Answer: D) 1 lone pair, square pyramidal
Explanation:
Formula used : ![\text{Number of electrons}=\frac{1}{2}[V+N-C+A]](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20electrons%7D%3D%5Cfrac%7B1%7D%7B2%7D%5BV%2BN-C%2BA%5D)
where,
V = number of valence electrons present in central atom = 7
N = number of monovalent atoms bonded to central atom = 5
C = charge of cation = 0
A = charge of anion = 0
Now we have to determine the hybridization of the 
molecule.
![\text{Number of electrons}=\frac{1}{2}\times [7+5+0-0]=6](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20electrons%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5B7%2B5%2B0-0%5D%3D6)
Bond pair electrons = 5
Lone pair electrons = 6-5 = 1
The number of electrons are 6 that means the hybridization will be 
and the electronic geometry of the molecule will be octahedral .
But as there are five atoms around the central iodine atom, the sixth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be square pyramidal.
D. liquid water vaporizes.
I have attached a phase diagram for water. As you can see, if kPa is 50, and the temperature is 100c. it is in the vapor phase.
The pure substance of water mixes with the waste. The waste does not dissolve into the water, therefore it is a mixture.
Answer:
Cl⁻, Na⁺, OH⁻
Explanation:
The titration is:
CuCl₂(aq) + 2 NaOH(aq) → Cu(OH)₂(s) + 2 NaCl(aq)
In solution, before the reaction, the ions are Cu²⁺ and Cl⁻. The addition of NaOH (Na⁺ + OH⁻) produce the precipitation of Cu²⁺ forming Cu(OH)₂(s). When you reach the equivalence point, there is no Cu²⁺ because precipitates completely. All OH⁻ ions reacts when are added but when Cu²⁺ is finished, excess OH⁻ ions still in solution helping to detect the equivalence point.
Thus, ions present after the equivalence point are:<em> Cl⁻, Na⁺</em> (Don't react, spectator ions), and <em>OH⁻</em>.
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<span>
4 multicellular yes autotrophic It is unable to move around its environment.
</span><span>D. kingdom protista </span>
