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Sagot :
1) List the known and unknown quantities.
Solution 1: KBr.
Molarity: 1.10 KBr.
Volume: 38.0 mL = 0.038 L.
Solution 2: KBr.
Molarity: 0.570 M.
Volume: 57.0 mL = 0.057 L.
Total volume: 75.0 mL = 0.075 L.
Molarity: unknown.
2) Molarity of KBr after mixing the two solutions.
2.1-Moles of KBr from solution 1
[tex]mol\text{ }KBr=1.10\text{ }M\text{ }KBr*0.038\text{ }L=0.0418\text{ }mol\text{ }KBr[/tex]2.2-Moles of KBr from solution 2
[tex]mol\text{ }KBr=0.570\text{ }M\text{ }KBr*0.057\text{ }L=0.03249\text{ }mol\text{ }KBr[/tex]Total moles of KBr: 0.0418 + 0.03249 = 0.07429 mol KBr.
2.3-New solution
[tex]M=\frac{0.0418+0.03249}{0.075\text{ }L}=0.9905\text{ }M\text{ }KBr[/tex]3) Moles of KBr needed to react to silver nitrate.
3.1-Write the chemical equation and balance it
[tex]KBr+AgNO_3\rightarrow AgBr+KNO_3[/tex]3.2-Moles of silver nitrate.
The molar ratio between KBr and AgNO3 is 1 mol KBr: 1 mol AgNO3.
[tex]mol\text{ }AgNO_3=0.07429\text{ }KBr*\frac{1\text{ }mol\text{ }AgNO_3}{1\text{ }mol\text{ }KBr}=0.07429\text{ }mol\text{ }AgNO_3[/tex]4) Mass of silver nitrate.
The molar mass of AgNO3 is 169.8731 g/mol.
[tex]g\text{ }AgNO_3=0.07429\text{ }mol\text{ }AgNO_3*\frac{169.8731\text{ }g\text{ }AgNO_3}{1\text{ }mol\text{ }AgNO_3}=12.62\text{ }g\text{ }AgNO_3[/tex]The mass of silver nitrate (AgNO3) required to precipitate out silver bromide is 12.62 g AgNO3.
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