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:
Cannot be determined
Explanation:
Hi there,
Could you please add an attachment of the table? We cannot tell what the errors are if there is no table.
Answer:
[Co(NH3)5CO3]I3
Explanation:
The naming of coordination compounds follows certain rules specified by IUPAC. Usually, the name of the complex makes it quite easy to deduce its structure.
"Pentaamine" means that there are five NH3 ligands as shown in the structure. The ligand carbonato is CO3^2-. It has no prefix attached to it in the IUPAC name of the complex hence there is only one carbonato ligand present(recall that the complex has a coordination number of six). I did not enclose it within parenthesis as required in the question.
Lastly the III that appeared after the metal name "cobalt" shows its oxidation state. The iodide counter ions must then be 3 in number in order to satisfy this primary valency of the metal hence the inclusion of I3 in the structure of the complex.