Answer:
The van't hoff factor of 0.500m K₂SO₄ will be highest.
Explanation:
Van't Hoff factor was introduced for better understanding of colligative property of a solution.
By definition it is the ratio of actual number of particles or ions or associated molecules formed when a solute is dissolved to the number of particles expected from the mass dissolved.
a) For NaCl the van't Hoff factor is 2
b) For K₂SO₄ the van't Hoff factor is 3 [it will dissociate to give three ions one sulfate ion and two potassium ions]
Out of 0.500m and 0.050m K₂SO₄, the van't hoff factor of 0.500m K₂SO₄ will be more.
c) The van't Hoff factor for glucose is one as it is a non electrolyte and will not dissociate.
Answer:
Francium (Fr)
Explanation:
From the given choices, francium will have the lowest ionization energy.
Ionization energy is the energy required to remove the most loosely held electron within an atom.
The magnitude of the ionization energy depends on the characteristics of the atom in relation to its nuclear charge, atomic radius, stability etc.
- Generally on the periodic table, ionization energy increases from left to right on the table
- As you go from metals to non-metals and to gases, the value of the ionization energy increases steadily.
- Down the group, the value reduces.
- Since Francium is the most metallic of all the given choices, it has the highest ionization energy.
Answer:
10.2 mg
Explanation:
Step 1: Calculate the total amount of water she drank
1 year has 365 days and she lived in Chicago for 2 years = 2 × 365 days = 730 days.
If she drank 1.4 L of water per day, the total amount of water she drank is:
730 day × 1.4 L/day = 1022 L
Step 2: Calculate the amount of Pb in 1022 L of water
The concentration of Pb is 10 ppb (10 μg/L).
1022 L × 10 μg/L = 10220 μg
Step 3: Convert 10220 μg to milligrams
We will use the conversion factor 1 mg = 1000 μg.
10220 μg × 1 mg/1000 μg = 10.2 mg