Benzene is the compound name and it is a covalent compound
Give the positional isomers of 2 halopropane
(CH3)2CH
answer:
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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:
Substance 4
Explanation:
Density is defined as mass per unit volume. A substance floats in another when its density is less than that of the substance on which it floats.
If an object is immersed in a denser liquid, it will sink
Since the density of water is 1 g/cm^3, any substance whose density is less than that of water will float in it.
If we look at the table, substance 4 has a density of 0.5g/cm^3. Hence, substance 4 is expected to float in water.
Answer:
Explanation:
The possible energy states for the particles are 0, 10 and 20 J.
The constraint in the system is that the total energy of the particles must be 20 J.
One given configuration where the total energy is 20 J is if both the particles occupy the 10 J state.
Hence, (10;10) is the given configuration.
Another possibility is if one of the particle is in 0 J state and another is in 20 J state. Hence, the system has a total energy of 0+20 = 20 J.
Hence, the possible configuration can be written as (0;20) or (20;0) which are energetically equivalent to the given configuration. Note that if the circles are indistinguishable, then the configuration (0,20) and (20,0) is the same thing.
