1. how many grams of cacl2 would be required to produce a 3.5 m (molar) solution with a volume of 2.0 l? . 2. what is the molarity of a 5.00 x 102 ml solution containing 2490 g of ki? . 3. how many moles of lif would be required to produce a 2.5 m solution with a volume of 10.5 l? .

Explanation:

1.) Moles of calcium chloride = n

Molarity of the solution = 3.5 M

Volume of the solution =V= 2.0 L

n = 7 moles

Mass of 7 moles of calcium chloride :

777 grams of calcium chloride would be required.

2.) Mass of the KI = 2490 g

Moles of KI = n =

Volume of the solution,V =

Molarity of the KI solution is 30 mol/L.

3.)Molarity of the LiF solution = 2.5 M

Volume of the solution = 10.5 L

Moles of LiF = n

n = 26.25 moles

26.25 moles of LiF would be required.

thats how its supposed to be

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )