order these units from smallest to largest:cm,um,km,mm,m,nm,dm,pm. then give each a measurement in terms of meters.

1 answer:

4 0

km kilometer 1000 meters
m meter 1 meter
dm decimeter 1/10 meter
cm centimeter 1/100 meter
mm millimeter 1/1000 meter
um micrometer 1/1000000 (one millionth) meter
nm nanometer 1/1000000000 (one billionth) meter
pm picometer 1/1000000000000 (one trillionth) meter

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A 150.0 mL sample of an aqueous solution at 25°C contains 15.2 mg of an unknown nonelectrolyte compound. If the solution has an

inysia [295]

Answer: The molar mass of the unknown compound is 223.2 g/mol

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

where,

$\pi$ = osmotic pressure of the solution = 8.44 torr

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = ?

R = Gas constant = $62.3637\text{ L torr }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$8.44torr=1\times M\times 62.3637\text{ L. torr }mol^{-1}K^{-1}\times 298K\\\\M=\frac{8.44}{1\times 62.3637\times 298}=4.54\times 10^{-4}M$

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = $4.54\times 10^{-4}M$

Given mass of unknown compound = 15.2 mg = 0.0152 g (Conversion factor: 1 g = 1000 mg)

Volume of solution = 150.0 mL

Putting values in above equation, we get:

$4.54\times 10^{-4}M=\frac{0.0152\times 1000}{\text{Molar mass of unknown compound}\times 150.0}\\\\\text{Molar mass of unknown compound}=\frac{0.0152\times 1000}{150.0\times 4.54\times 10^{-4}}=223.2g/mol$