According to table 1.3, what volume of iron would be needed to balance a 1.00cm3 sample of lead on a two pan laboratory balance? Table 1.3 shows densities g/cm3 such as iron=7.87g/cm3 and lead is 11.4g/cm3
density= m/v
7.87 g/cm3 = m/ 1.00cm3
7.87g / 11.36 g/cm3
answer is 0.69 cm3
Mass/Density = Volume
_11.4_g_ = v
7.87g/cm3
1.45cm3 = v
What volume of iron (density 7.87 g/cm3) would be required to balance a 1.45 cm3 sample of lead (density 11.4 g/cm3) on a two-pan laboratory balance
Let's review the solution...
What volume of iron (density 7.87 g/cm3) would be required to balance a 2.90 cm3 sample of lead (density 11.4 g/cm3) on a two-pan laboratory balance?
Lead has a density of 11.4 g/cm3. A 2.90 cm3 sample of lead would have a mass of
ρ = m/V therefore m = ρV
m = (11.4 g/cm3)(2.90 cm3)
= 11.4 × 2.90g/cm3 × cm3
= 33.1 g
Iron has a density of 7.87 g/cm3. The volume of iron required to balance a mass of 33.1 g of lead is:
ρ = m/V therefore V = m/ρ
V = 33.1 g/7.87g/cm3 = 33.1g/ 7.87 1g x cm3/g
= 4.200762 cm3
= 4.20 cm3
To determine the volume of iron needed to balance a 1.00cm3 sample of lead on a two pan laboratory balance, we can use the densities of iron and lead to calculate the mass of both substances.
First, let's calculate the mass of the lead sample. We know that the density of lead is 11.4g/cm3, and the volume of the lead sample is 1.00cm3. So, the mass of the lead sample can be calculated as:
Mass of lead = Density of lead x Volume of lead
= 11.4g/cm3 x 1.00cm3
= 11.4g
Now, to balance the lead sample with iron on the other pan of the balance, we need to determine the volume of iron required, assuming its density is 7.87g/cm3.
We can calculate the volume of iron required using the equation:
Volume of iron = Mass of iron / Density of iron
Since the mass of iron required is equal to the mass of lead, which is 11.4g, we can substitute this value into the equation:
Volume of iron = 11.4g / 7.87g/cm3
≈ 1.45 cm3
Therefore, approximately 1.45 cm3 of iron would be needed to balance a 1.00 cm3 sample of lead on a two pan laboratory balance, according to the information provided in Table 1.3.
We have access to no textbook and therefore can not see table 1.3.
Sra