The type of colloid that is represented by each of the substance has to be given – (a) Rain cloud (b) milk of magnesia (c) soap suds (e) silt in water Concept Introduction: Based on the particle size of the components, solutions can be of three types – true solutions, colloidal solutions and suspension. Colloidal solutions have the particle size of the range 1 - 1 0 0 0 n m . A colloidal solution is made of dispersed phase and dispersion medium which corresponds to solute and solvent respectively. Depending upon the physical state of dispersed phase and dispersion medium, there are various types of colloids such as gel, foam, alloy, sol etc.
The type of colloid that is represented by each of the substance has to be given – (a) Rain cloud (b) milk of magnesia (c) soap suds (e) silt in water Concept Introduction: Based on the particle size of the components, solutions can be of three types – true solutions, colloidal solutions and suspension. Colloidal solutions have the particle size of the range 1 - 1 0 0 0 n m . A colloidal solution is made of dispersed phase and dispersion medium which corresponds to solute and solvent respectively. Depending upon the physical state of dispersed phase and dispersion medium, there are various types of colloids such as gel, foam, alloy, sol etc.
Solution Summary: The author explains the type of colloid that is represented by each of the substances — aerosol, milk of magnesia, and silt in water.
The type of colloid that is represented by each of the substance has to be given –
(a) Rain cloud (b) milk of magnesia (c) soap suds (e) silt in water
Concept Introduction:
Based on the particle size of the components, solutions can be of three types – true solutions, colloidal solutions and suspension. Colloidal solutions have the particle size of the range 1-1000nm. A colloidal solution is made of dispersed phase and dispersion medium which corresponds to solute and solvent respectively. Depending upon the physical state of dispersed phase and dispersion medium, there are various types of colloids such as gel, foam, alloy, sol etc.
Calculate the free energy of formation of 1 mol of Cu in cells where the electrolyte is 1 mol dm-3 Cu2+ in sulfate solution, pH 0. E° for the Cu2+/Cu pair in this medium is +142 mV versus ENH.Assume the anodic reaction is oxygen evolution.Data: EH2 = -0.059 pH (V) and EO2 = 1.230 - 0.059 pH (V); 2.3RT/F = 0.059 V
If the normal potential for the Fe(III)/Fe(II) pair in acid at zero pH is 524 mV Hg/Hg2Cl2 . The potential of the saturated calomel reference electrode is +246 mV versus the NHE. Calculate E0 vs NHE.
Given the galvanic cell whose scheme is: (-) Zn/Zn2+ ⋮⋮ Ag+/Ag (+). If we know the normal potentials E°(Zn2+/Zn) = -0.76V and E°(Ag+/Ag) = 0.799 V. Indicate the electrodes that are the anode and the cathode and calculate the E0battery.
Chapter 12 Solutions
Bundle: General Chemistry, Loose-Leaf Version, 11th + LabSkills PreLabs v2 for Organic Chemistry (powered by OWLv2), 4 terms (24 months) Printed ... for Ebbing/Gammon's General Chemistry, 11th