1 ) The atomic weights of the three components of potassium chlorate (K, Cl, and O[3]) are in the ratio 39:35:16*3 or 39:35;48.

So we can set up an equation to find the relative weight of each component as follows:

39x+35x+48x = 6832 gm

122x = 6832 gm Divide both sides by 122.

x = 56, so the weight of each component in the sample of 6832 grams is:

K = 39(56) = 2184 gm

Cl = 35(56)= 1960 gm

O[3] = 48(56) = 2688 gm.

2 ) The sum of the measures of the interior angles is (n - 2)180° where n is the number of sides. Let n = 16, so:

(16 - 2)180°

= (14)(180°)

= 2520° [A]