A real gas behaves least like an ideal gas under the conditions of
1 answer:
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Explanation:
The given data is as follows.
Concentration = 0.1
= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions
=
T = = (30 + 273) K = 303 K
Formula for electric double layer thickness () is as follows.
=
where, = concentration =
Hence, putting the given values into the above equation as follows.
=
=
= m
or, =
= 1 nm (approx)
Also, it is known that =
Hence, we can conclude that addition of 0.1 of KCl in 0.1
of NaBr "
" will decrease but not significantly.
Answer:
Decomposition of aluminium oxide forms aluminium atoms and oxygen atoms.
Explanation:
<u>Decomposition reaction:</u>
When a single compound break down into two or more simpler products.
For example "AB" reactant undergoes decomposition to form "A" and "B" products.
The chemical reaction is as follows.
The given compound is aluminium oxide.
The decomposition reaction of aluminium oxide is a follows.
The balanced equation is as follows.
Therefore, Decomposition of aluminium oxide forms aluminium atoms and oxygen atoms.
Answer:
The correct answer is option a.
Explanation:
Equilibrium concentration cadmium ions =
Equilibrium concentration fluoride ions =
Molar solubility is the maximum concentration of salt present in water in ionic form beyond that no more salt will exist in its ionic form and will settle down in bottom of the solution.
The molar solubility of the solid cadmium fluoride = 0.0585 M
..[1]
Due to addition of sodium fluoride will increase concentration of fluoride in the solution.And due to common ion effect the equilibrium will shift in backward direction in [1], that is precipitation of more cadmium fluoride.
Hence, decrease in solubility will be observed.