<span>The answer is 4. The molecules of each material entice each other over dispersion (London) intermolecular forces. Whether a substance is a solid, liquid, or gas hinge on the stability between the kinetic energies of the molecules and their intermolecular magnetisms. In fluorine, the electrons are firmly apprehended to the nuclei. The electrons have slight accidental to stroll to one side of the molecule, so the London dispersion powers are comparatively weak. As we go from fluorine to iodine, the electrons are far from the nuclei so the electron exhausts can more effortlessly misrepresent. The London dispersion forces developed to be increasingly stronger.</span>
Answer:
Solution of isopropanol is 10.25 molal
Explanation:
615 g of isopropanol (C3H7OH) per liter
We gave the information that 615 g of solute (isopropanol) are contained in 1L of water. We need to find out the mass of solvent, so we use density.
Density of water 1g/mL → Density = Mass of water / 1000 mL of water
Notice we converted the L to mL
Mass of water = 1000 g (which is the same to say 1kg)
Molality are the moles of solute in 1kg of solvent, so let's convert the moles of isopropanol → 615 g . 1mol / 60g = 10.25 moles
Molality (mol/kg) = 10.25 moles / 1kg = 10.25 m
Answer:
Molar concentration of the weak acid solution is 0.0932
Explanation:
Using the formula: 
Where Ca = molarity of acid
Cb = molarity of base = 0.0981 M
Va = volume of acid = 25.0 mL
Vb = volume of base = 23.74 mL
na = mole of acid
nb = mole of base
Since the acid is monopromatic, 1 mole of the acid will require 1 mole of NaOH. Hence, na = nb = 1
Therefore, 
Ca = 0.0981 x 23.74/25.0
= 0.093155 M
To 4 significant figure = 0.0932 M
Answer:
Molecular formula for the gas is: C₄H₁₀
Explanation:
Let's propose the Ideal Gases Law to determine the moles of gas, that contains 0.087 g
At STP → 1 atm and 273.15K
1 atm . 0.0336 L = n . 0.082 . 273.15 K
n = (1 atm . 0.0336 L) / (0.082 . 273.15 K)
n = 1.500 × 10⁻³ moles
Molar mass of gas = 0.087 g / 1.500 × 10⁻³ moles = 58 g/m
Now we propose rules of three:
If 0.580 g of gas has ____ 0.480 g of C _____ 0.100 g of C
58 g of gas (1mol) would have:
(58 g . 0.480) / 0.580 = 48 g of C
(58 g . 0.100) / 0.580 = 10 g of H
48 g of C / 12 g/mol = 4 mol
10 g of H / 1g/mol = 10 moles
Answer:
The freezing point will be 
Explanation:
The depression in freezing point is a colligative property.
It is related to molality as:
Where
Kf= 
the molality is calculated as:
Depression in freezing point = 
The new freezing point = 
