Answer:
1) potential energy of the bond.
2) Linear
3) The electrons are transferred from K to Cl.
4) ClF
5) Oxygen
6) Electrolysis
7) Double displacement
Explanation:
As two atoms approach each other in a bonding situation, the potential energy of the bond is minimized as the internuclear distance of the bonding atoms decreases.
BeH2 has two electron domains and the central beryllium atom is sp2 hybridized. According to valence shell electron pair repulsion theory. A molecule having two regions of electron density will lead to a linear molecule.
KCl is an ionic compound hence there is a transfer of electrons from K(metal) to Cl(nonmetal).
ClF has partial charges because it contains a polar covalent bond. The partial charges arise from the dipole within the molecule. LiF is a pure ionic compound formed by transfer of electrons from Li to F. The species possess full and not partial charges.
When an oxygen atom bonds with another oxygen atom, what has been formed is a homonuclear covalent bond. Since the electro negativity of the both atoms is exactly the same, a pure covalent bond is formed. Recall that polar covalent bonds are formed when there is a significant electro negativity difference between the bonding atoms.
When direct current is passed through certain salt solutions during electrolysis, gases may be evolved and collected at the appropriate electrodes.
A double-replacement reaction is a reaction in which the cations and anions present in two different ionic compounds that are reacting together exchange their positions to form two new compounds on the product side. For instance, look at the reaction shown in question 7 as a typical example of this;
AgNO3 (s) + NaCl (s) → AgCl (s) + NaNO3 (s).
Answer:
H2O<en<phen
Explanation:
The degree of d- splitting is observed from the intensity of colour. The order of d splitting from least to greatest is H2O<en<phen. Phen shows the greatest d-splitting. The degree of splitting of d- orbitals by ligands depends on their relative positions in the spectrochemical series. The spectrochemical series is an experimentally determined series. The series separates the ligands into strong field and weak field ligands. Strong field ligands are found towards the end of the series. Strong field ligands such as en and phen can participate in metal to ligand or ligand to metal pi-bonding. Hence they cause more d-splitting. Ethylendiamine and phenanthroline occur towards the end of the spectrochemical series hence the higher order of d-splitting.
Explanation:
As the given reaction is as follows.
So, according to the balanced equation, it can be seen that rate of formation of 
will be twice the rate of disappearance of 
.
And, it is known that rate of disappearance of reactant will be negative and rate of formation of products will be positive value.
This means that,
Rate of the reaction = -Rate of disappearance of
= ![k[H_{2}S][Cl_{2}]](https://tex.z-dn.net/?f=k%5BH_%7B2%7DS%5D%5BCl_%7B2%7D%5D)
= 
= 
M/s
Therefore, calculate the rate of formation of as follows.
Rate of formation of = 
= 
M/s
Thus, we can conclude that the rate of formation of is M/s.
Total number of atoms in the sample is the sum of number of atoms of all the elements present in the sample.
Number of gold, 
atoms in 
= 
atoms (given)
From the formula of compound that is it is clear that the number of potassium and gold are same whereas those of carbon and nitrogen are 2 times of them.
So, the number of atoms of each element is:
Number of potassium, 
atoms in = atoms
Number of carbon, 
atoms in = 
atoms = 
Number of nitrogen, 
atoms in = atoms =
Total number of atoms in = Number of gold, atoms+Number of potassium, atoms +Number of carbon, atoms + Number of nitrogen, atoms
Total number of atoms in = 
Total number of atoms in = 
atoms
Hence, the total number of atoms in is 
atoms.
