#### Answer

a) $5.4\times10^{-2}g$, 2
b) $5.462\times10^3g$, 4
c) $7.92\times10^{-4}g$, 3
d) $1.6\times10^3mL$, 2

#### Work Step by Step

a) For the first significant figure to be before the decimal, the number needs to be shifted twice to the left, with a -2 power of ten to make it equal.
$0.054 g=5.4\times10^{-2}g$, only the 5 and the 4 are significant figures.
b) For there to be only one digit before the decimal, the number must be shifted three times to the right, with a 3 power of ten to make it equal.
$5462g = 5.462\times10^3g$, all digits are significant figures.
c) For the first significant figure to be before the decimal, the number needs to be shifted fourth times to the left, with a -4 power of ten to make it equal.
$0.000792 g=7.92\times10^{-4}g$: the 7 the 9 and the 2 are significant figures.
d) For there to be only one digit before the decimal, the number must be shifted three times to the right, with a 3 power of ten to make it equal.
$1600mL= 1.6\times10^3mL$ This is the ambiguous scenario, but by the book's definition, only the 1 and the 6 are significant figures.