multiple - step problem\nin chemistry lab, you are attempting to determine the density of iron. you mass a…

multiple - step problem\nin chemistry lab, you are attempting to determine the density of iron. you mass a sample of iron on a balance, as 58.4 grams. due to the irregular shape of the iron you must determine the volume with a graduated cylinder. with an initial volume (v_i) of 40.1 ml, and a final volume (v_f) of 47.8 ml, what is the volume of the iron? show your work.\nwhat is the density of the iron, in g/cm³? show all work. remember correct significant figures.\ncalculate the percent error of your determined density.\nusing your determined density, convert the value into kg/m³. use the factor label method. remember that cm³ represents (cm*cm*cm) so you will need to convert all of the cm to m (one cubic meter is not the same as 100 cubic centimeters).

multiple - step problem\nin chemistry lab, you are attempting to determine the density of iron. you mass a sample of iron on a balance, as 58.4 grams. due to the irregular shape of the iron you must determine the volume with a graduated cylinder. with an initial volume (v_i) of 40.1 ml, and a final volume (v_f) of 47.8 ml, what is the volume of the iron? show your work.\nwhat is the density of the iron, in g/cm³? show all work. remember correct significant figures.\ncalculate the percent error of your determined density.\nusing your determined density, convert the value into kg/m³. use the factor label method. remember that cm³ represents (cm*cm*cm) so you will need to convert all of the cm to m (one cubic meter is not the same as 100 cubic centimeters).

Answer

Explanation:

Step1: Calculate volume of iron

The volume of the iron is the difference between the final and initial volumes of water in the graduated - cylinder. $V = V_f - V_i$ $V=47.8\ mL - 40.1\ mL=7.7\ mL$ Since $1\ mL = 1\ cm^3$, the volume of the iron $V = 7.7\ cm^3$.

Step2: Calculate density of iron

The density formula is $\rho=\frac{m}{V}$, where $m$ is mass and $V$ is volume. Given $m = 58.4\ g$ and $V = 7.7\ cm^3$ $\rho=\frac{58.4\ g}{7.7\ cm^3}\approx7.6\ g/cm^3$ (rounded to two significant figures)

Step3: Calculate percent error

The accepted density of iron is $\rho_{accepted}=7.87\ g/cm^3$. The percent - error formula is $\text{Percent Error}=\left|\frac{\rho_{experimental}-\rho_{accepted}}{\rho_{accepted}}\right|\times100%$ $\text{Percent Error}=\left|\frac{7.6\ g/cm^3 - 7.87\ g/cm^3}{7.87\ g/cm^3}\right|\times100%$ $=\left|\frac{- 0.27\ g/cm^3}{7.87\ g/cm^3}\right|\times100%\approx3.4%$

Step4: Convert density to $kg/m^3$

We know that $1\ g = 10^{- 3}\ kg$ and $1\ cm = 10^{-2}\ m$, so $1\ cm^3=(10^{-2}\ m)^3 = 10^{-6}\ m^3$ Starting with $\rho = 7.6\ g/cm^3$, we use the factor - label method: $\rho=7.6\frac{g}{cm^3}\times\frac{1\ kg}{1000\ g}\times\frac{1\ cm^3}{(10^{-2}\ m)^3}$ $=7.6\times\frac{1\ kg}{1000\ g}\times\frac{1\ cm^3}{10^{-6}\ m^3}$ $=7.6\times10^{3}\ kg/m^3$

Answer:

Volume of iron: $7.7\ cm^3$ Density of iron: $7.6\ g/cm^3$ Percent error: $3.4%$ Density in $kg/m^3$: $7.6\times10^{3}\ kg/m^3$