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.
I believe the answer is compound B may have a lower molecular weight compared to compound A.
At the same temperature, lighter particles of a compound have a higher average speeds than do heavier particles of another compound. Thus, particles of compound B are lighter than those of compound A and thus they have a higher average speed, hence evaporating faster compared to compound A.
The enthalpy of vaporization of H2O is higher than the enthalpy of fusion of H2O, therefore vaporizing the same mass of H2O would require more heat/energy than melting the same mass of H2O.
Answer: The correct option is The properties of a noble gas.
Explanation: There are 7 periods in the periodic table.
The last element of each period are Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe), Radon (Rn) and Ununoctium (Uuo).
- The electronic configuration for Helium is
. For He, The outermost electrons are 2.
- The electronic configuration for all the other elements is
( where, n = 2, 3, 4, 5, 6 and 7 respectively). For all the other gases, the outermost electrons are 8.
All these elements have stable electronic configuration and are not reactive in nature. Hence, they are considered as noble gases.
Therefore, the last element of each period always have the properties of a noble gas.
Molar mass Na = 23g/mol
46g = 456/2 = 2mol
1mol = 6.022*10^23 atoms
2mol = 2*6.022*10623
= 1.204*10^24 atoms
