6. write the following number as a decimal: 6.3×10⁻⁴\n7. write the following number as a decimal…
6. write the following number as a decimal: 6.3×10⁻⁴\n7. write the following number as a decimal: 9.2×10⁷\n8. write the following number in scientific notation: 219,000,000\n9. write the following number in scientific notation: 0.0000507\n10. how is experimental uncertainty related to precision. if an instrument is listed to have an uncertainty of (±0.1 g) and another is (±0.001 g). which instrument is more precise?\n11. explain how the law of conservation of energy applies to electrical production in a coal plant.\n12. explain how the state of matter relates to the amount of energy a substance has.\n13. determine which object has the greatest density.\nobject mass volume\n#1 53.5 g 138 ml\n#2 218 g 985 ml\n#3 154 g 526 ml\nc: the most dense object is ___, it has a density of ___.\ne:(calculate density of each object. show work and include units)\nobject #1 object #2 object #3\nr: since object ___ has a density of ___ compared to ......
Answer
Explanation:
Step1: Recall density formula
The density formula is $\rho=\frac{m}{V}$, where $\rho$ is density, $m$ is mass and $V$ is volume.
Step2: Calculate density of Object #1
For Object #1, $m = 53.5\ g$ and $V=138\ mL$. So $\rho_1=\frac{53.5\ g}{138\ mL}\approx 0.39\ g/mL$.
Step3: Calculate density of Object #2
For Object #2, $m = 218\ g$ and $V = 985\ mL$. So $\rho_2=\frac{218\ g}{985\ mL}\approx 0.22\ g/mL$.
Step4: Calculate density of Object #3
For Object #3, $m = 154\ g$ and $V = 526\ mL$. So $\rho_3=\frac{154\ g}{526\ mL}\approx 0.29\ g/mL$.
Step5: Compare densities
Comparing $\rho_1\approx 0.39\ g/mL$, $\rho_2\approx 0.22\ g/mL$ and $\rho_3\approx 0.29\ g/mL$, we find that Object #1 has the greatest density.
Answer:
The most dense object is #1, it has a density of approximately $0.39\ g/mL$.