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

the answer is alpha particles hope i you

answer:

the answer is 6 valence electrons.

explanation:

valence is the # of electrons on the outer most shell of an atom.

oxygen has 6.

although, the valency of oxygen is 2, because it only needs to make 2 covalent bonds, or gain 2 electrons, to achieve the stable octet configuration of a noble gas.

that's it!

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