What is the number of particles in 27.5 grams of chromium (iii) sulfite ?

= 4.81 × 10^22 particles

The molar mass of Cr2(SO3)3 is 344.18 g

1 mole of a substance contains 6.02 10^23 particles

But; 1 mole of chromium (iii) sulfite = 344.18 g

Therefore;

344.18 g = 6.02×10^23 particles

27.5 g = ?

Hence;

= (27.5 × 6.02 × 10^23)/344.18

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.