A0.2121g sample of an organic compound was burned in a stream of oxygen and the carbon dioxide produced was collected in a stream of barium hydroxide . calculate the percentage carbon in the sample if 0.6006g barium carbonate was formed

The answer to your question is: 17.26% of carbon

Explanation:

Data

CxHy = 0.2121 g

BaCO₃ = 0.6006 g

Molecular mass BaCO₃ = 137 + 12 + 48 = 197 g

Reaction

                    CO₂  +  Ba(OH)₂  ⇒   BaCO₃  +  H₂O

Process

1.- Find the amount of carbon in BaCO₃

                              197 g of BaCO₃    12 g of Carbon

                               0.6006 g              x

                               x = (0.6006 x 12) / 197

                               x = 0.0366 g of carbon

2.- Calculate the percentage of carbon in the organic compound

           0.2121 g of organic compound    100%

           0.0366g                                          x

                              x = (0.0366 x 100) / 0.2121

                              x = 17.26%

(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