Answer:
The wavelength for the transition from n = 4 to n = 2 is<u> 486nm</u> and the name name given to the spectroscopic series belongs to <u>The Balmer series.</u>
Explanation
lets calculate -
Rydberg equation- 
where ,
is wavelength , R is Rydberg constant (
),
and
are the quantum numbers of the energy levels. (where
)
Now putting the given values in the equation,


Wavelength 
=
= 486nm
<u> Therefore , the wavelength is 486nm and it belongs to The Balmer series.</u>
You can use the equation ΔS(surr)=q(surr)/T or ΔS(surr)=-q(rxn)/T.
the two equations are equal since we know that the energy the system (reactoin) puts out just goes into the surroundings.
(In other words q(surr)=-q(rxn))
Using the equation, <span>ΔS(surr)=-(-283kJ/298K)=0.9497kJ/K or 949.7J/K
This answer makes sense since the reaction is exothermic which means it released energy into the system which usually causes the entropy to increase.
I hope that helps.</span>
Answer:
ΔS = -661.0J/mol is the entropy change for the system
ΔS = -842J/mol.K is the entropy change for the surroundings
Explanation:
From the relationship between ΔG, T, ΔH and ΔS,
Mathematically, ΔG = ΔH - TΔS
TΔS = ΔH - ΔS
ΔS = ΔH - ΔS / T
but ΔG = -54 kJ/mol, ΔH = -251 kJ/mol and T = 25 °C (298 K)
plugging into the equation,
ΔS = -251 kJ/mol - ( -54 kJ/mol) / 298
ΔS = -0.6610KJ/mol or in J.mol
ΔS = -661.0J/mol is the entropy change for the system
- For entropy change for the surroundings = ΔS = ΔH/T
- ΔS = -0.84KJ/mol.K or -842J/mol.K is the entropy change for the surroundings
The atomic radius increases down a column (group) and decreases along a row
The correct answer would be the third option. There would be two atoms that can form hydrogen bonds with the water molecules from the molecule NH2CH2CH2OH. The atoms O and N could make hydrogen bonds with H. Hydrogen bond is an intermolecular force that is a dipole-dipole interaction between a hydrogen and an atom of O, F and N.