QUESTION IMAGE
Question
applying scientific methods
a chemistry student is given the task of analyzing three unknown samples. her data is listed in data table 1. use data table 1 to answer the questions below.
data table 1
| trial | sample a | sample b | sample c | |||
|---|---|---|---|---|---|---|
| 1 | 80.72 | 10.01 | 95.41 | 10.72 | 72.28 | 10.00 |
| 2 | 80.64 | 10.00 | 92.33 | 10.51 | 72.32 | 9.99 |
| 3 | 80.91 | 10.05 | 93.78 | 10.62 | 72.34 | 9.95 |
| avg | 80.76 | 10.02 | 93.84 | 10.62 | 72.30 | 9.98 |
- based on the data given, what is the density of each sample? follow the rules for significant figures and rounding for your answers. show your work
sample a ____________
sample b ____________
sample c ____________
- compare the data for each trail and each sample in data table 1. which sample(s) did the student measure precisely? explain your answer.
Step1: Recall density formula
The density formula is $
ho=\frac{m}{V}$, where $
ho$ is density, $m$ is mass and $V$ is volume.
Step2: Calculate density of Sample A
Using the average values, $m = 80.76$ g and $V=10.02$ mL. So, $
ho_A=\frac{80.76}{10.02}\approx8.06$ g/mL.
Step3: Calculate density of Sample B
Using the average values, $m = 93.84$ g and $V = 10.62$ mL. So, $
ho_B=\frac{93.84}{10.62}\approx8.84$ g/mL.
Step4: Calculate density of Sample C
Using the average values, $m = 72.30$ g and $V=9.98$ mL. So, $
ho_C=\frac{72.30}{9.98}\approx7.24$ g/mL.
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Sample A: 8.06 g/mL
Sample B: 8.84 g/mL
Sample C: 7.24 g/mL