The density of water is 1.0 grams per milliliter then it will be sink in water
Density is a word we use to describe how much space an object or substance takes up in the volume and in relation to the amount of matter in that object or substance its mass) and another way to put it is that density is the amount of mass per unit of volume
Here given density is 1.0 grams per milliliter and it will be sink in water because the density of water in 1.0 g/ml and this object is more dense than water and the density of an object determines whether it will float or sink in another substance and an object will float if it is less dense than the liquid it is placed in and an object will sink if it is more dense than the liquid it is placed
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Answer:
2.7 °C.kg/mol
Explanation:
Step 1: Calculate the freezing point depression (ΔT)
The normal freezing point of a certain liquid X is-7.30°C and the solution freezes at -9.9°C instead. The freezing point depression is:
ΔT = -7.30 °C - (-9.9 °C) = 2.6 °C
Step 2: Calculate the molality of the solution (b)
We will use the following expression.
b = mass of solute / molar mass of solute × kilograms of solvent
b = 102. g / (162.2 g/mol) × 0.650 kg = 0.967 mol/kg
Step 3: Calculate the molal freezing point depression constant Kf of X
Freezing point depression is a colligative property. It can be calculated using the following expression.
ΔT = Kf × b
Kf = ΔT / b
Kf = 2.6 °C / (0.967 mol/kg) = 2.7 °C.kg/mol
Answer:
The mean free path = 2.16*10^-6 m
Explanation:
<u>Given:</u>
Pressure of gas P = 100 kPa
Temperature T = 300 K
collision cross section, σ = 2.0*10^-20 m2
Boltzmann constant, k = 1.38*10^-23 J/K
<u>To determine:</u>
The mean free path, λ
<u>Calculation:</u>
The mean free path is related to the collision cross section by the following equation:
where n = number density
Substituting for P, k and T in equation (2) gives:
Next, substituting for n and σ in equation (1) gives:
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