Answer:
filter the hot mixture.
Explanation:
Solid is stayed undissolved since the arrangement is gotten super saturated. On the off chance that solid molecule is available recrysallization won't happen in this way we need expel the solid molecule by filtarion in hot condition itself . Subsequently, arrangement become totally homogenous and recrysallization item will shaped by moderate cooling
Answer:
18.84 g of silver.
Explanation:
We'll begin by calculating the number atoms present in 5.59 g of sulphur. This can be obtained as follow:
From Avogadro's hypothesis,
1 mole of sulphur contains 6.02×10²³ atoms.
1 mole of sulphur = 32 g
Thus,
32 g of sulphur contains 6.02×10²³ atoms.
Therefore, 5.59 g of sulphur will contain = (5.59 × 6.02×10²³) / 32 = 1.05×10²³ atoms.
From the calculations made above, 5.59 g of sulphur contains 1.05×10²³ atoms.
Finally, we shall determine the mass of silver that contains 1.05×10²³ atoms.
This is illustrated below:
1 mole of silver = 6.02×10²³ atoms.
1 mole of silver = 108 g
108 g of silver contains 6.02×10²³ atoms.
Therefore, Xg of silver will contain 1.05×10²³ atoms i.e
Xg of silver = (108 × 1.05×10²³)/6.02×10²³
Xg of silver = 18.84 g
Thus, 18.84 g of silver contains the same number of atoms (i.e 1.05×10²³ atoms) as 5.59 g of sulfur
Elements with three p-electrons....
That would be N, P, As, Sb, and Bi -- elements in group 15
For example, energy diagram showing "empty" orbitals up through the 3p.
.....3p __ __ __
3s __
.....2p __ __ __
2s __
1s __
Energy diagram of phosphorous showing three unpaired electrons in 3p-sublevel
.....3p ↑_ ↑_ ↑_
3s ↑↓
.....2p ↑↓ ↑↓ ↑↓
2s ↑↓
1s ↑↓
According to Hund's rule, the electrons singly occupy the p-orbitals, and all have the same spin.
<u>Answer:</u> The molarity of Iron (III) chloride is 0.622 M.
<u>Explanation:</u>
Molarity is defined as the number of moles present in one liter of solution. The equation used to calculate molarity of the solution is:

Or,

We are given:
Mass of iron (III) chloride = 1.01 g
Molar mass of iron (III) chloride = 162.2 g/mol
Volume of the solution = 10 mL
Putting values in above equation, we get:

Hence, the molarity of Iron (III) chloride is 0.622 M.
It will melt
This is because the molecules in the metal begin to move faster and faster turning from solid to liquid