- Sedimentation and Decantation is the process of setting of solid particles from a liquid either to produce a concentrated slurry or to clarify a liquid containing solid particles . The solid must not be mixed with liquid. For example- sand insoluble in water
- Sublimation - In this process solid phase is directly transformed into vapour form without passing through liquid phase.This method is used in the purification of those substances which have sublimation property. At first the mixture should be sublimed before dissolution in liquid.
- EX- Iodine , naphthalene, ammonium chloride etc. Here, NH4Cl is seperated by sublimation.
- Evaporation is a technique used to separate out homogeneous mixtures where there is one or more dissolved salts. The process convert the liquid components from the solid components. The process typically involves heating the mixture until no more liquid remains.
- so, now by evaporation of the liquid solvent we finally get the salt that was dissolved in it that is NaCl.
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HA -----> H(+) + A(-)
0,2-x.........x..........x
100% - 99,4% = 0,6% = 0,006
x = 0,6%×0,2 = 0,006 × 0,2 = 0,0012
HA --------------> H(+) + A(-)
0,2-x.....................x..........x
0,2-0,0012.....0,0012..0,0012
0,1988.............0,0012..0,0012
K = [H+][A-]/[HA]
K = (0,0012)^2/0,1988
K = 0,0000072434
pKa = -logK= -log0,0000072434 = 5,14
:)
Answer:
Physical properties: Sodium chloride is a white crystalline solid with a density of 2.16 g/mL, and a melting point of 801 °C. It is also available as aqueous solutions of different concentrations, called saline solutions. Chemical properties: Sodium chloride is readily soluble in water and other polar solvents.
Explanation:
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Answer:
it's a measure of how far down light can penetrate through the water column. ... Because water clarity is closely related to light penetration, it has important implications for the diversity and productivity of aquatic life that a system can support
<span>a. Use PV = nRT and solve for n = number of mols O2.
mols NO = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols O2 to mols NO2. Do the same for mols NO to mols NO2. It is likely that the two values will not be the same which means one is wrong; the correct value in LR (limiting reagent) problems is ALWAYS the smaller value and the reagent producing that value is the LR.
b.
Using the smaller value for mols NO2 from part a, substitute for n in PV = nRT, use the conditions listed in part b, and solve for V in liters. This will give you the theoretical yield (YY)in liters. The actual yield at these same conditions (AY) is 84.8 L.
</span>and % will be 60%.