[50 Points] In the experiment below, Beaker 1 contains 12.0 M HCl. Each subsequent beaker is diluted by 50%. What is the [OH-] in Beaker 6?

2.67x10⁻¹⁴M = [OH-]

Explanation:

To solve this question we need first to find the HCl concentration of the beaker 6. Then, using:

Kw = [OH-][H+]

Where Kw = 1x10⁻¹⁴

[OH-] is our incognite

[H+] is = [HCl]

Beaker 2 = 12.0M / 2 = 6.0M

Beaker 3 = 6.0M / 2 = 3.0M

Beaker 4 = 3.0M / 2 = 1.5M

Beaker 5 = 1.5M / 2 = 0.75M

Beaker 6 = 0.75M / 2 = 0.375M

HCl = 0.375M = [H+]

1x10⁻¹⁴ / [H+] = [OH-]

1x10⁻¹⁴ / 0.375M = 2.67x10⁻¹⁴M

b. accuracy of the data matters alot more than its number of digits, organization or amount of detail. quantitative observation is more accepted than the qualitative observation.

(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