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)?
Answer:
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
Answer:
125000
Explanation:
Because it is halved and halved again.
Explanation:
Vapor pressure is defined as the pressure exerted by vapors or gas on the surface of a liquid.
It is known that at standard condition, vapor pressure is 760 mm Hg.
And, it is given that methanol vapor pressure in air is 88.5 mm Hg.
Hence, calculate the volume percentage as follows.
Volume percentage = 
= 
= 11.65%
Thus, we can conclude that the maximum volume percent of Methanol vapor that can exist at standard conditions is 11.65%.
207 is the mass number. 82 would be the atomic number
Answer:
34.02 g.
Explanation:
Hello!
In this case, since the gas behaves ideally, we can use the following equation to compute the moles at the specified conditions:

Now, since the molar mass of a compound is computed by dividing the mass over mass, we obtain the following molar mass:

So probably, the gas may be H₂S.
Best regards!