Answer:
Most similar - Lithium
Least similar - Phosphorus
Explanation:
Rubidium is an element in group 1A of the periodic table. It is a metal and forms an ionic compound with chlorine. The formula of the compound is RbCl.
If we look at the options, Lithium is also a group 1A element and forms an ionic compound with chlorine having the formula LiCl which is very much similar to RbCl chemically.
Phosphorus is a nonmetal. Its compounds with chlorine, PCl3 and PCl5 are covalent and does not resemble RbCl in any way.
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 = ![\frac{mass}{volume}](https://tex.z-dn.net/?f=%5Cfrac%7Bmass%7D%7Bvolume%7D)
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:
mass of CO = 210.42 g
mass in three significant figures = 210. g
Explanation:
Given data:
mass of Fe2O3 = 0.400 Kg
mass of CO= ?
Solution:
chemical equation:
Fe2O3 + 3CO → 2Fe + 3CO2
Now we will calculate the molar mass of Fe2O3 and CO.
Molar mass of Fe2O3 = (55.845 × 2) + (16 × 3) = 159.69 g/mol
Molar mass of CO = 12+ 16 = 28 g/mol
now we will convert the kg of Fe2O3 in g.
mass of Fe2O3 = 0.400 kg × 1000 = 400 g
number of moles of Fe2O3 = 400 g/ 159.69 g/mol = 2.505 mol
mass of CO = moles of Fe2O3 × 3( molar mass of CO)
mass of CO = 2.505 mol × 84 g/mol
mass of CO = 210.42 g
mass in three significant figures = 210. g
Answer:
97.78% KCl in the original sample
Explanation: