How many grams are in 3 mol of KBr?

The grams in 3 mol of KBr are: 357 grams.

Explanation:

We calculate the weight of 1 mol of KBr

Weight KBr = Weight K + Weight Br= 39,098grams +79,904grams= 119, 002 grams/mol

1 mol KBr119, 002 grams

3 mol KBrx= (3 mol KBr x119, 002 grams)/ 1 mol KBr= 357, 006 grams

(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

the answer your looking for is c).