The number of moles present in 29.5 grams of argon is 0.74 mole.
The atomic mass of argon is given as;
Ar = 39.95 g/mole
The number of moles present in 29.5 grams of argon is calculated as follows;
39.95 g ------------------------------- 1 mole
29.5 g ------------------------------ ?

Thus, the number of moles present in 29.5 grams of argon is 0.74 mole.
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Argon = Ar
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Answer:

Explanation:
Hello,
In this case, given the balanced reaction:

We can see a 2:4 mole ration between permanganate ion (118.9 g/mol) and manganese (IV) oxide (86.9 g/mol), that is why the resulting mas of this last one turns out:

Best regards.
Volume = 15.5 g × (1 cm³/0.789 g) = 19.6 cm³