Posted by Ivy on Monday, December 6, 2010 at 5:28pm.
40 ml of 0.4M solution contains 0.04*0.4
=0.016 mole of AgNO3.
RMM of AgNO3 = 107.87+14+3*16=169.87
So 0.016*169.87=2.72 g
Related Questions
Chemistry - Silver(I) nitrate and sodium sulfate react to form silver(I) sulfate...
chemistry - An aqueous solution of silver nitrate contains 5.00 g of silver ...
chemistry - The addition of a dilute hydrobromic acid would clearly distinguish ...
Chemistry - Calculate the mass of silver nitrate, AgNO3, FW = 169.9 g/mole, ...
chem. - Silver(I) nitrate and sodium sulfate react to form silver(I) sulfate and...
Chem - Solutions of sodium carbonate and silver nitrate react to form solid ...
chem - a 6.54 g sample consisting of a mixture of silver nitrate and sodium ...
chemistry - The traditional method of analyzing the amount of chloride ion ...
chemistry - An aqueous solution is 22.0% by mass silver nitrate, AgNO3, and has ...
Chemistry - it takes 37.50 mL of 0.152 M sodium chromate to titrate 25.00 mL of ...
For Further Reading
