Answer:
can only be determined experimentally.
Explanation:
In the early days of inorganic chemistry, the structure of complex ions remained a mystery hence the name ''complex''.
These ions appear to have structures that defied accurate elucidation. However, by diligent laboratory investigation, Alfred Werner was able to accurately determine the structure of cobalt complexes. As a result of this, he is regarded as a pathfinder in coordination chemistry.
Hence, the structure of complex ions can only be determined experimentally.
Answer:
Binary compound
Explanation:
Binary compounds:
The compounds which are made up of the atoms of only two elements are called binary compounds.
For example:
The following compounds are binary:
HCl
H₂O
NH₃
HCl is binary because it is composed of only hydrogen and chlorine. Ammonia is also binary compound because it is made up of only two elements nitrogen and hydrogen.
water is also binary because it is also made up of only two elements hydrogen and oxygen.
SF₆ is binary compound because it consist of atoms of only two elements i.e, sulfur and fluorine.
Answer: The bond between boron and hydrogen in boron trihydride is covalent bond.
Explanation:
The type of bonding between the atoms forming a compound is determined by using the electronegativity difference between the atoms. According to the pauling's electronegativity rule:
- If
, then the bond is non-polar. - If
, then the bond will be covalent. - If
, then the bond will be ionic.
We are given:
Electronegativity for boron = 2.0
Electronegativity for hydrogen = 2.1
As, 
is less than 1.7 and not equal to 0. Hence, the bond between boron and hydrogen is covalent bond.
Answer:
Water to move into the cell
Answer:
![[I_2]=[Br]=0.31M](https://tex.z-dn.net/?f=%5BI_2%5D%3D%5BBr%5D%3D0.31M)
Explanation:
Hello there!
In this case, according to the given information, it is possible for us to set up the following chemical equation at equilibrium:
Now, we can set up the equilibrium expression in terms of x (reaction extent) whereas the initial concentration of both iodine and bromine is 0.5mol/0.250L=2.0M:
![K=\frac{[IBr]^2}{[I_2][Br_2]} \\\\1.2x10^2=\frac{(2x)^2}{(2.0-x)^2}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BIBr%5D%5E2%7D%7B%5BI_2%5D%5BBr_2%5D%7D%20%5C%5C%5C%5C1.2x10%5E2%3D%5Cfrac%7B%282x%29%5E2%7D%7B%282.0-x%29%5E2%7D)
Thus, we solve for x as show below:
Therefore, the concentrations of both bromine and iodine are:
![[I_2]=[Br]=2.0M-1.69M=0.31M](https://tex.z-dn.net/?f=%5BI_2%5D%3D%5BBr%5D%3D2.0M-1.69M%3D0.31M)
Regards!
