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:
Yes, yield.
Explanation:
N2(g) + 3 H2(g) → 2 NH3 (g) balanced equation
First, find limiting reactant:
Moles H2 = 1.83 g x 1 mole/2 g = 0.915 moles H2
Moles N2 = 9.84 g N2 x 1 mole/28 g = 0.351 moles N2
The mole ratio of H2: N2 is 3:1, so H2 is limiting (0.915 is less than 3 x 0.351)
Theoretical yield of NH3 = 0.915 mol H2 x 2 mol NH3/3 mol H2 = 0.61 moles NH3
Answer:
Option E. 2.04 L
Explanation:
Data obtained from the question include:
Molarity of NaCl = 2.25 M
Mole of NaCl = 4.58 moles
Volume =..?
Molarity is simply defined as the mole of solute per unit litre of the solution. It is represented mathematically as:
Molarity = mole /Volume
With the above formula, we can obtain the volume of the solution as follow:
Molarity = mole /Volume
2.25 = 4.58/volume
Cross multiply
2.25 x volume = 4.58
Divide both side by 2.25
Volume = 4.58/2.25
Volume = 2.04 L
Therefore, the volume of the solution is 2.04 L