Question

an american scientist reports the density of a metal to be 0.94000 $\frac{lb}{in^{3}}$. express the density of this metal in si units. 16.087 0.000033930 26,019 none of these choices formula 10 points perform the following calculation: 169.29 - 25.222 the solution contains how many significant figures? your answer should be a positive, whole number (no words). answer

Explanation:

Step1: Convert density units

We know that 1 lb = 0.453592 kg and 1 in³=(2.54 cm)³ = (0.0254 m)³= 1.63871×10⁻⁵ m³.
The given density is $
ho = 0.94000\frac{lb}{in^{3}}$.
Substitute the conversion - factors:
$
ho=0.94000\times\frac{0.453592 kg}{1.63871\times 10^{- 5}m^{3}}$.
$
ho=\frac{0.94000\times0.453592}{1.63871\times 10^{-5}}\frac{kg}{m^{3}}$.
$0.94000\times0.453592 = 0.42637648$ and $\frac{0.42637648}{1.63871\times 10^{-5}}=\frac{0.42637648}{1.63871}\times10^{5}$.
$\frac{0.42637648}{1.63871}\approx0.26019$, and $0.26019\times10^{5}=26019\frac{kg}{m^{3}}$.

Step2: Calculate significant - figures in subtraction

First, perform the subtraction $169.29−25.222 = 144.068$.
For addition and subtraction, the result should be rounded to the same number of decimal places as the number with the fewest decimal places in the values being added or subtracted. 169.29 has two decimal places and 25.222 has three decimal places. So, we round 144.068 to two decimal places, getting 144.07.
The significant - figures in 144.07 are 5.

Answer:

1. C. 26,019
2. 5