Practice Problems - Dimensional Analysis Blank | PDF - Free Printable

It looks like you've uploaded a page from a chemistry or physics worksheet titled "CHM 111 Dimensional Analysis Unit 1 Math Skills Toolbox", which contains several problems related to unit conversions, significant figures, and dimensional analysis.

Since I can't view images directly, I’ll solve each problem based on the text you provided. Let's go through them one by one and solve them step-by-step with explanations.

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Problem 5:

The acidic part of vinegar is acetic acid. If a 10.0 mL vinegar sample contains 483 mg acetic acid, what is the mass in micrograms (μg)?

#### Step-by-step:

We are given:
- Mass of acetic acid = 483 mg
- We need to convert mg → μg

We know:
- 1 mg = 1000 μg

So:
$$
483\ \text{mg} \times \frac{1000\ \mu\text{g}}{1\ \text{mg}} = 483,000\ \mu\text{g}
$$

Now consider significant figures:
- 483 mg has 3 significant figures.
- The conversion factor (1000) is exact.

So answer should have 3 significant figures.

$$
\boxed{4.83 \times 10^5\ \mu\text{g}}
$$

✔ Answer: $ 4.83 \times 10^5\ \mu\text{g} $

---

Problem 6:

Convert 3.45 mi to cm. Remember to express your answer with the correct number of significant figures. Note that there are multiple ways (routes) to solve this problem.

#### Step-by-step:

We need to convert miles (mi) → centimeters (cm).

Use known conversion factors:

1. 1 mile = 5280 feet
2. 1 foot = 12 inches
3. 1 inch = 2.54 cm

So:
$$
3.45\ \text{mi} \times \frac{5280\ \text{ft}}{1\ \text{mi}} \times \frac{12\ \text{in}}{1\ \text{ft}} \times \frac{2.54\ \text{cm}}{1\ \text{in}}
$$

Calculate step by step:

1. $ 3.45 \times 5280 = 18,264 $
2. $ 18,264 \times 12 = 219,168 $
3. $ 219,168 \times 2.54 = 556,511.52\ \text{cm} $

Now round to correct significant figures:
- 3.45 has 3 sig figs
- All other numbers are exact (definitions)

So final answer: $ \boxed{5.57 \times 10^5\ \text{cm}} $

✔ Answer: $ 5.57 \times 10^5\ \text{cm} $

---

Problem 7:

Convert 3.45 μg/μL to g/mL. Remember to express your answer with the correct number of significant figures. Note that there are multiple ways (routes) to solve this problem.

We want to convert:
$$
\frac{3.45\ \mu\text{g}}{\mu\text{L}} \to \frac{\text{g}}{\text{mL}}
$$

We know:
- 1 μg = $ 1 \times 10^{-6} $ g
- 1 μL = $ 1 \times 10^{-3} $ mL

So:
$$
\frac{3.45\ \mu\text{g}}{\mu\text{L}} = \frac{3.45 \times 10^{-6}\ \text{g}}{10^{-3}\ \text{mL}} = 3.45 \times 10^{-3}\ \frac{\text{g}}{\text{mL}}
$$

Significant figures: 3.45 has 3 sig figs.

✔ Answer: $ \boxed{3.45 \times 10^{-3}\ \text{g/mL}} $

---

Problem 8:

For a given printer, 1.0 mL of ink is used in printing 50 pages of text. How many pages could be printed using 100 gallons of ink?

We need to convert gallons → mL → pages.

First, convert 100 gallons to mL.

We know:
- 1 gallon (US) = 3.78541 L
- 1 L = 1000 mL

So:
$$
100\ \text{gal} \times 3.78541\ \frac{\text{L}}{\text{gal}} \times 1000\ \frac{\text{mL}}{\text{L}} = 378,541\ \text{mL}
$$

Now, 1.0 mL prints 50 pages → so:

$$
378,541\ \text{mL} \times \frac{50\ \text{pages}}{1.0\ \text{mL}} = 18,927,050\ \text{pages}
$$

But we must consider significant figures.

