From electronic configuration valence electron of Nitrogen is 5, oxygen 6x2 which 12 since it involve two molecules , that of is frulorine is 7, and that No2F is 24 which is gotten form adding (5,12,7 ).All resonance structure are as follows
F
.. I ..
: O : N :O:
..
OR : F:
I
N .. : F:
/ \ or I
.. .. N
:O : :O: / / \\
/ / \\
:O : : O:
Answer:
T2 =21.52°C
Explanation:
Given data:
Specific heat capacity of sample = 1.1 J/g.°C
Mass of sample = 385 g
Initial temperature = 19.5°C
Heat absorbed = 885 J
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = Final temperature - initial temperature
885J = 385 g× 1.1 J/g.°C×(T2 - 19.5°C )
885 J = 423.5 J/°C× (T2 - 19.5°C )
885 J / 423.5 J/°C = (T2 - 19.5°C )
2.02°C = (T2 - 19.5°C )
T2 = 2.02°C + 19.5°C
T2 =21.52°C
Answer is attached below ( sorry for mistake )
Nitrogen I believe . I need 20 characters.
Above question is incomplete. Complete question is attached below
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Solution:
Reduction potential of metal ions are provided below. Higher the value to reduction potential, greater is the tendency of metal to remain in reduced state.
In present case,
reduction potential of Au is maximum, hence it is least prone to undergo oxidation. Hence, it is
least reactive.
On other hand,
reduction potential of Na is minimum, hence it is most prone to undergo oxidation. Hence, it is
most reactive.