Answer:
molecular weight (Mb) = 0.42 g/mol
Explanation:
mass sample (solute) (wb) = 58.125 g
mass sln = 750.0 g = mass solute + mass solvent
∴ solute (b) unknown nonelectrolyte compound
∴ solvent (a): water
⇒ mb = mol solute/Kg solvent (nb/wa)
boiling point:
- ΔT = K*mb = 100.220°C ≅ 373.22 K
∴ K water = 1.86 K.Kg/mol
⇒ Mb = ? (molecular weight) (wb/nb)
⇒ mb = ΔT / K
⇒ mb = (373.22 K) / (1.86 K.Kg/mol)
⇒ mb = 200.656 mol/Kg
∴ mass solvent = 750.0 g - 58.125 g = 691.875 g = 0.692 Kg
moles solute:
⇒ nb = (200.656 mol/Kg)*(0.692 Kg) = 138.83 mol solute
molecular weight:
⇒ Mb = (58.125 g)/(138.83 mol) = 0.42 g/mol
Answer:
that are formed from hybridized orbitals
Explanation:
The chemical concept of resonance is when a change in the position of the electrons occurs, without changing the position of the atoms.
The structure obtained in the resonance will not be any of the previous ones, but a hybrid of resonance between those structures.
Answer:
P = 20.1697 atm
Explanation:
In this case we need to use the ideal gas equation which is:
PV = nRT (1)
Where:
P: Pressure (atm)
V: Volume (L)
n: moles
R: universal gas constant (=0.082 L atm / K mol)
T: Temperature
From here, we can solve for pressure:
P = nRT/V (2)
According to the given data, we have the temperature (T = 20 °C, transformed in Kelvin is 293 K), the moles (n = 125 moles), and we just need the volume. But the volume can be calculated using the data of the cylinder dimensions.
The volume for any cylinder would be:
V = πr²h (3)
Replacing the data here, we can solve for the volume:
V = π * (17)² * 164
V = 148,898.93 cm³
This volume converted in Liters would be:
V = 148,898.93 mL * 1 L / 1000 mL
V = 148.899 L
Now we can solve for pressure:
P = 125 * 0.082 * 293 / 148.899
<h2>
P = 20.1697 atm</h2>
Answer:
Explanation:
In 150 ml of .06 g / ml solution , gram of iodine = 150 x .06 g = 9 g
Let volume of given concentration of .12 g / ml required be V
In volume V , gram of iodine = V x .12 g
According to question
V x .12 = 9 g
V = 9 / .12 = 75 ml
So, 75 ml of .12 g/ml will be taken and it is diluted to the volume of 150 ml to get the solution of required concentration .
