Answer:
(4) concentrated and supersaturated
Explanation:
At 50.°C, 90g of KNO3 lies above the solubility curve [on the Regents Reference Table G]. This indicates that the solution is supersaturated, meaning it contains more solute than will naturally dissolve, and was formed when a saturated solution cooled. Furthermore, the percent concentration of this solution is 90% KNO3 making this solution concentrated. This can be calculated using the formula for mass percent concentration.
Percent Mass = <u>Mass of Solute (g)</u> x 100
Mass of Solution (g)
Answer:- 3.84 grams
Solution:- Volume of the sample is 44.8 mL and the density is 1.03 gram per mL.
From the density and volume we calculate the mass as:
mass = volume*density

= 46.1 g
From given info, potassium bromide solution is 8.34% potassium bromide by mass. It means if we have 100 grams of the solution then 8.34 grams of potassium bromide is present in. We need to calculate how many grams of potassium bromide are present in 46.1 grams of the solution.
The calculations could easily be done using dimensional analysis as:

= 3.84 g KBr
Hence, 3.84 grams of KBr are present in 44.8 mL of the solution.
Density = mass/volume
so rearranged mass = volume x density
mass = 8.920 x 45 = 401.4g
rearrange (there are 1000grams in 1kg)
volume = mass/density
volume = 1000/8.920
volume = 112.1076233cm3
<h2>C</h2>
Explanation:
The atomic number of S is 
So,number of electrons in S is 
The electronic configuration of S is 
The orbital electronic configuration of S is 
So,the number of orbitals involved is 9.