Overharvesting that would be your answer!
Answer:
+1
Explanation:
Na₂O₂
NOTE: the oxidation number of oxygen is always –2 except in peroxides where it is –1.
Thus, we can obtain the oxidation number of sodium (Na) in Na₂O₂ as illustrated below:
Na₂O₂ = 0 (oxidation number of ground state compound is zero)
2Na + 2O = 0
O = –1
2Na + 2(–1) = 0
2Na – 2 = 0
Collect like terms
2Na = 0 + 2
2Na = 2
Divide both side by 2
Na = 2/2
Na = +1
Thus, the oxidation number of sodium (Na) in Na₂O₂ is +1
Answer:
I think it is AM and frequency
Explanation:
Sorry if i'm wrong ;)
Answer:
the wavelength of radiation emitted is 
Explanation:
The energy of the Bohr's hydrogen atom can be expressed with the formula:

For n = 7:


For n = 4


The electron goes from the n = 7 to the n = 4, then :


Wavelength of the radiation emitted:

where;
hc = 1242 eV.nm

