Answer:
691.29 K or 418.14 °C
Explanation:
Hello, at first the moles of oxygen gas are required:
[tex]n_{O_2}=1.75 g * \frac{1mol O_2}{32 g O_2} =0.0547 mol[/tex]
Now, based on the ideal gas equation, we solve for the temperature:
[tex]PV=nRT\\T=\frac{PV}{nR}\\T=\frac{1atm * 3.10 L}{0.0547 mol*0.082 \frac{atm*L}{mol*K} }\\T=691.29 K[/tex]
Best regards.
