This problem involves a gas undergoing a change in pressure while its temperature remains constant. This scenario is described by Boyle's Law, which states that for a fixed mass of gas at constant temperature, the pressure ($P$) and volume ($V$) are inversely proportional. Mathematically, this is expressed as: $$P_1 V_1 = P_2 V_2$$ Where: $P_1$ is the initial pressure $V_1$ is the initial volume $P_2$ is the final pressure $V_2$ is the final volume Step 1: Identify the given values and make necessary assumptions. Initial pressure, $P_1 = 740 \text{ torr}$ Initial volume, $V_1 = 7.0 \text{ mL}$ (Assuming "7.0 millimetres" is a typo for "7.0 milliliters", as millimeters is a unit of length, not volume, in this context.) Final pressure, $P_2 = 760 \text{ torr}$ (Standard pressure is commonly taken as 1 atmosphere, which is equal to 760 torr.) Temperature remains constant. We need to find the final volume, $V_2$. Step 2: Rearrange Boyle's Law to solve for $V_2$. $$V_2 = \frac{P_1 V_1}{P_2}$$ Step 3: Substitute the known values into the equation. $$V_2 = \frac{(740 \text{ torr}) \times (7.0 \text{ mL})}{760 \text{ torr}}$$ Step 4: Calculate the final volume. $$V_2 = \frac{5180 \text{ torr} \cdot \text{mL}}{760 \text{ torr}}$$ $$V_2 \approx 6.815789 \text{ mL}$$ Step 5: Round the answer to an appropriate number of significant figures. Since the initial volume (7.0 mL) has two significant figures, we round our answer to two significant figures. $$V_2 \approx 6.8 \text{ mL}$$ The volume of the oxygen gas at standard pressure (760 torr) and constant temperature is approximately $\boxed{\text{6.8 mL}}$.