What volume of coppr ( density 8.96g/cm3)would be needed to balance a 2.28 cm3sample of lead (density 11.4g/cm30 on a two-pan laboratory balance

2.9 cm³

Step-by-step explanation:

Note: For the copper and the lead to balance, they must have thesame mass.

Applying,

D' = m'/V Equation 1

Where D' = density of the lead, m' = mass of the lead, V = Volume of the lead.

make m' the subject of the equation

m' = D'V Equation 2

From the question,

Given: D' = 11.4 g/cm³, V = 2.28 cm³

Substitute these values into equation 2

m' = (11.4×2.28)

m' = 25.992 g

Similarly,

for copper,

D = m/V Equation 3

Where D = density of  copper, m = mass of copper, V = volume of copper.

make V the subject of the equation

V = m/D Equation 4

where, m = m' = 25.992 g

Given: D = 8.96 g/cm³

Substitute these values into equation 4

V = 25.992/8.96

V = 2.9 cm³

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°