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
