How many milliliters of a 0.266 M CsNO3 solution are required to make 150.0 mL of 0.075 M CsNO3 solution

The required milliliters of CsNO3 in solution 1 = 42.29 mL

Explanation:

Given that:

The molarity of CsNO3 in solution 1 M₁= 0.266 M

The Volume of CsNO3 in solution 1 V₁  =   ???

The Volume of CsNO3 in solution 2 V₂ = 150.0 mL

The molarity of CsNO3 in solution 2 M₂= 0.075 M

By the application of the titrimetric concept:

M₁V₁ = M₂V₂

We have;

0.266 M ×  V₁  = 0.075 M × 150.0 mL

V₁  = ( 0.075 M × 150.0 mL ) / 0.266 M

V₁  = (11.25 / 0.266 ) mL

V₁  = 42.29 mL

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.