Why is the atomic radius of sodium much smaller than the atomic radius of potassium
1 answer:
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Answer:
Due to the specific heat capacity of iron, 0.444 J/(g·°C), is more than the specific heat capacity for lead, 0.160 J/(g·°C)
Explanation:
The given parameters are;
The metals provided to melt the ice and their temperature includes;
One kg (1000 g) of iron;
Specific heat capacity = 0.444 J/(g·°C)
Temperature = 100°C
1 kg (1000 g) of lead
Specific heat capacity = 0.160 J/(g·°C)
Temperature = 100°C
Therefore, the heat provided to the ice of mass m, and latent heat of 334 J/g at 0°C by the metals are as follows;
For iron, we have;
ΔQ = Mass × Specific heat capacity × Temperature change
ΔQ = Heat obtained from the iron by the ice
ΔQ = 0.444 m × 1000 × (100 - 0) = 44400 J
Heat absorbed by the ice for melting, H = Heat obtained from the iron
∴ Heat absorbed by the ice for melting, H = Mass of ice × Latent heat of ice
H = Mass of ice × 334 J/g = 44400 J
Mass of ice melted by the iron = 44400 J/334 (J/g) ≈ 132.9 g
Mass of ice melted by the iron ≈ 132.9 g
For lead, we have;
ΔQ = Mass × Specific heat capacity × Temperature change
ΔQ = Heat obtained from the iron by the ice
ΔQ = 0.160 m × 1000 × (100 - 0) = 16000 J
Heat absorbed by the ice for melting, H = Heat obtained from the iron
∴ Heat absorbed by the ice for melting, H = Mass of ice × Latent heat of ice
H = Mass of ice × 334 J/g = 16000 J
Mass of ice melted by the lead = 16000 J/334 (J/g) ≈ 47.9 g
Mass of ice melted by the lead ≈ 47.9 g
Therefore, mass of ice melted by the iron, approximately 132.9 g, is more than mass of ice melted by the lead, approximately 47.9 g.
Answer:
The [SO₃²⁻]
Explanation:
From the first dissociation of sulfurous acid we have:
H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)
At equilibrium: 0.50M - x x x
The equilibrium constant (Ka₁) is:
With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:
Similarly, from the second dissociation of sulfurous acid we have:
HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)
At equilibrium: 7.94x10⁻²M - x x x
The equilibrium constant (Ka₂) is:
Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:
Therefore, the final concentrations are:
[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M
[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M
[SO₃²⁻] = 7.07x10⁻⁵M
[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M
So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.
I hope it helps you!
Explanation:
Dipole moment is defined as the measurement of the separation of two opposite electrical charges.
is a bent shaped molecule with a dipole moment of 1.87.
is also a bent shaped molecule with a dipole moment of 1.10.
is a also a bent shaped molecule and has a negligible dipole moment.
has a dipole moment of 0.29.
Therefore, given molecules are arranged according to their increasing dipole moment as follows.
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