gdm^-3/gmol^-1
g/g=1 dm^-3/1=dm^3 1/mol^-1=mol^1 1 x dm^-3 x mol^1 mol/dm^3
To convert from g/dm^3 to mol/dm^3 you have to divide g/dm^3 by g/mol or the molecular mass.
Answer:
Explanation:
Henry's law states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid
number of moles of cyclopropane = 1.84 g / 42.08 g/mol = 0.044 mol
number of mole of water = 1010000 g / 18.02 g/mol = 56048.83 mol
mole fraction of cyclopropane = mole of cyclopropane / ( mole of water + mole of cyclopropane) = 0.044 mol / 56048.87 = 7.85 × 10⁻⁷
P cyclopropane = k henry × mole fraction of cyclopropane = 4273 atm × 7.85 × 10⁻⁷ = 0.00335 atm
b) number of cyclopropane molecules present in each cubic centimeter = ( 0.044 mol / 1010) × 6.022 × 10 ²³ / 1000 cm³ = 2.623 × 10 ¹⁶ molecules in cm³
Answer
Naphthalene is a non electrolyte
If the unknown compound is an electrolyte it gives 2 or more ions in solution
( NaCl >> Na+ + Cl- => 2 ions
Ca(NO3)2 >> Ca2+ + 2 NO3- => 3 ions)
the f.p. lowering is directly proportional to the molal concentration of dissolved ions in the solution )
For naphthalene
delta T = 1.86 x m
for a salt that gives 2 ions
delta T = 1.86 x m x 2
hence the lowering in freezion point of unkown is greater then napthalene
Answer:
1.014×10²³ atoms
Explanation:
Given data:
Volume of tungsten = 1.6 cm³
Density of tungsten = 19.35 g/cm³
Number of atoms of tungsten = ?
Solution:
First of all we will calculate the mass of tungsten.
density = mass/volume
by putting values,
19.35 g/cm³ = mass/1.6 cm³
Mass = 19.35 g/cm³× 1.6 cm³
Mass = 30.96 g
Number of moles of tungsten:
Number of moles = mass/molar mass
Number of moles = 30.96 g/ 183.84 g/mol
Number of moles = 0.1684 mol
1 mole contain 6.022×10²³ atoms.
0.1684 mol × 6.022×10²³ atoms / 1mol
1.014×10²³ atoms
Answer:
E) A, B, and C
Explanation:
Syn addition refers to the addition of two substituents on the same face or side of a double bond. This differed from anti addition which a occurs across opposite face of the double bond.
Hydrogenation, hydroboration and dihydroxylation all involve syn addition to the double bond, hence the answer chosen above.