If the concentration of the stock solution is 25 m, what would be the dilution factor and concentration of sample ‘c’ with respect to original stock solution?

The dilution factor = 5 and concentration of sample ‘c’ = 5M

Dilution is the process of adding solvents (generally water) which aims to obtain a solution with a concentration that is smaller than the initial concentration of a more concentrated substance.

So the principle of dilution is to get lower concentration.

In dilution the number of dissolved moles before and after dilution is the same

The dilution formula is as follows:

M1 = Molarity of the solution before dilution

V1 = volume of the solution before dilution

M2 = Molarity of the solution after dilution

V2 = Molarity volume of the solution after dilution

Dilution factors are factors that must be included / multiplied in the calculation formula, usually to find the percentage of substances

So the dilution factor is the division of the final volume by the initial volume.

Formula used:

Vf = final volume

Vi = initial volume

The dilution factor is used as the denominator of a fraction.

For example, a DF of 10 means a 1:10 dilution.

This is different from the term "diluted 10 x or 100 x" then what is multiplied is the initial volume before dilution, so if it is diluted 10 x it means the concentration after dilution = initial concentration divided by 10

If the concentration of the stock solution is 25 m, the dilution factor:

20 ml of sample

100 ml final volume

So DF = 100/20 = 5

Then

concentration of solution C = 1/5 x 25 = 5 M

The molarity of the stock solution of luminol

link

NaOH solution

link

Delta H solution

link

Keywords: stock solution, Dilution factors, dilution, concentration

(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