Determining density and using density to determine volume or mass

(a) calculate the density of mercury if 1.00 × 10 g occupies a volume of 7.36 cm
(b) calculate the volume of 65.0 g of liquid methanol (wood alcohol) if its density is 0.791 g/ml.
(c) what is the mass in grams of a cube of gold (density = 19.32 g/cm) if the length of the cube is 2.00 cm?
(d) calculate the density of a 374.5-g sample of copper if it has a volume of 41.8 cm a student needs 15.0 g of ethanol for an experiment. if the density of ethanol is 0.789 g/ml, how many milliliters of ethanol are needed? what is the mass, in grams, of 25.0 ml of mercury (density = 13.6 g/ml)?

corrected question:

Determining Density and Using Density to Determine Volume or Mass

(a) Calculate the density of mercury if 1.00 × 10 g occupies a volume of 7.36 cm³

(b) Calculate the volume of 65.0 g of liquid methanol (wood alcohol) if its density is 0.791 g/mL.

(c) What is the mass in grams of a cube of gold (density = 19.32 g/cm) if the length of the cube is 2.00 cm?

(d) Calculate the density of a 374.5-g sample of copper if it has a volume of 41.8 cm³ A student needs 15.0 g of ethanol for an experiment. If the density of ethanol is 0.789 g/mL, how many milliliters of ethanol are needed? What is the mass, in grams, of 25.0 mL of mercury (density = 13.6 g/mL)?

density =

ρ=m/v ,m=ρv,    v=m/ρ

(a)m=1*10g  , v=7.36cm³

    ρ=10/7.36 =1.36g/cm³

(b) m=65g, ρ=0.791 g/mL.

   v= 65/0.791 =82.17g/mL

(c) ρ=19.32g/cm³, l=2cm, v=l³=8cm³

    m=19..32*8=154.56g/cm³

(d) mass of copper=374.5g , v=41.8cm³

   ρ=374.5/41.8 =8.96g/cm³

 mass of ethanol=15g,  density of ethanol=0.789g/mL

v=15/0.789 =19.01mL

volume of mecury=25mL, density of mercury=13.6g/mL

m=25*13.6=340g

Explanation:

A) Given mass of mercury =

Volume of mercury =

Hence,

B) Given mass =

Density =

Hence,

C) Given density =

Side of the cube =

Hence,

Thus,

.

D)

For more information on this visit

the transfer of heat by electromagnetic radiation is called thermal radiation . it is usually carried out by electromagnetic waves in an energy range just below the visible spectrum, which is the infrared range.

so the answer would be radiation.

: )