The nucleus, that dense central core of the atom, contains both protons and neutrons. Electrons are outside the nucleus in energy levels. Protons have a positive charge,neutrons have no charge, and electrons have a negative charge. A neutral atom contains equal numbers of protons and electrons.
<u>Answer:</u> The value of <em>i</em> is 1.4 and 40% dissociation of 100 particles of zinc sulfate will yield 60 undissociated particles.
<u>Explanation:</u>
The equation used to calculate the Vant' Hoff factor in dissociation follows:
where,
= degree of dissociation = 40% = 0.40
i = Vant' Hoff factor
n = number of ions dissociated = 2
Putting values in above equation, we get:
The equation used to calculate the degee of dissociation follows:
Total number of particles taken = 100
Degree of dissociation = 40% = 0.40
Putting values in above equation, we get:
This means that 40 particles are dissociated and 60 particles remain undissociated in the solution.
Hence, 40% dissociation of 100 particles of zinc sulfate will yield 60 undissociated particles.
There are several ways of expressing concentration of solution. Few of them are listed below
1) mass percentage
2) volume percentage
3) Molarity
4) Normality
5) Molality
In most of the drugs, concentration is expressed either in terms of mass percentage or volume percentage. For, solid in liquid type systems, mass percentage is convenient way of expressing concentration, while for liquid in liquid type solutions, expressing concentration in terms of volume percentage is preferred. Present system is an example of liquid in liquid type solution
Here, concentration of H2O2 is given antiseptic = 3.0 % v/v
It implies that, 3ml H2O2 is present in 100 ml of solution
Thus, 400 ml of solution would contain 4 X 3 = 12 ml H2O2
FILTRATION When the substances in a mixture have different particle sizes, they are separated by filtration. The mixture is poured through a sieve or filter. The smaller particles slip through the holes, but the larger particles do not.p
Ans: The entropy change for the given reaction is 93.3 J/K
Given reaction:
Br2(l) → Br2(g)
ΔS = ∑n(products)S⁰(products) - ∑n(reactants)S⁰(reactants)
= 1 mole* S°(Br2(g)) - 1 mole*S°(Br2(l))
= 1 mole *245.5 J/mol-K - 1 mole*152.2 J/mol-K
= 93.3 J/K
