Question

1. if you are asked to calculate percent error, you should know that percent error = (\frac{vert\text{measured value}-\text{accepted value}vert}{\text{accepted value}}\times100%) a certain piece of metal has an accepted mass of 65.0 grams. its mass was recorded in the laboratory as 55.0 grams. calculate the percent error in this measurement. 13. the following information concerns the metric system and other unit conversions. you should definitely know these numbers. 1 l = ____ ml 1 kg = __ g (^{circ}c+)____ = k 14. students often forget how to determine how many significant figures are in a given measurement. see if you remember how to do this.

number	significant figures

|25.7|
|100.62|
|5.00|
|200|
|200.0|
|0.075|
|0.0050|

1. when you multiply or divide two numbers, the rule is that the final answer should be rounded so that it has the same number of sig figs as the measurement with the fewest sig figs. if the mass of an object is 2.7 g and the volume is 3.5 ml, calculate the density and round your answer to the proper number of sig figs. 16. report the average of these three measurements using the correct number of significant figures.

trial 1	trial 2	trial 3	average

Explanation:

Step1: Calculate percent - error for problem 12

Percent error formula is $\text{Percent Error}=\frac{\vert\text{Measured Value}-\text{Accepted Value}\vert}{\text{Accepted Value}}\times100\%$. Given measured value = 55.0 g and accepted value = 65.0 g. First, find the difference: $\vert55.0 - 65.0\vert=\vert- 10.0\vert = 10.0$. Then, $\text{Percent Error}=\frac{10.0}{65.0}\times100\%$.
Percent Error=$\frac{1000}{65.0}\% \approx 15.4\%$

Step2: Fill in unit - conversion for problem 13

1 L = 1000 mL, 1 kg = 1000 g, $^{\circ}C+273.15 = K$

Step3: Determine significant figures for problem 14

• For 25.7, there are 3 significant figures.
• For 100.62, there are 5 significant figures.
• For 5.00, there are 3 significant figures.
• For 200, there is 1 significant figure (if no decimal point, trailing zeros are not significant).
• For 200.0, there are 4 significant figures (trailing zeros after decimal point are significant).
• For 0.075, there are 2 significant figures (leading zeros are not significant).
• For 0.0050, there are 2 significant figures (leading zeros are not significant, but the trailing 0 is significant).

Step4: Calculate density for problem 15

Density formula is $
ho=\frac{m}{V}$, where $m = 2.7$ g and $V = 3.5$ mL. $
ho=\frac{2.7}{3.5}\text{g/mL}\approx0.77\text{g/mL}$ (2 significant figures since both 2.7 and 3.5 have 2 significant figures)

Step5: Calculate average for problem 16

Average formula is $\text{Average}=\frac{\text{Trial 1}+\text{Trial 2}+\text{Trial 3}}{3}$. $\text{Average}=\frac{85.2 + 84.9+85.4}{3}=\frac{255.5}{3}\approx85.2$

Answer:

1. Approximately 15.4%
2. 1000, 1000, 273.15
3. 3, 5, 3, 1, 4, 2, 2
4. 0.77 g/mL
5. 85.2