Volume = 4/3 * pi * r3 Volume = 4/3 * pi * (7.5)3 = 1767 ( units ) ----------------------------though the significant figures constraint may show 1800
figures with the same volume does not have the same surface area.
The core sample is a cylinder. The volume of any cylinder is (pi) x (radius)2 x (length).
You calculate the volume of three-dimensional figures, not of numbers like pi.
The volume of the sample whose mass is 20 g and density is 4 g/ml is 5 milliliters.
It varies. Volume may be reported with more or less significant figures. However, in general the result should not have more significant figures than the underlying data - otherwise, it would look more accurate than it really is.
The question is based on a fallacy. Volumes can be reported in any number of significant figures.
Because density expressed in two significant figures depends on your accuracy of your measurements of mass and volume to calculate as well as any variables that you are expected to use.
The accuracy of the measurement device determines the number of significant figures that should be retained in recording measurements.
No, the units are independent of the accuracy. If you are measuring volume, how accurate the measurement is (or isn't) will not affect what you are measuring - it will always be volume.
If you know the density of mercury, you can determine the mass of a specific volume of mercury. Mercury has a density of 13.534g/cm3. 1cm3 = 1mL, so we can restate its density as 13.534g/mL. Density = mass/volume. If we know any two variables, we can manipulate the density equation to find the third variable. In this case, we know volume and density, so to find the mass, do the following calculation: Mass = density x volume Mass Hg = 13.534g/mL x 136mL = 1.84g Hg* *The answer is limited to 3 significant figures, because 136mL has only 3 significant figures, even though the density has 5 significant figures. When multiplying or dividing, the answer is limited to the same number of significant figures as the measurement with the fewest significant figures used in the calculation.
The density of the ore sample is 1.97 g/cm3. This value is calculated by dividing the mass of the ore (57 g) by its volume (29 cm3).
The material has a density of about 0.848 g/cm3
The density of mercury is found by dividing its mass by volume. In this case, the density is 13.53 g/mL.
Density = mass/volume Density = 22.4g/8cm3 = 2.8g/cm3 or 3g/cm3 if significant figures are considered.
The answer should have three significant figures since the divisor (3.42) has three significant figures. Therefore, the result of the division, 13.3669590643 kg/L, should be rounded to three significant figures, which is 13.4 kg/L.
The volume of 12.0000 mL would be recorded as 12.00 mL when measured from a 50-mL graduated cylinder because the cylinder has markings in increments of 1 mL. It is standard practice to record the volume to two decimal places for greater accuracy.