Question

question 10
5 points
using your data table 1 step 2, from your procedure document: input your answers here. (use correct significant figures)
a) 52.30 ounces is equal to 1547 milliliters
b) 52 cm is equal to 20.5 inches
c) 16.90 pounds is equal to 7666 grams.
d) (140.567 + 12.209) / 13.0 = blank 4
blank 1 1547
blank 2 20.5
blank 3 7666
blank 4 11.8

question 11
5 points
using your data table 1 step 3, from your procedure document: input your answers here. (use correct significant figures)
a) initial volume of the graduated cylinder before solid is added is blank 1 ml.
b) final volume of the graduated cylinder after solid is added is blank 2 ml.
c) the volume of the solid is blank 3 ml.
blank 1 add your answer
blank 2 add your answer
blank 3 add your answer

Explanation:

Question 10d Solution:

Step 1: Add the numbers in the numerator

First, we calculate the sum of \( 140.567 \) and \( 12.209 \).
\( 140.567 + 12.209 = 152.776 \)

Step 2: Divide the result by 13.0

Now, we divide the sum from Step 1 by \( 13.0 \).
\( \frac{152.776}{13.0} \approx 11.752 \)
Considering significant figures, the least number of decimal places in the division (or the number of significant figures in the divisor \( 13.0 \) which has 3 significant figures, and the dividend \( 152.776 \) has more, so we go by the divisor's significant figures for the result's precision. Wait, actually, when adding \( 140.567 \) (5 sig figs) and \( 12.209 \) (5 sig figs), the sum is \( 152.776 \) (which we can consider as having the precision of the least precise decimal place, but both have 3 decimal places, so the sum is precise to the thousandth place, but when dividing by \( 13.0 \) (3 sig figs), the result should have 3 sig figs. Wait, \( 13.0 \) has 3 sig figs (the trailing zero after the decimal is significant). The sum \( 140.567 + 12.209 = 152.776 \) (let's check the number of sig figs: 140.567 has 6, 12.209 has 5, sum is 152.776 which we can take as 6 sig figs for the intermediate step). Then dividing by 13.0 (3 sig figs) gives \( 152.776 \div 13.0 \approx 11.752 \), which rounds to 11.8 (3 sig figs, since 13.0 has 3, and the sum's precision doesn't limit it here as the division is the last operation. Wait, actually, the rule is: for multiplication/division, the result has the same number of sig figs as the least precise measurement. \( 13.0 \) has 3 sig figs, so the result should have 3. \( 152.776 \div 13.0 = 11.752 \), which rounds to 11.8 (3 sig figs: 1, 1, 8).

Answer:

11.8

Question 11 (Note: Since no data table is provided, we can't calculate the exact values, but here's the method for each part):

Part a:

To find the initial volume of the graduated cylinder before adding the solid, you would read the volume from the graduated cylinder (using the meniscus) and record it with the correct significant figures based on the cylinder's graduation (e.g., if it's a 100 mL cylinder with 1 mL graduations, you can estimate to the tenths place, so something like 50.0 mL, but this depends on the actual cylinder used).

Part b:

After adding the solid, read the final volume from the graduated cylinder (again, using the meniscus) and record it with the correct significant figures. For example, if the initial was 50.0 mL and the solid displaced some volume, the final might be 60.5 mL (depending on the solid's volume).

Part c:

The volume of the solid is calculated by subtracting the initial volume (Blank 1) from the final volume (Blank 2). So, \( \text{Volume of solid} = \text{Final Volume (Blank 2)} - \text{Initial Volume (Blank 1)} \). The number of significant figures in the result is determined by the least precise measurement (the one with the least number of decimal places or significant figures between the initial and final volumes).

For example, if Blank 1 (initial) is \( 50.0 \) mL (3 sig figs) and Blank 2 (final) is \( 60.5 \) mL (3 sig figs), then the volume of the solid is \( 60.5 - 50.0 = 10.5 \) mL (3 sig figs).

But since the actual data from the procedure document (Data Table 1 Step 3) is needed, you would use those values to compute each blank.