- 100 gallons: ambiguous — if it’s 100 with 1 sig fig, then answer would be $ 2 \times 10^7 $
- But typically in such contexts, 100 may be considered as having 3 sig figs if written as 100., but here it's just "100"
- However, 1.0 mL has 2 sig figs, and 50 pages might be exact (since it's a count), but probably not.

Let’s assume:
- 100 gallons has 3 sig figs (common assumption)
- 1.0 mL has 2 sig figs
- 50 pages is exact

So limiting sig figs = 2

Thus:
$$
1.9 \times 10^7\ \text{pages}
$$

✔ Answer: $ \boxed{1.9 \times 10^7\ \text{pages}} $

---

Problem 9:

A clerk sorts 375 sheets of paper in one hour. How many minutes would it take for the clerk to sort 325 reams of sheets? (There are 250 sheets in each 1 inch thick stack of sheets.)

Wait — let's clarify:

> "There are 250 sheets in each 1 inch thick stack of sheets."

But “ream” is a standard term: 1 ream = 500 sheets

So:
- 325 reams = $ 325 \times 500 = 162,500 $ sheets

Clerk sorts 375 sheets per hour.

How many hours?
$$
\frac{162,500}{375} = 433.\overline{3}\ \text{hours}
$$

Now convert to minutes:
$$
433.333... \times 60 = 25,999.98 \approx 26,000\ \text{minutes}
$$

Significant figures:
- 375 sheets/hour: 3 sig figs
- 325 reams: 3 sig figs
- 500 sheets/ream: exact (definition)

So answer should have 3 sig figs.

$$
\boxed{2.60 \times 10^4\ \text{minutes}}
$$

✔ Answer: $ \boxed{2.60 \times 10^4\ \text{min}} $

---

Problem 10:

Convert $ 10.3\ \text{g} \cdot \text{cm}^3/\text{s}^2 $ to kg / m² / s²

We are converting units of mass × length³ / time² → kg·m⁻²·s⁻²

Given:
- $ 10.3\ \text{g} \cdot \text{cm}^3/\text{s}^2 $

We need to convert:
- g → kg: $ 1\ \text{kg} = 1000\ \text{g} $
- cm → m: $ 1\ \text{m} = 100\ \text{cm} \Rightarrow 1\ \text{cm}^3 = (10^{-2})^3\ \text{m}^3 = 10^{-6}\ \text{m}^3 $

So:
$$
10.3\ \frac{\text{g} \cdot \text{cm}^3}{\text{s}^2} = 10.3 \times \frac{10^{-3}\ \text{kg} \cdot 10^{-6}\ \text{m}^3}{\text{s}^2} = 10.3 \times 10^{-9}\ \frac{\text{kg} \cdot \text{m}^3}{\text{s}^2}
$$

But the target unit is kg / m² / s², which is $ \text{kg} \cdot \text{m}^{-2} \cdot \text{s}^{-2} $

Wait! There’s a mismatch in dimensions.

Original: $ \text{g} \cdot \text{cm}^3 / \text{s}^2 $ → $ \text{mass} \cdot \text{length}^3 / \text{time}^2 $

Target: $ \text{kg} / (\text{m}^2 \cdot \text{s}^2) $ → $ \text{mass} \cdot \text{length}^{-2} \cdot \text{time}^{-2} $

These are not dimensionally equivalent.

So unless there's a typo, this conversion is invalid.

But perhaps it's meant to be g·cm/s² → kg·m/s², but it says cm³.

Alternatively, maybe it's supposed to be g/cm³/s²? That would make more sense.

But as written: $ 10.3\ \text{g} \cdot \text{cm}^3/\text{s}^2 $

This is not convertible to $ \text{kg}/\text{m}^2/\text{s}^2 $ because the dimensions don’t match.

Possibility: It's a typo, and they mean $ \text{g} \cdot \text{cm}^{-3} / \text{s}^2 $? Or perhaps it's $ \text{g} \cdot \text{cm}^{-2} / \text{s}^2 $?

But based on what's written, conversion is impossible.

