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3 years ago

A steel calorimeter has a volume of 75.0 mL and is charged with Oxygen gas to a

Chemistry

1 answer:

EastWind [94]3 years ago

7 0

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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Density is a quantity derived from the mass and volume

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