What volume of 8.00 M HCI is needed to make 206 mL of 4.50 M HCI
1 answer:
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Answer:
The value of entropy change for the process
Explanation:
Mass of the ideal gas = 0.0027 kilo mol
Initial volume = 4 L
Final volume = 6 L
Gas constant for this ideal gas ( R ) =
Where = Universal gas constant = 8.314
⇒ Gas constant R = 8.314 × 0.0027 = 0.0224
Entropy change at constant temperature is given by,
Put all the values in above formula we get,
This is the value of entropy change for the process.
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.
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