Answer:
2.39 atm
Explanation:
We'll begin by calculating the number of mole in 0.41 g of neon (Ne). This can be obtained as follow:
Mass of Ne = 0.41 g
Molar mass of Ne = 20 g/mol
Mole of Ne =.?
Mole = mass / molar mass
Mole of Ne = 0.41 / 20
Mole of Ne = 0.0205 mole
Next we shall convert 200 mL to L.
1000 mL = 1 L
Therefore,
200 mL = 200 mL × 1 L / 1000 mL
200 mL = 0.2 L
Next, we shall convert 11 °C to Kelvin temperature.
T(K) = T(°C) + 273
T(°C) = 11 °C
T(K) = 11 + 273
T (K) = 284 K
Finally, we shall determine the pressure. This can be obtained as follow:
Mole of Ne (n) = 0.0205 mole
Volume (V) = 0.2 L
Temperature (T) = 284 K
Gas constant (R) = 0.0821 atm.L/Kmol
Pressure (P) =?
PV = nRT
P × 0.2 = 0.0205 × 0.0821 × 284
P × 0.2 = 0.4779862
Divide both side by 0.2
P = 0.4779862 / 0.2
P = 2.49 atm
Therefore, the pressure of the gas is 2.39 atm
Idk i’m on that question now
When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.
The heat of combustion (∆Hc) for an unknown hydrocarbon is -8.21 kJ/mol. The heat released by the combustion of 0.424 moles of the hydrocarbon is:
According to the law of conservation of energy, the sum of the heat released by the combustion (Qc) and the heat absorbed by the bomb calorimeter (Qb) is zero.
Given the heat absorbed by the bomb calorimeter (Qb) and the heat capacity of the bomb calorimeter (C), we can calculate the temperature change (ΔT) using the following expression.
When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.
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