Response:
V1 = 20.3L
Clarification:
P2 = 811.4Pa
V2 = 25.6L
P1 = 1023.6Pa
V1 =?
To answer this query, we will utilize Boyle's law, which states that the volume of a gas at constant temperature is inversely related to its pressure.
In mathematical terms,
V = k / P, where k = PV
The relationship can be defined as P1 × V1 = P2 × V2 = P3 × V3 =......=Pn × Vn
This simplifies to P1 × V1 = P2 × V2
Let’s rearrange for V1
V1 = (P2 × V2) / P1
Substituting values gives
V1 = (811.4 × 25.6) / 1023.6Pa
So, V1 = 20771.84 / 1023.6
This results in V1 = 20.29L, rounded to 20.3L
15.9 KJ/mol Explanation: Given data: Temperature = T1 = 307 K, Temperature = T2 = 343 K, Gas constant R = 8.314 J/(mol • K), rate constant = k2/K1 = 89. To determine: Activation energy (in kJ/mol) = Ea =? Formula: The Arrhenius equation establishes the relationship between temperature and reaction rates. Here, in this equation, k = the rate constant, Ea = the activation energy, R = the Universal Gas Constant, T = the temperature. Solution: ln 89 = Ea / 8.314 J/mol.K * (0.0325 - 0.00291). then ln 89 = Ea / 8.314 J/mol.K * (2.95 x 10^2). Resulting in 4.488 = Ea / 8.314 J/mol.K * (2.95 x 10^2). Therefore, Ea = 4.488 * (2.95 x 10^2) / 8.314 J/mol.K which simplifies to Ea = 0.1324 / 8.314. Thus, Ea = 0.0159 and finally, Ea = 1.59 x 10^2 J/mol or 15.9 KJ/mol.
<span>The partial pressure of oxygen is 438.0 mmHg. The ideal gas equation is expressed as PV = nRT where P represents pressure, V denotes volume, n is the number of moles, R is the ideal gas constant (8.3144598 (L*kPa)/(K*mol)), and T signifies absolute temperature. To convert from Celsius to Kelvin, we have 43.4 + 273.15 = 316.55 K. For the pressure conversion from mmHg to kPa: 675.9 mmHg * 0.133322387415 = 90.11260165 kPa. When solving for n using the ideal gas equation, we derive n = PV / (RT) which provides n = 90.11260165 kPa * 16.2 L / (8.3144598 (L*kPa)/(K*mol) * 316.55 K)= 1459.824147 L*kPa / 2631.94225 (L*kPa)/(mol), resulting in n = 0.554656603 mol. Thus, we have 0.554656603 moles of gas particles. Next, we determine the contribution from oxygen. The atomic weight of oxygen is 15.999 g/mol, while argon is 39.948 g/mol, and the molar mass of O2 is 31.998 g/mol. We establish the relationships where M is the number of moles of O2, and 0.554656603 - M gives the number of moles of Ar. Setting up the equation: M * 31.998 + (0.554656603 - M) * 39.948 = 19.3, we solve for M resulting in 0.359424148 moles of oxygen out of 0.554656603 total moles. This leads to oxygen providing 0.359424148 / 0.554656603 = 0.648012024 or 64.8012024% of the total pressure of 675.9 mmHg. The partial pressure therefore calculates to 675.9 * 0.648012024 = 437.9913271 mmHg, rounded to 438.0 mmHg</span>
The resulting temperature, following the change in volume and pressure, is -27.26°C. To find this temperature, we apply the combined gas law equation—a formulation where initial and final pressures, volumes, and temperatures are compared. Given the initial conditions and transformations, when we input the stipulated values, we reach the conclusion that the resultant temperature is -27.26°C.
