Explanation:
It is known that in a simple cubic unit cell the atoms are only present at the corner of the unit cell. This means that there are in total 8 atoms present in a simple cubic unit cell.
Therefore, in one simple cubic unit cell sharing of one atom is only
.
Hence, the total number of atoms in a unit cell will be as follows.
= 1
Thus, we can conclude that there is 1 calcium atom present in each unit cell.
Answer:
Saturated solution
We should raise the temperature to increase the amount of glucose in the solution without adding more glucose.
Explanation:
Step 1: Calculate the mass of water
The density of water at 30°C is 0.996 g/mL. We use this data to calculate the mass corresponding to 400 mL.

Step 2: Calculate the mass of glucose per 100 g of water
550 g of glucose were added to 398 g of water. Let's calculate the mass of glucose per 100 g of water.

Step 3: Classify the solution
The solubility represents the maximum amount of solute that can be dissolved per 100 g of water. Since the solubility of glucose is 125 g Glucose/100 g of water and we attempt to dissolve 138 g of Glucose/100 g of water, some of the Glucose will not be dissolved. The solution will have the maximum amount of solute possible so it would be saturated. We could increase the amount of glucose in the solution by raising the temperature to increase the solubility of glucose in water.
Answer:

for the balanced equation

The mole fraction of KCl in the solution is 0.1051
calculation
mole fraction of KCl in solution = moles of KCl / total number of moles(moles of KCl +moles of H2O)
moles=mass/molar mass
mass of KCl=32.7g
molar mass of KCl= 39 +35.5
moles of KCl is therefore= 32.7g/74.5 g/mol=0.439 moles
find the moles of H2O= mass of H2O/molar mass
mass of H2O=100-32.7=67.3g
molar mass of H2O=( 1 x2) +16=18 g/mol
moles = 67.3/18 =3.739 moles
total moles=3.739+0.439=4.178 moles
mole fraction is therefore=0.439/4.178=0.1051