37. three students were asked to determine the volume of a liquid by a method of their choosing. each…
37. three students were asked to determine the volume of a liquid by a method of their choosing. each performed three trials. the table below shows the results. the actual volume of the liquid is 24.8 ml.\n| |trial 1 (ml)|trial 2 (ml)|trial 3 (ml)|\n|----|----|----|----|\n|student a|24.8|24.8|24.4|\n|student b|24.2|24.3|24.3|\n|student c|24.6|24.8|25.0|\n\na. considering the average of all three trials, which students measurements show the greatest accuracy?\nb. which students measurements show the greatest precision?\n\nproblems. show all your work in the space provided.\n38. a single atom of platinum has a mass of 3.25×10^(-22) g. what is the mass of 6.0×10^(23) platinum atoms?\n\n39. a sample thought to be pure lead occupies a volume of 15.0 ml and has a mass of 160.0 g.\na. determine its density.\nb. the density of pure lead is 11.35 g/cm³. is the sample pure lead?\nc. determine the percentage error, based on the accepted value for the density of lead.
Answer
37.
a.
Explanation:
Step1: Calculate average for Student A
$\text{Average}_A=\frac{24.8 + 24.8+24.4}{3}=\frac{74}{3}\approx24.67$ mL
Step2: Calculate average for Student B
$\text{Average}_B=\frac{24.2 + 24.3+24.3}{3}=\frac{72.8}{3}\approx24.27$ mL
Step3: Calculate average for Student C
$\text{Average}_C=\frac{24.6 + 24.8+25.0}{3}=\frac{74.4}{3}=24.8$ mL
Step4: Compare with actual value
The actual volume is 24.8 mL. Student C's average is exactly 24.8 mL, so Student C has the greatest accuracy.
Answer:
Student C
b.
Explanation:
Precision refers to how close the measurements are to each other. Student B's measurements (24.2, 24.3, 24.3) are the closest to each other compared to Student A and Student C.
Answer:
Student B
38.
Explanation:
Step1: Use multiplication
If one - atom mass is $m = 3.25\times10^{-22}$ g and number of atoms is $N=6.0\times 10^{23}$, then the total mass $M$ is given by $M = m\times N$. $M=(3.25\times 10^{-22})\times(6.0\times 10^{23})$
Step2: Apply exponent rule
Using the rule $a^m\times a^n=a^{m + n}$, we have $M=(3.25\times6.0)\times10^{-22 + 23}=19.5\times10^{1}=195$ g
Answer:
195 g
39.
a.
Explanation:
Step1: Recall density formula
The density formula is $\rho=\frac{m}{V}$, where $m$ is mass and $V$ is volume. Given $m = 160.0$ g and $V = 15.0$ mL. Since 1 mL = 1 $cm^3$, $V=15.0$ $cm^3$.
Step2: Calculate density
$\rho=\frac{160.0}{15.0}=\frac{32}{3}\approx10.67$ g/$cm^3$
Answer:
$10.67$ g/$cm^3$
b.
Explanation:
The density of pure lead is 11.35 g/$cm^3$. The calculated density of the sample is 10.67 g/$cm^3$. Since 10.67 g/$cm^3\neq11.35$ g/$cm^3$, the sample is not pure lead.
Answer:
No
c.
Explanation:
Step1: Recall percentage - error formula
The percentage - error formula is $\text{Percentage error}=\left|\frac{\text{Experimental value}-\text{Accepted value}}{\text{Accepted value}}\right|\times100%$ The experimental value $\rho_{exp}=10.67$ g/$cm^3$ and the accepted value $\rho_{acc}=11.35$ g/$cm^3$
Step2: Calculate percentage error
$\text{Percentage error}=\left|\frac{10.67 - 11.35}{11.35}\right|\times100%=\left|\frac{- 0.68}{11.35}\right|\times100%\approx6.0%$
Answer:
Approximately 6.0%