Calculate the volume of 1.0 m hcl required to react this amount of calcium carbonate as described in equation 1 in the background section. c) compare the volume of hcl calculated with the volume described in the procedure. will the amount of hcl used allow for calcium carbonate to be the limiting reactant?

sorry idk the answer well to you i have a explaination

explanation: a formula which shows the arrangement of atoms in the molecule of a compound

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied