Transcribed image text: Four liquids are described in the table below. Use the second column of the table to explain the order of their freezing points, and the third column to explain the order of their boiling points. For example, select '1' in the second column next to the liquid with the lowest freezing point. Select '2' in the second column next to the liquid with the next higher freezing point, and so on. In the third column, select '1' next to the liquid with the lowest boiling point, '2' next to the liquid with the next higher boiling point, and so on. Note: the density of water is 1.00g/mL .
One of the differences I can think of is that hydrogen is no longer listed as a group I element.
According to the mendeleev tables that I looked up, hydrogen is catorgrized as a group I element, along with Lithium, sodium, Potassium etc. However, nowadays, hydrogen does not belong to any groups in the periodic table. This is because there are arguments about whether hydrogen belongs to group I. Group I elements are all alkali metals, while hydrogen is not. However, some people says that hydrogen only have one outer shell electron so it should be in group I. Some people even say hydrogen should belong to group VII because it only needs one more electron in order to achieve the duplet of electrons.
Therefore as you may notice, hydrogen in modern periodic tables are put in the center of the periodic table on the top.
B. Conducts heat readily.
Can i have brainlyest?
Answer:
(D) Na₂SO₄•10H₂O (M = 286).
Explanation:
- The depression in freezing point of water by adding a solute is determined using the relation:
<em>ΔTf = i.Kf.m,</em>
Where, ΔTf is the depression in freezing point of water.
i is van't Hoff factor.
Kf is the molal depression constant.
m is the molality of the solute.
- Since, Kf and m is constant for all the mentioned salts. So, the depression in freezing point depends strongly on the van't Hoff factor (i).
- van't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.
(A) CuSO₄•5H₂O:
CuSO₄ is dissociated to Cu⁺² and SO₄²⁻.
So, i = dissociated ions/no. of particles = 2/1 = 2.
B) NiSO₄•6H₂O:
NiSO₄ is dissociated to Ni⁺² and SO₄²⁻.
So, i = dissociated ions/no. of particles = 2/1 = 2.
(C) MgSO₄•7H₂O:
MgSO₄ is dissociated to Mg⁺² and SO₄²⁻.
So, i = dissociated ions/no. of particles = 2/1 = 2.
(D) Na₂SO₄•10H₂O:
Na₂SO₄ is dissociated to 2 Na⁺ and SO₄²⁻.
So, i = dissociated ions/no. of particles = 3/1 = 3.
∴ The salt with the high (i) value is Na₂SO₄•10H₂O.
So, the highest ΔTf resulted by adding Na₂SO₄•10H₂O salt.
