Answer:
Mass of oxygen = 2.2 g
Explanation:
Given data:
Mass of benzoic acid= 8.20 g
Mass of oxygen= ?
Solution:
Molar mass of oxygen = 16×2 g/mol
Molar mass of C₆H₅COOH = 7×12 + 1×6 + 2×16
Molar mass of C₆H₅COOH = 84 + 6 + 32
Molar mass of C₆H₅COOH = 122g/mol
Mass of oxygen in 8.20 g of C₆H₅COOH :
Mass of oxygen = 32 g.mol⁻¹/122 g.mol⁻¹ × 8.20 g
Mass of oxygen = 2.2g
Answer:
20.79 kilojoules
Explanation:
Using Q = m×c×∆T
Where;
Q = Quantity of heat (J)
c = specific heat capacity of solid DMSO (1.80 J/g°C)
m = mass of DMSO
∆T = change in temperature
According to the provided information, m= 50g, initial temperature = 19.0°C, final temperature= 250.0°C
Q = m×c×∆T
Q = 50 × 1.80 × (250°C - 19°C)
Q = 90 × 231
Q = 20790 Joules
To convert Joules to kilojoules, we divide by 1000 i.e.
20790/1000
= 20.79 kilojoules
Hence, 20.79 kilojoules of energy is required to convert 50.0 grams of solid DMSO to gas.
Answer:
Four (4) hydrogen atoms
Explanation:
C=C-C(triple bond)C
only four hydrogen atoms can bond with the vacant space
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.
