Answer:
V of Sulfur tetrafluoride is 17.2 L
Step-by-step explanation:
Given data;
T = -6°C = 267K [1° C = 273 K]
n = 786 mmol of SF4 which is 0.786 mol
P = 1 atm
from ideal gas law we have
PV = nRT
where n is mole, R is gas constant, V is volume
![V = (nRT)/(P)](https://img.qammunity.org/2020/formulas/chemistry/college/o5m9aa3m769d9ckuf750asan1yn3t4am52.png)
![V = (0.786 mol * 0.082 atmL/mol K ** 267K)/(1atm) = 17.2 L](https://img.qammunity.org/2020/formulas/chemistry/college/nc3u8xfuhvrhuo0vl24p36phiglufuqmgx.png)
V of Sulfur tetrafluoride is 17.2 L