Sulfur dioxide gas reacts with oxygen to form sulfur trioxide gas, as represented by the equation2 SO2(g) + O2(g) <>2 SO3(g) ∆H = −197.8 kJ In order to obtain the equilibrium system above, 2.60 mol of SO2(g) and 2.30 mol of O2(g) are injected into a 1.00 L container. When the system reaches equilibrium, the concentration of the remaining SO2(g) is 1.32 mol/L. The concentration of O2(g) at equilibrium is mol/L. (do NOT write units, simply give a 3-digit numeric answer

1.66mol/L remain

Explanation:

Based on the reaction:

2 SO2(g) + O2(g) ⇄ 2 SO3(g) ∆H = −197.8 kJ

2 moles of SO2 react per mole of oxygen. To find the concentration of SO2 we need to see the moles of SO2 that react to find thus moles of oxygen that react and the moles and concentration of O2 that remains in the equilibrium:

Moles SO2 that remain:

1.32mol /L = 1.32moles of SO2 remains

2.60 initial moles - 1.32moles that remains = 1.28 moles of SO2 react

Moles O2 that reacted:

1.28 moles SO2  * (1mol O2 / 2mol SO2) = 0.64 moles O2

Moles O2 that remain:

2.30 moles O2 - 0.64 moles = 1.66 moles remain

And concentration is:

1.66moles / 1L =

answer: they produce more slow than unicellular organisms.

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mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b