Answer:
Anhydrous calcium chloride dissolves and becomes liquid
Anhydrous copper (ii) sulphate will produce crystal particles
Explanation:
Anhydrous calcium chloride is deliquescent and hence when it is exposed to air, it absorbs water from air. After absorbing water, it dissolves and after some time a pool of clear liquid appears.
Anhydrous copper (ii) sulphate will form crystal structures and the following reaction will takes place
CuSO4 + 5 H20 --> CuSO4.5H2O
Answer:
Percent ionic character of HI bond is 4.91%.
Explanation:
<h3>
Given Data:</h3>
Measured Dipole = 0.380D
bond distance = d = 161pm = 1.61*10^-8 cm
<h3>
Calculation:</h3>
% ionic character is determined by following equation:
% ionic= (dipole measured/dipole calculated)*100
Now,

(In above step 3*10^8 is multiplied to convert coulomb into esu)

As,

So,

Now we can % ionic character using above equation:
%ionic=(0.380D/7.728D)*100
% ionic character=4.91%
Answer:
The electrons are found on the atomic nucleus.
Explanation: