Do you want to evaluate 2sin 18 cos 18? There are a lot of apporaches for this problem. Here I show you one approach: Let x = 18 degrees. sin 2x = sin 36 = cos 54 = cos 3x, using co-function property Expand, 2sin x cos x = cos x cos 2x - sin x sin 2x Cancel cos x and let u = sin x = sin 18, 2u = 1 - 2u^2 - 2u^2 = 0, sinc cos 2x = 1 - 2sin^2 x Collect all terms in one side, 4u^2 + 2u - 1 = 0 Apply quadratic formula, u = (1/8)[-2+sqrt(4+16)] = (1/4)(-1+sqrt(5)) cos 18 = sqrt[1-u^2] = (1/4)sqrt[10+2sqrt(5)] 2sin 18 cos 18 = 2(1/4)(-1+sqrt(5))(1/4)sqrt[10+2sqrt(5)] which can be simplified further. Solution Do you want to evaluate 2sin 18 cos 18? There are a lot of apporaches for this problem. Here I show you one approach: Let x = 18 degrees. sin 2x = sin 36 = cos 54 = cos 3x, using co-function property Expand, 2sin x cos x = cos x cos 2x - sin x sin 2x Cancel cos x and let u = sin x = sin 18, 2u = 1 - 2u^2 - 2u^2 = 0, sinc cos 2x = 1 - 2sin^2 x Collect all terms in one side, 4u^2 + 2u - 1 = 0 Apply quadratic formula, u = (1/8)[-2+sqrt(4+16)] = (1/4)(-1+sqrt(5)) cos 18 = sqrt[1-u^2] = (1/4)sqrt[10+2sqrt(5)] 2sin 18 cos 18 = 2(1/4)(-1+sqrt(5))(1/4)sqrt[10+2sqrt(5)] which can be simplified further..

1. Moles of Be = mass of Be/molar mass of Be
= (0.33 g)/(9.012 g/mol)
= (0.33 g) x (1 mol/9.012 g)
= 0.037 mol
Solution
Moles of Be = mass of Be/molar mass of Be
= (0.33 g)/(9.012 g/mol)
= (0.33 g) x (1 mol/9.012 g)
= 0.037 mol