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Doss [256]
4 years ago
13

If the density of Mercury is 1.36 × 10 by 4 Kgm - 3 at 0 degrees. Calculate its value at 100 degrees and at 22 degrees. Take cub

ic expansivity of mercury as equal to 180 × 10-6 K - 1
Chemistry
1 answer:
Tasya [4]4 years ago
4 0

Answer:

1.35 × 10⁴ kg/m³ at 22 °C; 1.34 × 10⁴ kg/m³ at 100 °C

Explanation:

The cubic expansivity (γ) of a liquid is the fractional change in volume per unit change in temperature.

\gamma = (\frac{\Delta V }{ V_0} )(\frac{1 }{ \Delta T} )   Multiply by V₀ΔT and transpose

ΔV = γV₀ΔT  

and

V = V₀ + ΔV

===============

<em>At 0 °C </em>

Assume you have 1 m³ of Hg

ρ = m/V     Multiply by V and transpose

m = ρV

ρ = 1.36 × 10⁴ kg/m³

m = 1.36 × 10⁴ × 1 = 1.36 × 10⁴ kg

===============

<em>At 22 °C </em>

Assume that you have 1 m³ of Hg

γ = 180 × 10⁻⁶ K⁻¹

ΔT = 22 °C – 0 °C = 22 °C

ΔV = 180 × 10⁻⁶ × 22

ΔV = 3.96 × 10⁻³ m³      Calculate volume

V = 1 + 0.00396

V = 1.00396 m³             Calculate density

ρ = 1.36 × 10⁴/1.00396

ρ = 1.35 × 10⁴ kg/m³

===============

<em>At 100 °C </em>

ΔT = 100 °C – 0 °C = 100 °C

ΔV = 180 × 10⁻⁶ × 100

ΔV = 0.0180 m³      Calculate volume

V = 1 + 0.0180

V = 1.0180 m³          Calculate density

ρ = 1.36 × 10⁴/1.0180

ρ = 1.34 × 10⁴ kg/m³

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The combustion of 0.570 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera
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Answer:

The temperature change from the combustion of the glucose is 6.097°C.

Explanation:

Benzoic acid;

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

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Moles of benzoic acid = \frac{0.570 g}{122.12 g/mol}=0.004667 mol

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Q=3,228 kJ/mol \times 0.004667 mol=15.0668 kJ=15,066.8 J

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Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=2.900 g

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Energy released by the 0.016097 moles of calorimeter  combustion:

Q'=2,780 kJ/mol \times 0.016097 mol=44.7491 kJ=44,749.1 J

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

Q'=C\times \Delta T'

44,749.1 J=7,338.92 J/^oC\times \Delta T'

\Delta T'=6.097^oC

The temperature change from the combustion of the glucose is 6.097°C.

6 0
3 years ago
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