So, when two different air masses meet, a boundary is formed. The boundary between two air masses is called a front. Weather at a front is usually cloudy and stormy. There are four different fronts- Cold, Warm, Stationary, and Occluded.
MO Diagram of C₂⁻ is shown below,
Bond order is calculated as,
Bond Order = [# of e⁻s in BMO]-[#of e⁻s in ABMO] / 2
Where,
BMO = Bonding Molecular Orbital
ABMO = Anti-Bonding Molecular Orbital
Putting values,
Bond Order = [9]-[4] / 2
Bond Order = 5 / 2
Bond Order = 2.5
Initially, weigh the correct amount of 
calculated from the formula.
Number of moles =
As 1 mole of is given and molar mass of is 
, then
1 mole of = 
Mass of in g = 
Thus,
1 mole of = 58.44 g.
Now, the weighed amount of i.e. 58.44 g is added to a 1 liter container and then add small amount of water in the container to dissolve the salt. After that, fill the container with distilled water to the graduation mark or until the total volume reaches 1 liter.
Thus, option (c) is correct i.e. first put one mole of salt in the container then add the water with stirring till the total volume reaches 1 liter.
Answer:
No, It would be a unsaturated solution
Explanation:
The solubility of a compound gives us information about how a compound may dissolve or not in a determinate solvent.
In this case we have Z, which in 25 °C the solubility of this compound is 40g/100 mL water. This means that if we have 60 g of Z and try to dissolve it in 100 mL of water, only 40 g of Z will solve and the remaining 20 g will be in the water as precipitate or remaining solid.
Now if you just put 40 g of Z in 100 mL water, it will dissolve completely in water, and in this case, we have a saturated solution. A saturated solution is when you dissolve a determinated quantity of a solute in a determinated quantity of solvent, without remaining of solid or excess of solvent.
According to this explanation, we now have 120 g of Z. To make a saturated solution of Z with this quantity, well, let's do math. If 40 g dissolves in 100 mL, then 80 g would be 200 mL and 120 g would have to be 300 mL of water. But in this case, we have 450 mL of water, we have more than 300 mL, an excess of water, so, the 120 g will dissolve but it's dissolved in more than the needed quantity to be a saturated solution, therefore, we have an unsaturated solution of Z (more solvent than the needed).
Hope this helps.
