Question

1. —- a measuring cup that is accurate but not precise.
2. a student in lab weighed three samples of 0.25 g. when he measures the mass of his three times, he noticed he would have the same value of 0.250 g each time. what term is used to describe measurements? include significant figures and units.
3. rock salt with has density of 2.18 g/cm³. what would the volume be of a 4.8 g sample of rock salt?
4. a school of 1956 mineral ore was measured before with “era”, and a mass of 890.0 g. gold has a density of... make sure you have the correct number of significant figures and units.
5. you calculated your grade to chemistry as a 89.7, but you actually value is 89.4. what is your percent error? name only your final the correct number of significant figures.
6. mount everest is 8848.7 m high. how many significant figures in the mountain? make sure you list the correct number of significant figures and units.

Explanation:

To solve the problem of finding the volume of a rock sample with a density of \(0.78 \, \text{g/cm}^3\) and a mass of \(24.8 \, \text{g}\), we use the formula for density:

\[

ho = \frac{m}{V}
\]

where:

• \(

ho\) is the density,

• \(m\) is the mass,
• \(V\) is the volume.

Step 1: Rearrange the density formula to solve for volume

We need to solve for \(V\). Rearranging the formula gives:

\[
V = \frac{m}{
ho}
\]

Step 2: Substitute the given values into the formula

We know the mass \(m = 24.8 \, \text{g}\) and the density \(
ho = 0.78 \, \text{g/cm}^3\). Substituting these values into the formula for volume:

\[
V = \frac{24.8 \, \text{g}}{0.78 \, \text{g/cm}^3}
\]

Step 3: Perform the division

Calculating the division:

\[
V \approx \frac{24.8}{0.78} \approx 31.79 \, \text{cm}^3
\]

Step 4: Consider significant figures

The given values are:

• Mass: \(24.8 \, \text{g}\) (3 significant figures)
• Density: \(0.78 \, \text{g/cm}^3\) (2 significant figures)

When dividing, the result should have the same number of significant figures as the least precise measurement (which is 2 significant figures here). However, let's check the calculation again. Wait, actually, \(24.8\) has 3 significant figures and \(0.78\) has 2. The rule for multiplication/division is that the result should have the same number of significant figures as the quantity with the least number of significant figures. So we need to round \(31.79\) to 2 significant figures? Wait, no, wait:

Wait, \(0.78\) is two significant figures, \(24.8\) is three. So the result should have two significant figures? Wait, no, let's check the calculation:

\(24.8 \div 0.78 \approx 31.79487\)

If we consider significant figures, since \(0.78\) has two, the result should be rounded to two significant figures: \(32 \, \text{cm}^3\)? Wait, no, \(31.79\) rounded to two significant figures is \(32\)? Wait, \(31.79\) is approximately \(32\) when rounded to two significant figures? Wait, no, \(31.79\) to two significant figures: the first two significant figures are \(3\) and \(1\), the next digit is \(7\), which is more than \(5\), so we round up the \(1\) to \(2\), so \(32\). But wait, maybe the problem expects us to use the given values as is without strict significant figures? Let's check the calculation again.

Wait, maybe I made a mistake. Let's recalculate:

\(24.8 \div 0.78\):

\(0.78 \times 31 = 24.18\)

\(24.8 - 24.18 = 0.62\)

\(0.62 \div 0.78 \approx 0.79\)

So total is \(31 + 0.79 = 31.79\), so approximately \(31.8 \, \text{cm}^3\) (if we take three significant figures) or \(32 \, \text{cm}^3\) (if we take two). But let's see the problem statement: it says "What would the volume be of a 24.8 g sample of rock salt?" with density \(0.78 \, \text{g/cm}^3\).

So the volume is:

\[
V = \frac{24.8}{0.78} \approx 31.8 \, \text{cm}^3
\] (if we consider three significant figures for the mass) or \(32 \, \text{cm}^3\) (if we consider two for the density). But maybe the problem expects the exact calculation without worrying about significant figures, so approximately \(31.8 \, \text{cm}^3\) or \(32 \, \text{cm}^3\).

