Nanoscale diagram for the contents of a beaker containing water and sugar has to be drawn. Concept introduction: Like dissolves like principle: ‘Like dissolves like’ principle says that polar compounds including ionic compound should be more soluble in polar solvent. Likewise, non–polar compound should be more soluble in non–polar solvent. Polar molecule: Polar molecules have large dipole moments. Non-Polar molecules: Non-polar molecules have bonded atoms with similar electronegativity results to have zero dipole moments. Solubility: is the quantity of solute (solid, liquid or gas) that will dissolve in a given amount of solvent (solid, liquid or gas) at definite temperature and pressure. Solubility depends on the physical and chemical property of solvent, solute and concentration of the solution.
Nanoscale diagram for the contents of a beaker containing water and sugar has to be drawn. Concept introduction: Like dissolves like principle: ‘Like dissolves like’ principle says that polar compounds including ionic compound should be more soluble in polar solvent. Likewise, non–polar compound should be more soluble in non–polar solvent. Polar molecule: Polar molecules have large dipole moments. Non-Polar molecules: Non-polar molecules have bonded atoms with similar electronegativity results to have zero dipole moments. Solubility: is the quantity of solute (solid, liquid or gas) that will dissolve in a given amount of solvent (solid, liquid or gas) at definite temperature and pressure. Solubility depends on the physical and chemical property of solvent, solute and concentration of the solution.
Solution Summary: The author explains the like dissolves like principle, which says that polar compounds including ionic compound should be more soluble, and non-polar molecules have bonded atoms with similar electronegativity results to have zero
Nanoscale diagram for the contents of a beaker containing water and sugar has to be drawn.
Concept introduction:
Like dissolves like principle: ‘Like dissolves like’ principle says that polar compounds including ionic compound should be more soluble in polar solvent. Likewise, non–polar compound should be more soluble in non–polar solvent.
Polar molecule: Polar molecules have large dipole moments.
Non-Polar molecules: Non-polar molecules have bonded atoms with similar electronegativity results to have zero dipole moments.
Solubility: is the quantity of solute (solid, liquid or gas) that will dissolve in a given amount of solvent (solid, liquid or gas) at definite temperature and pressure. Solubility depends on the physical and chemical property of solvent, solute and concentration of the solution.
(b)
Interpretation Introduction
Interpretation:
Nanoscale diagram for the contents of a beaker containing Carbon tetrachloride and sugar has to be drawn.
Concept introduction:
Like dissolves like principle: ‘Like dissolves like’ principle says that polar compounds including ionic compound should be more soluble in polar solvent. Likewise, non–polar compound should be more soluble in non–polar solvent.
Polar molecule: Polar molecules have large dipole moments.
Non-Polar molecules: Non-polar molecules have bonded atoms with similar electronegativity results to have zero dipole moments.
Solubility is the quantity of solute (solid, liquid or gas) that will dissolve in a given amount of solvent (solid, liquid or gas) at definite temperature and pressure. Solubility depends on the physical and chemical property of solvent, solute and concentration of the solution.
Using a cell of known pathlength b = 1.25115 x 10-3 cm, a water absorption spectrum was measured. The band at 1645 cm-1, assigned to the O-H bending, showed an absorbance, A, of 1.40.
a) Assuming that water density is 1.00 g/mL, calculate the water molar concentration c (hint: M= mole/L)
b) Calculate the molar absorptivity, a, of the 1645 cm-1 band
c) The transmitted light, I, can be written as I= Ioexp(-xb), where x is the absorption coefficient (sometimes designated as alpha), Io is the input light, and b is the cell pathlength. Prove that x= (ln10)*x*c
d) Calculate x for the 1645 cm-1 band
Convert 1.38 eV into wavelength (nm) and wavenumber (cm-1) (c = 2.998 x 108 m/s; h = 6.626 x 10-34 J*s).
Can you help me understand the CBC method on metal bridging by looking at this problem?
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