A certain FM radio wave has a frequency of 1.31 x 108 Hz. Given that radio waves travel at
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Answer: 4
Explanation:
Principle Quantum Numbers: This quantum number describes the size of the orbital. It is represented by n.
Azimuthal Quantum Number: This quantum number describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...
Magnetic Quantum Number: This quantum number describes the orientation of the orbitals. It is represented as . The value of this quantum number ranges from
. When l = 2, the value of
will be -2, -1, 0, +1, +2.
Given : a f subshell, thus l = 3 , Thus the subshells present would be 3, 2, 1, 0 and thus n will have a value of 4.
Also electrons give are 32.
The formula for number of electrons is .
Thus principal quantum no will be n= 4.
This is an incomplete question, here is a complete question.
Suppose we now collect hydrogen gas, H₂(g), over water at 21°C in a vessel with total pressure of 743 Torr. If the hydrogen gas is produced by the reaction of aluminum with hydrochloric acid:
what volume of hydrogen gas will be collected if 1.35 g Al(s) reacts with excess HCl(aq)? Express your answer in liters.
Answer : The volume of hydrogen gas that will be collected is 1.85 L
Explanation :
First we have to calculate the number of moles of aluminium.
Given mass of aluminium = 1.35 g
Molar mass of aluminium = 27 g/mol
The given chemical reaction is:
As, hydrochloric acid is present in excess. So, it is considered as an excess reagent.
Thus, aluminium is a limiting reagent because it limits the formation of products.
By Stoichiometry of the reaction:
2 moles of aluminium produces 3 moles of hydrogen gas
So, 0.005 moles of aluminium will produce = of hydrogen gas
Now we have to calculate the mass of helium gas by using ideal gas equation.
PV = nRT
where,
P = Pressure of hydrogen gas = 743 Torr
V = Volume of the helium gas = ?
n = number of moles of hydrogen gas = 0.075 mol
R = Gas constant =
T = Temperature of hydrogen gas =
Now put all the given values in above equation, we get:
Hence, the volume of hydrogen gas that will be collected is 1.85 L