Jasmine is healthy, but while outside, her internal body temperature rises above 38°C. What most likely happening to help her return to homeostasis?

Kc = [H2S]²*[O2]³ / [H2O]²*[SO2]²
Let x be the moles of H2S formed. Each mole of H2S takes one each mole of H2O and SO2 so after the reactions
[H2O] = 2.8 - x and [SO2] = 2.6 - x also for each mole of H2S, 1.5 moles of O2 are formed, so [O2] = 1.5*x
Kc = x²*(1.5*x)³ / (2.8 - x)²*(2.6 - x)²
Thus use 2.8 - x = 2.8 and 2.6 - x = 2.6 in the above equation for Kc:
Kc = x²*(1.5*x)³ / 2.8²*2.6² = 3.375x^5 / 2.8²*2.6² = 0.06368*x^5
x^5 = 1.3*10^-6 / 0.06368 = 2.0414*10^-5
x = 0.115M hope it helps