Wait — maybe it's asking to convert $ 10.3\ \text{g} \cdot \text{cm}^3/\text{s}^2 $ into kg·m³/s²?

That would make sense.

Let’s assume that the intended target is $ \text{kg} \cdot \text{m}^3 / \text{s}^2 $

Then:
$$
10.3\ \frac{\text{g} \cdot \text{cm}^3}{\text{s}^2} = 10.3 \times \frac{10^{-3}\ \text{kg} \cdot (10^{-2}\ \text{m})^3}{\text{s}^2} = 10.3 \times 10^{-3} \times 10^{-6} = 10.3 \times 10^{-9} = 1.03 \times 10^{-8}\ \frac{\text{kg} \cdot \text{m}^3}{\text{s}^2}
$$

But the question says kg / m² / s², which is not the same.

Unless it's a typo and meant to be kg·m³/s², then:

✔ Answer: $ \boxed{1.03 \times 10^{-8}\ \text{kg} \cdot \text{m}^3 / \text{s}^2} $

But since the question says kg / m² / s², and that’s $ \text{kg} \cdot \text{m}^{-2} \cdot \text{s}^{-2} $, we cannot convert $ \text{g} \cdot \text{cm}^3 / \text{s}^2 $ to that.

So likely a typo.

Assuming it was meant to be $ 10.3\ \text{g} \cdot \text{cm}^{-3} / \text{s}^2 $ → $ \text{kg} / \text{m}^3 / \text{s}^2 $, then:

$ 1\ \text{g/cm}^3 = 1000\ \text{kg/m}^3 $

So:
$$
10.3\ \frac{\text{g}}{\text{cm}^3 \cdot \text{s}^2} = 10.3 \times 1000 = 10,300\ \frac{\text{kg}}{\text{m}^3 \cdot \text{s}^2}
$$

But again, not matching.

Given ambiguity, I suspect the original expression is miswritten.

But as per the text: $ 10.3\ \text{g} \cdot \text{cm}^3/\text{s}^2 $ → $ \text{kg}/\text{m}^2/\text{s}^2 $

This is not dimensionally consistent.

So no valid conversion possible.

✘ Answer: Invalid conversion due to mismatched dimensions.

But if it's a typo and they meant $ \text{g} \cdot \text{cm}^{-3}/\text{s}^2 $, then:

$$
10.3\ \frac{\text{g}}{\text{cm}^3 \cdot \text{s}^2} = 10.3 \times 1000 = 1.03 \times 10^4\ \frac{\text{kg}}{\text{m}^3 \cdot \text{s}^2}
$$

But that’s still not $ \text{kg}/\text{m}^2/\text{s}^2 $

So unless the numerator is mass and denominator area×time², it doesn't work.

Perhaps it's $ \text{g} \cdot \text{cm}^{-2} / \text{s}^2 $? Then:

$ 1\ \text{g/cm}^2 = 1000\ \text{kg/m}^2 $, so:

$$
10.3\ \frac{\text{g}}{\text{cm}^2 \cdot \text{s}^2} = 10.3 \times 1000 = 1.03 \times 10^4\ \frac{\text{kg}}{\text{m}^2 \cdot \text{s}^2}
$$

That matches the requested unit!

So likely, the original was meant to be $ 10.3\ \text{g} \cdot \text{cm}^{-2}/\text{s}^2 $, not $ \text{cm}^3 $

Assuming typo: cm⁻² instead of cm³

Then:
$$
10.3\ \frac{\text{g}}{\text{cm}^2 \cdot \text{s}^2} = 10.3 \times \frac{10^{-3}\ \text{kg}}{(10^{-2}\ \text{m})^2 \cdot \text{s}^2} = 10.3 \times \frac{10^{-3}}{10^{-4}} = 10.3 \times 10^{1} = 103\ \frac{\text{kg}}{\text{m}^2 \cdot \text{s}^2}
$$

