The volume of a gas is 8.0 L at 9.9°C and {P3} atm. What volume will the gas occupy if the temperature is raised to 9.3°C at constant pressure?
Select one:
O a. 7.98
Ob.0.263
C. 8.02
O d. 228

Option A. 7.98 L.

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 8 L

Initial temperature (T1) = 9.9 °C

Final temperature (T2) = 9.3 °C

Final volume (V2) =..?

Next, we shall convert celsius temperature to Kelvin temperature. This can be achieved as shown below:

Temperature (K) = temperature (°C) + 273

T(K) = T (°C) + 273

Initial temperature (T1) = 9.9 °C

Initial temperature (T1) = 9.9 °C + 273 = 282.9 K

Final temperature (T2) = 9.3 °C

Final temperature (T2) = 9.3 °C + 273 = 282.3 K

Finally, we shall determine the new volume of the gas as follow:

Initial volume (V1) = 8 L

Initial temperature (T1) = 282.9 K

Final temperature (T2) = 282.3 K

Final volume (V2) =..?

V1/T1 = V2/T2

8/282.9 = V2/282.3

Cross multiply

282.9 x V2 = 8 x 282.3

Divide both side by 282.9

V2 = (8 x 282.3)/282.9

V2 = 7.98 L

Therefore, the new volume of the gas is 7.98 L

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%