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

Answer 1

We have access to no textbook and therefore can not see table 1.3.

Sra

Answer 2

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Answer 3

density= m/v
7.87 g/cm3 = m/ 1.00cm3
7.87g / 11.36 g/cm3
answer is 0.69 cm3

Answer 4

Mass/Density = Volume

_11.4_g_ = v
7.87g/cm3

1.45cm3 = v

Answer 5

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

Answer 6

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