Answer:
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100%.....50%.....25%......12.5%......6.25%......3.125%......1.5625%
...........1............2...........3..............4.................5................6..................
After six half-lives would be found 1.5625% of readioactive species.
Organic chemistry
biochemistry
inorganic chemistry
analytical chemistry
physical chemistry
<span>IF WE TAKE N=1 IT IS CALLED GROUND STATE. THEN THE OTHER FOLLOWING HIGHER STATES ARE CALLED EXCITED STATES. IF THE ELECTRON IN AN ATOM JUMPS FROM A STATE TO A LOWER STATE, IT LOSES ENERGY. FROM THE GIVEN STATEMENT, THE WAY TO FIND THE ENERGY RELEASED IS GIVEN BY THE FORMULA, E(n)=(-13.6 eV)/n^2. FIRST TO FIND E(5)=(-13.6 eV)/(5)^2, WE GET E(5)=-0.544 eV. E(3)=(-13.6 eV)/(3)^2, WE GET E(3)=-1.5111 eV. THEN WE HAVE TO FIND THE ENERGY TRANSITION LEVEL. ON SUBTRACTING WE GET 0.967eV. THIS ENERGY HAS TO BE CONVERTED IN JOULES. SO WE MAKE E=0.967*(1.60*10^(-19)) J/eV, WHICH CORRESPONDS TO 0.15472*10^(-18) J. WE NEED TO FIND TO THE WAVELENGTH. THE CORRESPONDING FORMULA E = hf = hc/λ, λ = hc/E. BY SUBSTITUTING THE KNOWN VALUES, WE GET THE ANSWER TO BE 1285.548 NM.</span>
We have to determine if the given compound is soluble or insoluble in water.
The given compound is copper (II) hydroxide,
. According to the solubility rules the hydroxides of all metal ions are insoluble except that of ammonium and the alkali metal ions. So copper(II) hydroxide will be insoluble in water. In aqueous solution, copper (II) hydroxide would exist as a sparingly soluble compound in which the undissociated compound remains in equilibrium with the ions
and
. The equilibrium would remain mostly towards undissociated form.

Therefore,
is insoluble in water.