In determining the boiling point
of solutions, always take note of the number of ions that will dissociate in
the solution. It does not depend on the nature of the substance. The greater
the number of ions dissociated, the greater is the boiling point of the solution.
The original concentration is 5.1 × 10⁹ CFU / ml
In order to attain a countable plate, the number of CFU must be present in between 10 and 200 per ml.
Let us take 0.1 ml and dilute it to 1 ml.
This minimizes the concentration to 5.1 × 10⁹ × 10⁻¹ = 5.1 × 10⁸ CFU/ml
In order to minimize the concentration in between 10 and 200, it can be reduced to 5.1 × 10¹
The final concentration = 5.1 × 10¹ CFU/ml
Initial concentration = 5.1 × 10⁹ CFU/ml
Volume of sample with 5.1 × 10¹ CFU = 5.1 × 10¹ CFU × (1 ml / 5.1 × 10⁹ CFU)
= 1.0 × 10⁻⁸ ml
This is the volume to be taken to obtain countable value, 51 CFU.
Changes in the appearance of emission and absorption spectrums are caused by the activities of electrons.
When you transmit an amount of light on atoms, the state of the energy of electrons will become higher. Why? Because it is absorbing the light's energy. Not all wavelengths will be absorbed by the atoms, only selective.
I hope that the explanation is enough to prove that it is the correct answer. Have a nice day!
Answer:
1: f 2: b 3: d 4: e 5:a 6:c
Explanation:
Answer:
Boiling point: 63.3°C
Freezing point: -66.2°C.
Explanation:
The boiling point of a solution increases regard to boiling point of the pure solvent. In the same way, freezing point decreases regard to pure solvent. The equations are:
<em>Boiling point increasing:</em>
ΔT = kb*m*i
<em>Freezing point depression:</em>
ΔT = kf*m*i
ΔT are the °C that change boiling or freezing point.
m is molality of the solution (moles / kg)
And i is Van't Hoff factor (1 for I₂ in chloroform)
Molality of 50.3g of I₂ in 350g of chloroform is:
50.3g * (1mol / 253.8g) = 0.198 moles in 350g = 0.350kg:
0.198 moles / 0.350kg = 0.566m
Replacing:
<em>Boiling point:</em>
ΔT = kb*m*i
ΔT = 3.63°C/m*0.566m*1
ΔT = 2.1°C
As boiling point of pure substance is 61.2°C, boiling point of the solution is:
61.2°C + 2.1°C = 63.3°C
<em>Freezing point:</em>
ΔT = kf*m*i
ΔT = 4.70°C/m*0.566m*1
ΔT = 2.7°C
As freezing point is -63.5°C, the freezing point of the solution is:
-63.5°C - 2.7°C = -66.2°C
