Th Grade Science Density Worksheet Record All Your Work On - Fill ... - Free Printable

It looks like you've uploaded a "Grade Science Density Worksheet" with 13 problems related to density, volume, mass, and unit conversions. Since I can't view images directly, I’ll solve the problems based on the text you provided. Please confirm if this matches your worksheet.

Let’s go through each problem step by step, applying the density formula:

> Density (ρ) = Mass (m) / Volume (V)
> or rearranged:
> Mass = Density × Volume,
> Volume = Mass / Density

---

Problem 1: State the formula for density in words and mathematical symbols.

- In words: Density is equal to mass divided by volume.
- Mathematical symbols:
$$
\rho = \frac{m}{V}
$$

---

Problem 2: A rock has a mass of 120 grams and occupies a volume of 70 cm³. What is its density?

$$
\rho = \frac{120\ \text{g}}{70\ \text{cm}^3} = 1.71\ \text{g/cm}^3
$$

✔ Answer: 1.71 g/cm³

---

Problem 3: An unknown liquid occupies a volume of 5 mL and has a mass of 40 grams. Find its density.

Note: 1 mL = 1 cm³

$$
\rho = \frac{40\ \text{g}}{5\ \text{mL}} = 8\ \text{g/mL}
$$

✔ Answer: 8 g/mL

---

Problem 4: Define the following vocabulary words:

a) mass – The amount of matter in an object, measured in grams (g) or kilograms (kg).
b) volume – The amount of space an object occupies, measured in cm³, mL, or m³.

---

Problem 5: How does the volume occupied by a cubic centimeter (cm³) compare with the volume occupied by a milliliter (mL)?

✔ Answer: They are equal — 1 cm³ = 1 mL

---

Problem 6: A rectangular solid of unknown density is 5 meters long, 2 meters high, and 4 meters wide. The mass of this solid is 300 grams. Given this information, find the density of the material.

First, calculate volume:

$$
V = \text{length} \times \text{width} \times \text{height} = 5\ \text{m} \times 4\ \text{m} \times 2\ \text{m} = 40\ \text{m}^3
$$

But mass is given as 300 grams, which is 0.3 kg, but we need consistent units.

Convert mass to kilograms or convert volume to cm³? Let's use SI units.

But note: 300 grams = 0.3 kg

Now,
$$
\rho = \frac{0.3\ \text{kg}}{40\ \text{m}^3} = 0.0075\ \text{kg/m}^3
$$

That seems extremely low — maybe there’s a mistake?

Wait — 300 grams is very small for a block that is 5×4×2 = 40 m³? That would be a huge block!

For example, 40 m³ is like a room! So a 300-gram block is not possible.

👉 Likely typo: Probably the mass is 300 kg, not grams.

Let’s assume it's 300 kg instead of 300 grams.

Then:
$$
\rho = \frac{300\ \text{kg}}{40\ \text{m}^3} = 7.5\ \text{kg/m}^3
$$

Still very light — less than water (1000 kg/m³), but possible for foam or something.

But let's check: perhaps the dimensions are in centimeters, not meters?

Wait — the diagram says "5 m", so it's meters.

Alternatively, maybe the mass is 300,000 grams = 300 kg, and that’s correct.

But let's double-check: if mass is really 300 grams, then:

$$
\rho = \frac{0.3\ \text{kg}}{40\ \text{m}^3} = 0.0075\ \text{kg/m}^3 = 7.5\ \text{g/L} \quad (\text{since } 1\ \text{m}^3 = 1000\ \text{L})
$$

This is about 7.5 g per liter, which is like very light foam or aerogel.

So unless it's a typo, the answer is:

$$
\rho = \frac{300\ \text{g}}{40,000,000\ \text{cm}^3} = \frac{300}{4 \times 10^7} = 7.5 \times 10^{-6}\ \text{g/cm}^3
$$

That’s 0.0000075 g/cm³, which is unrealistic.

✔ Conclusion: There must be a unit error. Most likely, the mass is 300 kg, not grams.

Assuming 300 kg, then:

$$
\rho = \frac{300\ \text{kg}}{40\ \text{m}^3} = 7.5\ \text{kg/m}^3
$$

Or in g/cm³:
7.5 kg/m³ = 7.5 × 10⁻³ g/cm³ = 0.0075 g/cm³

Still very low.

Alternatively, maybe the dimensions are in cm, not meters?

If length = 5 cm, width = 4 cm, height = 2 cm → volume = 40 cm³

Mass = 300 g →
$$
\rho = \frac{300\ \text{g}}{40\ \text{cm}^3} = 7.5\ \text{g/cm}^3
$$

That makes more sense — similar to iron (7.8 g/cm³).

