Answer:

Explanation:
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In this case, we can consider that the given heat of combustion is indeed the heat of reaction since it corresponds to the combustion of propane, which is computed by using the heat formation of all the involved species as shown below:

Thus, since the heat of formation of gaseous carbon dioxide is -393.5 kJ/mol, water -241.8 kJ/mol and oxygen 0 kJ/mol, the heat of formation of propane is:

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<span>It makes a longitudinal wave because it stretches and compresses while as it slithers foward.
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1. The molar mass of the unknown gas obtained is 0.096 g/mol
2. The pressure of the oxygen gas in the tank is 1.524 atm
<h3>Graham's law of diffusion </h3>
This states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass i.e
R ∝ 1/ √M
R₁/R₂ = √(M₂/M₁)
<h3>1. How to determine the molar mass of the gas </h3>
- Rate of unknown gas (R₁) = 11.1 mins
- Rate of H₂ (R₂) = 2.42 mins
- Molar mass of H₂ (M₂) = 2.02 g/mol
- Molar mass of unknown gas (M₁) =?
R₁/R₂ = √(M₂/M₁)
11.1 / 2.42 = √(2.02 / M₁)
Square both side
(11.1 / 2.42)² = 2.02 / M₁
Cross multiply
(11.1 / 2.42)² × M₁ = 2.02
Divide both side by (11.1 / 2.42)²
M₁ = 2.02 / (11.1 / 2.42)²
M₁ = 0.096 g/mol
<h3>2. How to determine the pressure of O₂</h3>
From the question given above, the following data were obtained:
- Volume (V) = 438 L
- Mass of O₂ = 0.885 kg = 885 g
- Molar mass of O₂ = 32 g/mol
- Mole of of O₂ (n) = 885 / 32 = 27.65625 moles
- Temperature (T) = 21 °C = 21 + 273 = 294 K
- Gas constant (R) = 0.0821 atm.L/Kmol
The pressure of the gas can be obtained by using the ideal gas equation as illustrated below:
PV = nRT
Divide both side by V
P = nRT / V
P = (27.65625 × 0.0821 × 294) / 438
P = 1.524 atm
Learn more about Graham's law of diffusion:
brainly.com/question/14004529
Learn more about ideal gas equation:
brainly.com/question/4147359
Answer:
1,15mL = V₂
Explanation:
Based on Charle's law the volume is directely proportional to the absolute temperature in a gas under constant pressure. The equation is:
V₁T₂ = V₂T₁
<em>Where V is volume and T absolute temperature of a gas where 1 is initial state and 2, final state.</em>
The V₁ is 1.23mL
T₁ = 32°C + 273.15 = 305.15K
T₂ = T₁ - 20°C = 285.15K
Replacing:
1.23mL*285.15K = V₂*305.15K
<h3>1,15mL = V₂</h3>
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Answer:
The answer is
Explanation:
- A nonpolar organic compound that is insoluble in water