Question

a 5.40 x 10⁴ gram sample of iron at 290 k has a density of 7.874 g/cm³. as it is 756.85 °c, its density changes. the diagram below shows the particles at 290 k.
express the mass of the sample in standard notation:

1. express the mass of the sample in kilograms. you must show your work to earn credit.
2. what is the starting temperature of the sample in celsius? show your work
3. what is the ending temperature of the sample in kelvin? show your work.

what is the volume of the sample of iron? show your work.

Explanation:

Step1: Convert mass to standard notation

The given mass is $5.40\times10^{4}$ grams. In standard - notation, we move the decimal point 4 places to the right. So, $5.40\times10^{4}=54000$ grams.

Step2: Convert mass to kilograms

We know that $1\ kg = 1000\ g$. To convert 54000 grams to kilograms, we divide by 1000. So, $m=\frac{54000\ g}{1000\ g/kg}=54\ kg$.

Step3: Convert starting temperature from Kelvin to Celsius

The conversion formula from Kelvin (K) to Celsius (C) is $C = K - 273.15$. Given $K = 290\ K$, then $C=290 - 273.15 = 16.85^{\circ}C$.

Step4: Convert ending temperature from Celsius to Kelvin

The conversion formula from Celsius to Kelvin is $K = C+273.15$. Given $C = 756.85^{\circ}C$, then $K=756.85 + 273.15=1030\ K$.

Step5: Calculate the volume of the iron sample

The density formula is $
ho=\frac{m}{V}$, where $
ho$ is density, $m$ is mass, and $V$ is volume. Rearranging for volume gives $V=\frac{m}{
ho}$. Given $m = 54000\ g$ and $
ho = 7.874\ g/cm^{3}$, then $V=\frac{54000\ g}{7.874\ g/cm^{3}}\approx6858.01\ cm^{3}\approx6.86\times 10^{3}\ cm^{3}$.

Answer:

1. 54000 grams
2. 54 kg
3. $16.85^{\circ}C$
4. 1030 K
5. $6.86\times 10^{3}\ cm^{3}$