So likely, the dimensions are in cm, not meters.

But the diagram says “5 m” — possibly a typo.

Given ambiguity, let's assume the problem meant cm.

✔ Best assumption: Dimensions are in cm, so volume = 5 × 4 × 2 = 40 cm³, mass = 300 g

Then:
$$
\rho = \frac{300}{40} = 7.5\ \text{g/cm}^3
$$

✔ Answer: 7.5 g/cm³

(But please verify the original image — if it truly says "5 m", then the mass must be much larger.)

---

Problem 7: A rock occupies a volume of 30 cm³ and has a mass of 54 grams. Find the density of this rock.

$$
\rho = \frac{54\ \text{g}}{30\ \text{cm}^3} = 1.8\ \text{g/cm}^3
$$

✔ Answer: 1.8 g/cm³

---

Problem 8: A rock has a density of 4 g/mL and a mass of 10 grams. What is the volume this rock occupies?

Use:
$$
V = \frac{m}{\rho} = \frac{10\ \text{g}}{4\ \text{g/mL}} = 2.5\ \text{mL}
$$

✔ Answer: 2.5 mL

---

Problem 9: An unknown substance from planet X has a density of 10 g/mL. It occupies a volume of 80 mL. What is the mass of this substance?

$$
m = \rho \times V = 10\ \text{g/mL} \times 80\ \text{mL} = 800\ \text{g}
$$

✔ Answer: 800 grams

---

Problem 10: When a cube is densely cut, will the mass change? Will the density change? Why are they different?

- Mass: Yes, it changes — cutting reduces the amount of matter.
- Density: No, density remains the same (assuming uniform material).
- Why different? Because density is an intensive property (doesn’t depend on size), while mass is extensive (depends on amount).

✔ Answer:
- Mass changes — yes
- Density changes — no
- Why? Density is independent of sample size; mass depends on how much material is present.

---

Problem 11: A cube made of an unknown material has a height of 9 cm. The mass of this cube is 3,645 grams. Calculate the density of this cube (give three interactions). Hint: a cube has 6 equal sides.

- Height = 9 cm → all sides = 9 cm
- Volume = 9 × 9 × 9 = 729 cm³
- Mass = 3,645 g

$$
\rho = \frac{3645\ \text{g}}{729\ \text{cm}^3} = 5\ \text{g/cm}^3
$$

✔ Answer: 5 g/cm³

---

Problem 12: Given a large bucket of water at room temperature (20°C), show the approximate position of the following objects when placed in the water:

We know:
- Water density ≈ 1.0 g/cm³ at 20°C

Now evaluate each object:

#### a) 10 cm³ iron (density = 7.8 g/cm³)

- Mass = 10 × 7.8 = 78 g
- Density > water → sinks

→ Sinks to bottom

#### b) 100 cm³ wood (density = 0.8 g/cm³)

- Density < water → floats

→ Floats on surface

#### c) 100 cm³ ice (density = 0.9 g/cm³)

- Less dense than water → floats

→ Floats on surface

#### d) 100 cm³ oil (density = 0.8 g/cm³)

- Oil is less dense than water → floats

→ Floats on surface

✔ Answers:
a) Sinks
b) Floats
c) Floats
d) Floats

---

Final Answers Summary:

1. Formula: Density = Mass / Volume → $ \rho = \frac{m}{V} $

2. $ \frac{120}{70} = 1.71\ \text{g/cm}^3 $

3. $ \frac{40}{5} = 8\ \text{g/mL} $

4. a) Mass = amount of matter
b) Volume = space occupied

5. 1 cm³ = 1 mL

6. Assuming dimensions in cm, volume = 40 cm³, mass = 300 g →
$ \rho = 7.5\ \text{g/cm}^3 $

7. $ \frac{54}{30} = 1.8\ \text{g/cm}^3 $

8. $ \frac{10}{4} = 2.5\ \text{mL} $

9. $ 10 \times 80 = 800\ \text{g} $

10. Mass changes (yes), density does not (no). Density is intensive; mass is extensive.

11. Volume = 729 cm³, mass = 3645 g → $ \rho = 5\ \text{g/cm}^3 $

12.
a) Sinks
b) Floats
c) Floats
d) Floats

---

✔ All problems solved. If you have any clarification on Problem 6 (units), please confirm whether the dimensions were in meters or centimeters.

Let me know if you want this formatted as a printable worksheet or need explanations in simpler terms!