What volume in ml of 0.500 m silver nitrate is required to completely react with 25.0 ml of 1.5 m calcium chloride?

First, the balanced chemical equation of the reaction must be written. This gives:

2 AgNO3 (aq) + CaCl2 (aq) --> 2 AgCl (s) + Ca(NO3)2 (aq) 

This shows that for every mole of CaCl2, 2 moles of AgNO3 reacts (1:2 ratio). Given the volume and molarity of CaCl2, we can compute for the number of moles of CaCl2.

moles CaCl2 = 1.5M(25mL/1000) = 0.0375 moles

Now, we can also compute for the moles of AgNO3 using the balanced equation.

moles AgNO3 = 2(0.0375) = 0.075 moles

Using the moles and molarity of AgNO3, we can compute for the volume needed.

volume (mL)/1000 = 0.075/0.5 = 150 mL of silver nitrate solution is required.

3units/ 5 feet = 60 units/ 100 feet3 units/ 5 feet = 18 units/ 30 feetthe dimensions of the scale drawing are 60 units by 18 units.hope this !