first calculate the number of moles of AgNO3 using the formula moles = Molarity x Volume (in liters). Then, use the balanced chemical equation to determine the mole ratio between AgNO3 and AgCl. Finally, convert the moles of AgCl to grams using the molar mass of AgCl (107.87 g/mol).
The chemical formula for aqueous silver nitrate is AgNO3, where Ag is the symbol for silver and NO3 is the polyatomic ion nitrate. When silver nitrate is dissolved in water, it dissociates into silver ions (Ag+) and nitrate ions (NO3-).
The chemical equation is:Fe + 2 AgNO3 = Fe(NO3)2 + 2 Ag
To find the mass of AgBr formed, first calculate the moles of AgNO3 in 35.5 mL of 0.184 M solution. Then, use the mole ratio from the balanced chemical equation between AgNO3 and AgBr to find the moles of AgBr formed. Finally, multiply the moles of AgBr by its molar mass to get the mass. Note that since HBr is in excess, AgNO3 will be the limiting reagent.
NaI­(aq)­­ + AgNO3(aq) à NaNO3(aq) + AgI(s)
For example silver nitrate: NaCl + AgNO3 = NaNO3 + AgCl(s)
The reaction is:LNaCl + AgNO3 = AgCl + NaNO3The white precipitate is silver chloride.
The reaction is:Cu + AgNO3 = Ag + CuNO3
Since the Ksp of AgBr is less than the Ksp of AgNO3, you can predict that the AgBr will precipitate out of solution and leave NO3- in the solution
When AgNo3 reacts with iodide ions, the precipitate of AgI is formed.AgI is insoluble in HNO3. The symbol of the cation os, I-.
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