Pressure of Butane in the container at 135°C = 1.07 atm
Given:
The H°vap of butane is 24.3 kJ/mol.
starting out at 25 °C
Temperature at the end: 135 °C
2.3 atm of pressure
To Find:
The container's pressure
The perpendicular force per unit area, also known as the stress at a point within a confined fluid, is known as pressure in the physical sciences.
Equation of Clausius-Clapeyron
P2 = 1.07 atm and ln (P2/2.3 atm)
= - 24.3*110/8.31
The pressure in the container at 135°C (ΔH°vap ) is 1.07 atm
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Answer:
45.0 L is the volume of gas will the balloon contain at 1.35 atm and 253 K.
Explanation:
Using Ideal gas equation for same mole of gas as
Given ,
V₁ = 25.0 L
V₂ = ?
P₁ = 2575 mm Hg
Also, P (atm) = P (mm Hg) / 760
P₁ = 2575 / 760 atm = 3.39 atm
P₂ = 1.35 atm
T₁ = 353 K
T₂ = 253 K
Using above equation as:
Solving for V₂ , we get:
<u>V₂ = 45.0 L</u>
45.0 L is the volume of gas will the balloon contain at 1.35 atm and 253 K.
Respuesta:
292 g / mol;
Por favor, consulte la explicación.
Explicación:
El número de átomos de cada elemento en el compuesto: Fe2 (co3) 3.
Fe = 2; C = 3; 0 = 3 * 3 = 9
Fe2 = 112 g / mol
C = 12 g / mol
O = 16 g /
Masa molecular = ((112 + (12 * 3) + (16 * 9)
= (112 + 36 + 144)
= 292 g / mol.
Total = 2 + 3 + 9 = 14
Fe2 = 112/292 * 100% = 38,356%
C = 36/292 * 100% = 12,328 * '
O = (16 * 9) / 292 * 100% = 49,32%
Answer:
hypothesis
Explanation:
a question that can be answered wity a hypothesis is a question a scientist can answer
Answer:
1.428 moles
Explanation:
If 0.0714 moles of N2 gas occupies 1.25 L space,
how many moles of N2 have a volume of 25.0 L?
Assume temperature and pressure stayed constant.
we experience it 0.0714 moles: 1.25L space
x moles : 25L of space
to get the x moles, cross multiply
(0.0714 x 25)/1.25
1.785/1.25 = 1.428 moles
