Asealed can with an internal pressure of 721 mmhg at
25degrees c is thrown into an incinerator operating at 755 degrees
c. what will be the pressure inside the heated can, assuming
thecontainer remains intact during incineration?

2486 mmHg

Explanation:

Gay-Lussac's Law states the pressure varies directly with temperature when volume remains constant:

P₁/T₁ = P₂/T₂

Where P₁ and T₁ are initial pressure and temperature and P₂ and T₂ are final pressure and temperatue.

The problem says initial pressure is 721 mmHg, initial temperature is 25°C and final temperature is 755°C. The question is final pressure.

°C must be converted to absolute temperature (K), thus:

25°C + 273,15 = 298,15 K

755°C + 273,15 = 1028,15 K

Thus, pressure P₂ is:

(T₂·P₁) / T₁ = P₂

1028,15K · 721mmHg / 298,15 K =  2486 mmHg

I hope it helps!

the dependant variable is spilled water

the independent is the heat

i don,t know so just answer mine