Wait — better:

$$
1\ \text{cm}^2 = 10^{-4}\ \text{m}^2 \\
1\ \text{g} = 10^{-3}\ \text{kg}
$$

So:
$$
\frac{10.3\ \text{g}}{\text{cm}^2 \cdot \text{s}^2} = \frac{10.3 \times 10^{-3}\ \text{kg}}{10^{-4}\ \text{m}^2 \cdot \text{s}^2} = 10.3 \times 10^{-3} \times 10^4 = 10.3 \times 10^{1} = 103\ \frac{\text{kg}}{\text{m}^2 \cdot \text{s}^2}
$$

With 3 sig figs: $ \boxed{103\ \text{kg}/\text{m}^2/\text{s}^2} $

✔ Answer (assuming typo): $ \boxed{103\ \text{kg} \cdot \text{m}^{-2} \cdot \text{s}^{-2}} $

But as written, the original problem is invalid.

---

Problem 11:

An air sample taken from a local city contains $ 3.5 \times 10^{-5}\ \text{g/L} $ of carbon monoxide. What is the concentration of carbon monoxide in units of mol/L?

We need to convert g/L → mol/L

Use molar mass of CO:
- C = 12.01 g/mol
- O = 16.00 g/mol
- CO = 28.01 g/mol

So:
$$
\frac{3.5 \times 10^{-5}\ \text{g}}{\text{L}} \times \frac{1\ \text{mol}}{28.01\ \text{g}} = \frac{3.5 \times 10^{-5}}{28.01} \approx 1.25 \times 10^{-6}\ \text{mol/L}
$$

Significant figures: 3.5 has 2 sig figs, 28.01 has 4 → so 2 sig figs.

$$
\boxed{1.3 \times 10^{-6}\ \text{mol/L}}
$$

✔ Answer: $ \boxed{1.3 \times 10^{-6}\ \text{mol/L}} $

---

Problem 12:

Convert $ 2.998 \times 10^8\ \text{m/s} $ to miles/day

We need to convert:
- meters → miles
- seconds → days

We know:
- 1 mile = 1609.34 m
- 1 day = 24 hours = 24 × 3600 = 86,400 seconds

So:
$$
2.998 \times 10^8\ \frac{\text{m}}{\text{s}} \times \frac{1\ \text{mile}}{1609.34\ \text{m}} \times \frac{86,400\ \text{s}}{1\ \text{day}}
$$

Compute:
1. $ \frac{2.998 \times 10^8}{1609.34} \approx 1.863 \times 10^5 $
2. $ 1.863 \times 10^5 \times 8.64 \times 10^4 = 1.610 \times 10^{10} $

So:
$$
\boxed{1.61 \times 10^{10}\ \text{miles/day}}
$$

(3 significant figures — 2.998 has 4, but 1609.34 and 86400 are precise; 2.998 × 10⁸ is close to speed of light, so 4 sig figs, but let's keep 4)

Actually:
- 2.998 × 10⁸ has 4 sig figs
- 1609.34 has 6
- 86400 has 3 (if not exact), but usually considered exact in such contexts

So use 4 sig figs.

Final calculation:
$$
\frac{2.998 \times 10^8 \times 86400}{1609.34} = \frac{2.591 \times 10^{13}}{1609.34} \approx 1.609 \times 10^{10}
$$

So:
$$
\boxed{1.609 \times 10^{10}\ \text{miles/day}}
$$

✔ Answer: $ \boxed{1.609 \times 10^{10}\ \text{miles/day}} $

---

✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 5 | $ 4.83 \times 10^5\ \mu\text{g} $ |
| 6 | $ 5.57 \times 10^5\ \text{cm} $ |
| 7 | $ 3.45 \times 10^{-3}\ \text{g/mL} $ |
| 8 | $ 1.9 \times 10^7\ \text{pages} $ |
| 9 | $ 2.60 \times 10^4\ \text{minutes} $ |
| 10 | Invalid (likely typo — see explanation) |
| 11 | $ 1.3 \times 10^{-6}\ \text{mol/L} $ |
| 12 | $ 1.609 \times 10^{10}\ \text{miles/day} $ |

If you can confirm the exact wording of Problem 10, I can revise it.

Let me know if you'd like these explained in a different format (e.g., step-by-step with cancellation).