Final Answer

The volume of the rock sample is approximately \(\boxed{32 \, \text{cm}^3}\) (or \(\boxed{31.8 \, \text{cm}^3}\) depending on significant figures).

Answer:

To solve the problem of finding the volume of a rock sample with a density of \(0.78 \, \text{g/cm}^3\) and a mass of \(24.8 \, \text{g}\), we use the formula for density:

\[

ho = \frac{m}{V}
\]

where:

• \(

ho\) is the density,

• \(m\) is the mass,
• \(V\) is the volume.

Step 1: Rearrange the density formula to solve for volume

We need to solve for \(V\). Rearranging the formula gives:

\[
V = \frac{m}{
ho}
\]

Step 2: Substitute the given values into the formula

We know the mass \(m = 24.8 \, \text{g}\) and the density \(
ho = 0.78 \, \text{g/cm}^3\). Substituting these values into the formula for volume:

\[
V = \frac{24.8 \, \text{g}}{0.78 \, \text{g/cm}^3}
\]

Step 3: Perform the division

Calculating the division:

\[
V \approx \frac{24.8}{0.78} \approx 31.79 \, \text{cm}^3
\]

Step 4: Consider significant figures

The given values are:

• Mass: \(24.8 \, \text{g}\) (3 significant figures)
• Density: \(0.78 \, \text{g/cm}^3\) (2 significant figures)

When dividing, the result should have the same number of significant figures as the least precise measurement (which is 2 significant figures here). However, let's check the calculation again. Wait, actually, \(24.8\) has 3 significant figures and \(0.78\) has 2. The rule for multiplication/division is that the result should have the same number of significant figures as the quantity with the least number of significant figures. So we need to round \(31.79\) to 2 significant figures? Wait, no, wait:

Wait, \(0.78\) is two significant figures, \(24.8\) is three. So the result should have two significant figures? Wait, no, let's check the calculation:

\(24.8 \div 0.78 \approx 31.79487\)

If we consider significant figures, since \(0.78\) has two, the result should be rounded to two significant figures: \(32 \, \text{cm}^3\)? Wait, no, \(31.79\) rounded to two significant figures is \(32\)? Wait, \(31.79\) is approximately \(32\) when rounded to two significant figures? Wait, no, \(31.79\) to two significant figures: the first two significant figures are \(3\) and \(1\), the next digit is \(7\), which is more than \(5\), so we round up the \(1\) to \(2\), so \(32\). But wait, maybe the problem expects us to use the given values as is without strict significant figures? Let's check the calculation again.

Wait, maybe I made a mistake. Let's recalculate:

\(24.8 \div 0.78\):

\(0.78 \times 31 = 24.18\)

\(24.8 - 24.18 = 0.62\)

\(0.62 \div 0.78 \approx 0.79\)

So total is \(31 + 0.79 = 31.79\), so approximately \(31.8 \, \text{cm}^3\) (if we take three significant figures) or \(32 \, \text{cm}^3\) (if we take two). But let's see the problem statement: it says "What would the volume be of a 24.8 g sample of rock salt?" with density \(0.78 \, \text{g/cm}^3\).

So the volume is:

\[
V = \frac{24.8}{0.78} \approx 31.8 \, \text{cm}^3
\] (if we consider three significant figures for the mass) or \(32 \, \text{cm}^3\) (if we consider two for the density). But maybe the problem expects the exact calculation without worrying about significant figures, so approximately \(31.8 \, \text{cm}^3\) or \(32 \, \text{cm}^3\).

Final Answer

The volume of the rock sample is approximately \(\boxed{32 \, \text{cm}^3}\) (or \(\boxed{31.8 \, \text{cm}^3}\) depending on significant figures).