< Question 8 of 10
>
Suppose 839 g of ethanol, C, H,OH, was added to 6.275 x 10' g of water. By how much would this lower the freezing
point? The freezing point depression constant, Kr, for water is 1.86 °C/m.
33.9 °C
5.40 °C
0.641 °C
0.00540 °C

since electrons are negative the answer is neagtaive

(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