Question

1. 28 g of iron shot is added to a graduated cylinder containing 40 ml of water. the water level rises to the 44 ml mark. from this information, calculate the density of iron.
2. the density of lead is 11.342 g/ml. what would be the volume of a 200.0 kg sample of this metal?
3. calculate the density in g/l if a block of silver (ag) has a mass of 34.6 ng with a volume of 1300.0 microliters.

Explanation:

Problem 6:

Step1: Find the volume of iron

The initial volume of water is 40 mL, and the final volume is 44 mL. The volume of iron ($V$) is the difference between the final and initial volumes.
$V = 44\ mL - 40\ mL = 4\ mL$
Since 1 mL = 1 $cm^3$, $V = 4\ cm^3$

Step2: Calculate the density of iron

Density ($
ho$) is given by the formula $
ho=\frac{m}{V}$, where $m$ is the mass and $V$ is the volume. The mass of iron ($m$) is 28 g.
$
ho=\frac{28\ g}{4\ cm^3}=7\ g/cm^3$
Since 1 $cm^3$ = 1 mL and 1000 mL = 1 L, but for density in $g/cm^3$ (or $g/mL$) to $g/L$, we know that 1 $g/cm^3$ = 1000 $g/L$. Wait, actually, let's check the units again. Wait, the volume here is in mL, and we can also calculate density in $g/mL$ first and then convert to $g/L$. But in this case, the volume of iron is 4 mL (which is 4 $cm^3$). So density in $g/mL$ is $\frac{28\ g}{4\ mL}=7\ g/mL$. To convert to $g/L$, since 1 L = 1000 mL, we multiply by 1000: $7\ g/mL\times1000\ mL/L = 7000\ g/L$. But actually, the standard density of iron is often expressed in $g/cm^3$ or $kg/m^3$, but following the calculation:

Wait, maybe I made a mistake. Wait, the volume of the iron is the displacement, so 44 mL - 40 mL = 4 mL. Mass is 28 g. So density $
ho=\frac{m}{V}=\frac{28\ g}{4\ mL}=7\ g/mL$. Since 1 mL = $10^{-3}$ L, so $7\ g/mL = 7\ g/(10^{-3}\ L)=7\times10^{3}\ g/L = 7000\ g/L$. But also, $g/cm^3$ is equal to $g/mL$ and $1\ g/cm^3 = 1000\ kg/m^3$ or $1000\ g/L$. So the density of iron here is 7 g/mL or 7000 g/L. But maybe the question just wants it in $g/cm^3$ (which is same as $g/mL$) or $g/L$. Let's confirm the steps again.

Step1: Convert mass to grams

The mass of the lead sample is 200.0 Kg. Since 1 Kg = 1000 g, we convert:
$m = 200.0\ Kg\times1000\ g/Kg = 200000\ g$

Step2: Use density formula to find volume

Density ($
ho$) is given by $
ho=\frac{m}{V}$, so we can rearrange to find volume $V=\frac{m}{
ho}$. The density of lead ($
ho$) is 11.342 g/mL.
$V=\frac{200000\ g}{11.342\ g/mL}\approx17634.99\ mL$
To convert to liters, since 1 L = 1000 mL, $V=\frac{17634.99\ mL}{1000\ mL/L}\approx17.63\ L$ (or we can keep it in mL if needed, but let's check the units. Wait, the density is in g/mL, so volume will be in mL. Let's do the calculation more accurately.

$V=\frac{200000}{11.342}\approx17634.99\ mL = 17.63\ L$ (approx)

Step1: Convert mass to grams

The mass of silver is 34.6 ng. Since 1 ng = $10^{-9}$ g, we convert:
$m = 34.6\ ng\times10^{-9}\ g/ng = 3.46\times10^{-8}\ g$

Step2: Convert volume to liters

The volume of silver is 1300.0 microliters. Since 1 microliter = $10^{-6}$ L, we convert:
$V = 1300.0\ \mu L\times10^{-6}\ L/\mu L = 1.3\times10^{-3}\ L$

Step3: Calculate density

Density ($
ho$) is given by $
ho=\frac{m}{V}$.
$
ho=\frac{3.46\times10^{-8}\ g}{1.3\times10^{-3}\ L}\approx2.66\times10^{-5}\ g/L$

Wait, that seems very low. Did I make a mistake in unit conversion? Wait, 1 microliter is $10^{-6}$ liters, so 1300 microliters is 1300 $\times10^{-6}$ L = $1.3\times10^{-3}$ L. Mass is 34.6 ng = 34.6 $\times10^{-9}$ g = $3.46\times10^{-8}$ g. Then density is $\frac{3.46\times10^{-8}}{1.3\times10^{-3}}\approx2.66\times10^{-5}\ g/L$. But that seems incorrect. Wait, maybe the volume is in microliters, but maybe we should convert to mL or $cm^3$? Wait, 1 microliter = $10^{-3}$ mL, so 1300 microliters = 1.3 mL = 1.3 $cm^3$. Then mass is 34.6 ng = 34.6 $\times10^{-9}$ g = $3.46\times10^{-8}$ g. Then density in $g/cm^3$ would be $\frac{3.46\times10^{-8}\ g}{1.3\ cm^3}\approx2.66\times10^{-8}\ g/cm^3$, which is also very low. Wait, maybe the mass is 34.6 mg? Or maybe the volume is 1300.0 mL? Wait, the problem says "1300.0 microliters". Maybe there's a typo, but following the given values:

Wait, let's recheck the unit conversions:

• 1 ng = $10^{-9}$ g, so 34.6 ng = 34.6 $\times10^{-9}$ g = $3.46\times10^{-8}$ g.

• 1 microliter ($\mu L$) = $10^{-6}$ L, so 1300.0 $\mu L$ = 1300.0 $\times10^{-6}$ L = $1.3\times10^{-3}$ L.

Then density $
ho=\frac{m}{V}=\frac{3.46\times10^{-8}\ g}{1.3\times10^{-3}\ L}\approx2.66\times10^{-5}\ g/L$.

But this is an extremely low density, which is not realistic for silver. Silver's density is about 10.49 g/cm^3. So maybe there's a mistake in the problem's values (e.g., mass is 34.6 mg instead of ng, or volume is 1300.0 mL instead of microliters). But assuming the problem's values are correct, the calculation is as above.

Answer:

The density of iron is $\boldsymbol{7\ g/cm^3}$ (or $\boldsymbol{7000\ g/L}$)

Problem 7: