Answer:
7.78×10¯³ mole
Explanation:
From the question given above, the following data were obtained:
Volume (V) = 75 mL
Pressure (P) = 255 kPa
Temperature (T) = 22.5 °C
Number of mole (n) =?
Next, we shall convert 75 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
75 mL = 75 mL × 1 L / 1000 mL
75 mL = 0.075 L
Next, we shall convert 22.5 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
Temperature (T) = 22.5 °C
Temperature (T) = 22.5 °C + 273
Temperature (T) = 295.5 K
Finally, we shall determine the amount of oxygen molecules present in steel calorimeter. This can be obtained as follow:
Volume (V) = 0.075 L
Pressure (P) = 255 kPa
Temperature (T) = 295.5 K
Gas constant (R) = 8.314 KPa.L/Kmol
Number of mole (n) =?
PV = nRT
255 × 0.075 = n × 8.314 × 295.5
19.125 = n × 2456.787
Divide both side by 2456.787
n = 19.125 / 2456.787
n = 7.78×10¯³ mole
Thus, amount of oxygen molecules present in steel calorimeter is 7.78×10¯³ mole
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energy <---------------------
Answer:
Another name of activation energy is "Needed energy" .
Explanation:
Activation energy can be imagined of as the measurement of the potential barrier (seldom described the energy barrier) distributing minima of the potential energy surface concerning the initial and ultimate thermodynamic phase. For a chemical reaction or distribution to advance at a reasonable rate, the specific temperature of the operation should be high satisfactory so that there subsists an apparent number of particles with translational energy equivalent to or higher than the activation energy. The word Activation Energy was founded in 1889 by the Swedish expert Svante Arrhenius.