11) A fixed amount of gas at 25.0 °C occupies a volume of 8.66 L when the pressure is 629 torr. Use Charles's law to calculate the volume (L) the gas will occupy when the temperature is increased to
112 °C while maintaining the pressure at 629 torr.
A) 1.93
B) 38.8
C) 11.2
D) 9.26
E) 6.70

Option C. 11.2 L

Explanation:

From the question given above, the following data were obtained:

Initial temperature (T₁) = 25.0 °C

Initial volume (V₁) = 8.66 L

Final temperature (T₂) = 112 °C

Pressure = constant

Final volume (V₂) =?

Next, we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 25 °C

Initial temperature (T₁) = 25 °C +273

Initial temperature (T₁) = 298 K

Final temperature (T₂) = 112 °C

Final temperature (T₂) = 112 °C + 273

Final temperature (T₂) = 385 K

Finally, we shall determine final volume of the gas. This can be obtained as follow:

Initial temperature (T₁) = 298 K

Initial volume (V₁) = 8.66 L

Final temperature (T₂) = 385 K

Pressure = constant

Final volume (V₂) =?

V₁/T₁ = V₂/T₂

8.66/298 = V₂/385

Cross multiply

298 × V₂ = 8.66 × 385

298 × V₂ = 3334.1

Divide both side by 298

V₂ = 3334.1 / 298

V₂ = 11.2 L

Thus, the final (i.e new) volume of the gas is 11.2 L

we need the questions , i may then you.

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.