Question

question 10 of 10 suppose a sample of water in this experiment had a mass of 95.4386 g. what would the mass of the sample be in the following units? use the listed equalities for your conversions. report your answers to the correct number of significant figures. equality 1 kg = 10³ g 1 g = 10³ mg 1 lb = 454 g 1 oz. = 28.35 g 0.0954386 kg 9.54386 ×10⁴ mg lb oz tools ×10ⁿ

Explanation:

Conversion to pounds (lb)

Step1: Identify conversion factor

We know that \( 1\ \text{lb} = 454\ \text{g} \), so the conversion factor to convert grams to pounds is \( \frac{1\ \text{lb}}{454\ \text{g}} \).

Step2: Set up the conversion

Multiply the mass in grams (\( 95.4386\ \text{g} \)) by the conversion factor:
\( 95.4386\ \text{g} \times \frac{1\ \text{lb}}{454\ \text{g}} \)

Step3: Perform the calculation

\( \frac{95.4386}{454} \approx 0.2102\ \text{lb} \) (rounded to the correct number of significant figures. The original mass has 6 significant figures, and the conversion factor \( 454\) has 3, so the result should have 3 significant figures? Wait, no: \( 95.4386\) has 6, \( 454\) has 3. When dividing, the result should have 3 significant figures? Wait, no, let's check: \( 95.4386 \div 454 \approx 0.210217 \). But \( 454\) is 3 sig figs, so the result should be 3 sig figs? Wait, no, the rule is that when multiplying or dividing, the result has the same number of significant figures as the least precise measurement. Here, \( 454\) has 3, \( 95.4386\) has 6. So the result should have 3? Wait, no, maybe I made a mistake. Wait, \( 1\ \text{lb} = 454\ \text{g} \) is an exact conversion? No, \( 454\) is a defined value? Wait, actually, the conversion factor \( 1\ \text{lb} = 453.59237\ \text{g} \) is more accurate, but here we are given \( 1\ \text{lb} = 454\ \text{g} \), which is 3 significant figures. So \( 95.4386\ \text{g} \) (6 sig figs) divided by \( 454\ \text{g/lb} \) (3 sig figs) gives a result with 3 sig figs? Wait, no, the number of significant figures in the result is determined by the least number of significant figures in the values used in the calculation. So \( 454\) has 3, so the result should have 3. Wait, but \( 95.4386 \div 454 \approx 0.2102 \), which would round to \( 0.210\ \text{lb} \)? Wait, no, let's calculate it properly: \( 95.4386 \div 454 = 0.2102171806 \). So with 3 significant figures, that's \( 0.210\ \text{lb} \)? Wait, no, 0.210 has 3 significant figures. Alternatively, maybe the problem expects us to use the given conversion factor as is, without worrying about significant figures beyond what's needed. Wait, the original mass is \( 95.4386\ \text{g} \), which has 6 significant figures. The conversion factor \( 454\) has 3. So when we do \( 95.4386 / 454 \), the result should have 3 significant figures. So \( 0.210\ \text{lb} \)? Wait, but let's check the calculation again. \( 95.4386 \div 454 = 0.210217 \), so rounding to 3 significant figures is \( 0.210\ \text{lb} \).

Conversion to ounces (oz)

Step1: Identify conversion factor

We know that \( 1\ \text{oz} = 28.35\ \text{g} \), so the conversion factor to convert grams to ounces is \( \frac{1\ \text{oz}}{28.35\ \text{g}} \).

Step2: Set up the conversion

Multiply the mass in grams (\( 95.4386\ \text{g} \)) by the conversion factor:
\( 95.4386\ \text{g} \times \frac{1\ \text{oz}}{28.35\ \text{g}} \)

Step3: Perform the calculation

\( \frac{95.4386}{28.35} \approx 3.366\ \text{oz} \) (Let's calculate: \( 95.4386 \div 28.35 \approx 3.366 \). The original mass has 6 significant figures, the conversion factor \( 28.35\) has 4, so the result should have 4 significant figures? Wait, \( 28.35\) has 4, \( 95.4386\) has 6, so the result should have 4. So \( 3.366\ \text{oz} \) (which is 4 significant figures).

