Erosols aren't aerosols at all. No, really, let's be clear about this. An aerosol is really the cloud<span> of </span>liquid and gas<span>that comes out of an aerosol can, not the can itself. In fact, to be strictly correct about it, an aerosol is a fine mist of liquid, or lots of solid particles, widely and evenly dispersed throughout a gas. So clouds, fog, and steam from your kettle are all examples of aerosols, because they're made up of </span>water<span> droplets dispersed through a much bigger volume of air. Smoke is an aerosol too, though unlike those other examples (which are liquids dispersed in gases) it's made up of </span>solid<span> particles of unburned carbon mixed through a cloud of warm, rising air. Even </span>candles<span> make aerosols: the smoky steam swirling above a candle flame consists of soot and water vapor dispersed through hot air.
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Answer : The incorrect option is, (d) The reactant that was the smallest given mass is the limiting reagent.
Explanation :
Limiting reagent : It is the reagent that is completely consumed in the chemical reaction when the chemical reaction is complete. No amount is left after the reaction is complete. The amount of product obtained is determined by the limiting reagent. A balanced equation is necessary to determine which reactant is limiting reagent.
Excess reagent : It is the reagent that are not completely consumed in the chemical reaction. That means the reagent is in excess amount. Some amount of the excess reagent is left over after the reaction is complete.
From this we conclude that the options, A, B and C are correct. While the option D is incorrect.
Option D is incorrect because it is not necessary the reactant that was the smallest given mass is the limiting reagent but it is judge by the number of moles present in the reaction.
Hence, the incorrect option is, (d)
Examples include carbonated water (i.e. soda water); honey; sugar syrup (used in confectionery); supersaturated drug delivery systems. "SDDS"; and sodium acetate solutions prepared from 160 g NaOAc and 30 mL water.
Answer:
a) pH will be 12.398
b) pH will be 4.82.
Explanation:
a) The moles of NaOH added = molarity X volume (L) = 2 X 0.01 = 0.02 moles
The total volume after addition of pure water = 0.780+0.01 = 0.79 L
The new concentration of /NaOH will be:
the [OH⁻] = 0.025
pOH = -log [OH⁻] = 1.602
pH = 14 -pOH = 12.398
b) The buffer has butanoic acid and butanoate ion.
i) Before addition of NaOH the pH will be calculated using Henderson Hassalbalch's equation:
![pH=pKa+log\frac{[salt]}{[acid]}](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D)
pKa=
ii) on addition of base the pH will increase.
