Is Scheme pure applicative order evaluation strategy?

[Nicholas Papadonis]

It seems that if the second argument to if is not evaluated.  Thanks!

(define (try a b)

  (if (= a 0) 1 b))

;Value: try

; Looks like applicative order evaluation:

(try 0 (/ 1 0))

;Division by zero signalled by /.

; Looks like normal order evaluation from the left

(if (= 1 1) 1 (/ 1 0))

;Value: 1

; How is it implemented?

(pp if)

;Syntactic keyword may not be used as an _expression_: #[keyword-value-item 16]