Calculate the volume (in mL) of 0.100 M Na, CO3 needed to produce 1.00 g of CaCO3(s)
. There is an excess of CaCl2.
Molar mass of calcium carbonate = 100.09 g/mol

10 ml

Explanation:

Molarity is equal to moles/volume (Liters) so you need to find how many moles are in 1.00 gram of Calcium carbonate

100.09 MM = 1.00/Xmol

move the equation so that mol is on the left

Xmol = 1.00 g/ 100.09 MM = .009991 mol (rounding = .01 mol)

then add that into your equation to find volume

M = molarity

mol= moles

V = volume in Liters

M= mol/V so 1.00M = .01 mol/ v

V = 0.01 L

since the answer is in Liters you need to multiply that answer by 1000 which is the conversion from Liters to mililiters

.01 L X1000 = 10 ml

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 : )

i think it's letter a.