unit 2 - measurements and calculations review\nhow does quantitative information differ from qualitative…
unit 2 - measurements and calculations review\nhow does quantitative information differ from qualitative information?\nround each of the following measurements to the number of significant figures indicated.\na) 67.029 g to three significant figures\nb) 0.15 l to one significant figure\nc) 52.8005 mg to five significant figures\nwhat is the volume, in cubic meters, of a rectangular solid that is 0.25 m long, 6.1 m wide, and 4.9 m high?\nfind the density of a material, given that a 5.03 g sample occupies 3.24 ml.\na sample of a substance that has a density of 0.824 g/ml has a mass of 0.451 g. calculate the volume of the sample.\nhow many grams are in 882 μg?\nthe density of gold is 19.3 g/cm³. what is the volume, in cubic centimeters, of a sample of gold that has a mass of 0.715 kg? if this sample of gold is a cube, what is the length of each edge in centimeters?\na student measures the mass of a sample as 9.67 g. calculate the percentage error, given that the correct mass is 9.82 g.\na book gives the density of calcium as 1.54 g/cm³. based on lab measurements, what is the percentage error for a density calculation of 1.25 g/cm³?
Answer
Explanation:
Step1: Define significant - figure rounding rules
When rounding to a certain number of significant figures, look at the digit to the right of the last significant digit. If it is 5 or greater, round up; if it is less than 5, round down.
Step2: Round 67.029 g to three significant figures
The fourth - digit 2 is less than 5, so we round down. The answer is 67.0 g.
Step3: Round 0.15 L to one significant figure
The second - digit 5 means we round up. The answer is 0.2 L.
Step4: Round 52.8005 mg to five significant figures
The sixth - digit 5 means we round up the fifth - digit 0 to 1. The answer is 52.801 mg.
Step5: Calculate volume of rectangular solid
The volume formula for a rectangular solid is $V = l\times w\times h$. Substitute $l = 0.25m$, $w = 6.1m$, and $h = 4.9m$ into the formula: $V=0.25\times6.1\times4.9 = 7.525m^{3}\approx7.5m^{3}$ (rounded to two significant figures).
Step6: Calculate density
The density formula is $D=\frac{m}{V}$. Given $m = 5.03g$ and $V = 3.24mL$, then $D=\frac{5.03}{3.24}\approx1.55g/mL$.
Step7: Calculate volume from density and mass
Given $D = 0.824g/mL$ and $m = 0.451g$, using the formula $V=\frac{m}{D}$, we get $V=\frac{0.451}{0.824}\approx0.547mL$.
Step8: Convert micrograms to grams
Since $1\mu g=1\times10^{- 6}g$, for $882\mu g$, $m = 882\times10^{-6}g = 0.000882g$.
Step9: Calculate volume of gold sample
Given $m = 0.715kg = 715g$ and $D = 19.3g/cm^{3}$, using $V=\frac{m}{D}$, we have $V=\frac{715}{19.3}\approx37.05cm^{3}\approx37.1cm^{3}$ (rounded to three significant figures). If it is a cube, and $V = s^{3}$ (where $s$ is the side - length), then $s=\sqrt[3]{V}=\sqrt[3]{37.05}\approx3.34cm$.
Step10: Calculate percentage error
The percentage error formula is $\text{Percentage Error}=\frac{\text{Measured Value}-\text{True Value}}{\text{True Value}}\times100%$. For measured value $m_1 = 9.67g$ and true value $m_2 = 9.82g$, $\text{Percentage Error}=\frac{9.67 - 9.82}{9.82}\times100%\approx - 1.53%$. The absolute value of the percentage error is approximately $1.53%$. For density, with measured density $D_1 = 1.25g/cm^{3}$ and true density $D_2 = 1.54g/cm^{3}$, $\text{Percentage Error}=\frac{1.25 - 1.54}{1.54}\times100%\approx - 18.8%$. The absolute value of the percentage error is approximately $18.8%$.
Answer:
a. 67.0 g b. 0.2 L c. 52.801 mg Volume of rectangular solid: 7.5 m³ Density of material: 1.55 g/mL Volume of substance: 0.547 mL Mass in grams for 882 μg: 0.000882 g Volume of gold sample: 37.1 cm³, side - length of[SSE Completed, Client Connection Error][SSE Completed, Client Connection Error][SSE Completed, Client Connection Error][LLM SSE On Failure]