Final Answers for Each Unit:

• kg: \( 0.0954386\ \text{kg} \) (already provided, but let's verify: \( 95.4386\ \text{g} \div 10^3\ \text{g/kg} = 0.0954386\ \text{kg} \), which is correct as \( 1\ \text{kg} = 10^3\ \text{g} \))
• mg: \( 9.54386 \times 10^4\ \text{mg} \) (verify: \( 95.4386\ \text{g} \times 10^3\ \text{mg/g} = 95438.6\ \text{mg} = 9.54386 \times 10^4\ \text{mg} \), correct)
• lb: \( \approx 0.210\ \text{lb} \) (or more precisely, using the calculation: \( 95.4386 \div 454 \approx 0.2102\), but with significant figures, maybe \( 0.210\ \text{lb} \))
• oz: \( \approx 3.366\ \text{oz} \) (calculation: \( 95.4386 \div 28.35 \approx 3.366\))

But since the problem has boxes for each unit, let's fill them properly:

For lb:

Step1: Conversion factor

\( 1\ \text{lb} = 454\ \text{g} \), so factor is \( \frac{1\ \text{lb}}{454\ \text{g}} \)

Step2: Multiply by mass

\( 95.4386\ \text{g} \times \frac{1\ \text{lb}}{454\ \text{g}} = \frac{95.4386}{454}\ \text{lb} \)

Step3: Calculate

\( \frac{95.4386}{454} \approx 0.210\ \text{lb} \) (or 0.2102, but let's check significant figures. The given mass has 6 sig figs, conversion factor 3, so 3 sig figs: 0.210)

For oz:

Answer:

Step1: Conversion factor

\( 1\ \text{oz} = 28.35\ \text{g} \), factor is \( \frac{1\ \text{oz}}{28.35\ \text{g}} \)

Step2: Multiply by mass

\( 95.4386\ \text{g} \times \frac{1\ \text{oz}}{28.35\ \text{g}} = \frac{95.4386}{28.35}\ \text{oz} \)

Step3: Calculate

\( \frac{95.4386}{28.35} \approx 3.366\ \text{oz} \)

Filling the Boxes:

• kg: \( 0.0954386 \) (already correct)
• mg: \( 9.54386 \times 10^4 \) (already correct)
• lb: \( \boldsymbol{0.210} \) (or more accurately, 0.2102, but with 3 sig figs, 0.210)
• oz: \( \boldsymbol{3.366} \) (or 3.366 as calculated)

Wait, but let's check the lb calculation again. \( 95.4386 \div 454 = 0.210217 \), which is approximately 0.210 when rounded to three significant figures. For oz, \( 95.4386 \div 28.35 = 3.36601 \), so 3.366 (four significant figures, since 28.35 has four).

So the answers are:

• kg: \( 0.0954386 \)
• mg: \( 9.54386 \times 10^4 \)
• lb: \( 0.210 \) (or 0.2102, but likely 0.210)
• oz: \( 3.366 \)

But since the problem might expect the exact calculation without rounding down too much, let's present the precise values:

For lb: \( \frac{95.4386}{454} = 0.2102171806 \approx 0.210\ \text{lb} \) (3 sig figs)

For oz: \( \frac{95.4386}{28.35} = 3.366010582 \approx 3.366\ \text{oz} \) (4 sig figs)

So the final answers for each box:

• kg: \( 0.0954386 \)
• mg: \( 9.54386 \times 10^4 \)
• lb: \( 0.210 \) (or 0.2102, but let's use the calculation result)
• oz: \( 3.366 \)

But maybe the problem expects the lb to be calculated as \( 95.4386 / 454 = 0.2102 \), so we can write \( 0.210 \) (3 sig figs) or \( 0.2102 \) (more precise). Similarly, oz is \( 3.366 \).

So to fill the boxes:

• lb: \( 0.210 \) (or 0.2102)
• oz: \( 3.366 \)

Final Answers:

• kg: \( \boxed{0.0954386} \)
• mg: \( \boxed{9.54386 \times 10^4} \)
• lb: \( \boxed{0.210} \) (or \( \boxed{0.2102} \) if more precision is needed)
• oz: \( \boxed{3.366} \)