Answer : The density of a sample of chlorine gas is, 12.59 g/L
Explanation :
To calculate the density of chlorine gas we are using ideal gas equation.
where,
P = pressure of chlorine gas = 4.5 atm
V = volume of chlorine gas = 12.6 L
n = number of moles of chlorine gas
w = mass of chlorine gas
R = gas constant = 0.0821 L.atm/mol.K
T = temperature of chlorine gas =
M = molar mass of chlorine gas = 71 g/mol
Now put all the given values in the above formula, we get:
Therefore, the density of a sample of chlorine gas is, 12.59 g/L
Answer:
4
Explanation:
Relationship between wavenumber and Rydberg constant (R) is as follows:
Here, Z is atomic number.
R=109677 cm^-1
Wavenumber is related with wavelength as follows:
wavenumber = 1/wavelength
wavelength = 253.4 nm
Z fro Be = 4
Therefore, the principal quantum number corresponding to the given emission is 4.
Answer:
We can not swim in 1.00 × 10²⁷ molecules of water
Explanation:
The given number of molecules of water = 1.00 × 10²⁷ molecules
The Avogadro's number, , gives the number of molecules in one mole of a substance
≈ 6.0221409 × 10²³ molecules/mol
Therefore
Therefore, we have;
The number of moles of water present in 1.00 × 10²⁷ molecules, n = (The number of molecules of water) ÷
∴ n = (1.00 × 10²⁷ molecules)/(6.0221409 × 10²³ molecules/mol) = 1,660.53902857 moles
The mass of one mole of water = The molar mass of water = 18.01528 g/mol
The mass, 'm', of water in 1,660.53902857 moles of water is given as follows;
Mass = (The number of moles of the substance) × (The molar mass of the substance)
∴ The mass of the water in the given quantity of water, m = 1,660.53902857 moles × 18.01528 g/mol ≈ 29.9150756 kg.
The density pf water, ρ = 997 kg/m³
Volume = Mass/Density
∴ The volume of the water present in the given quantity of water, v = 29.9150756 kg/(997 kg/m³) ≈ 30.0050909 liters
The volume of the water present in 1.00 × 10²⁷ molecules of water ≈ 30.0 liters
The average volume of a human body = 62 liters
Therefore, we can not swim in the given quantity of 1.00 × 10²⁷ molecules = 30.0 liters water
Answer: The standard entropy of vaporization of ethanol is 0.275 J/K
Explanation:
Using Gibbs Helmholtz equation:
For a phase change, the reaction remains in equilibrium, thus
Given: Temperature = 285.0 K
Putting the values in the equation:
Thus the standard entropy of vaporization of ethanol is 0.275 J/K