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:
Change in internal energy (ΔU) = -9 KJ
Explanation:
Given:
q = –8 kJ [Heat removed]
w = –1 kJ [Work done]
Find:
Change in internal energy (ΔU)
Computation:
Change in internal energy (ΔU) = q + w
Change in internal energy (ΔU) = -8 KJ + (-1 KJ)
Change in internal energy (ΔU) = -8 KJ - 1 KJ
Change in internal energy (ΔU) = -9 KJ
Answer:
Boron has a larger radius and the protons in carbon exert more pull.
Explanation:
Remember than elements have greater radius as they are closer to the bottom left corner, so boron would have the larger radius here. Carbon has a smaller radius, which makes it easier for the protons in carbon to exert more pull.
Explanation:
A chemical bond which is formed in between positively charged atoms when there is sharing of free electrons in a lattice of cations is known as a metallic bond.
In a pure metal, atoms are surrounded by free moving valence electrons which move from one part of metal to another.
Thus, we can conclude that pure metals are held together by metallic bonds due to attraction between mobile valence electrons and positively charged metal ions.
Explanation:
P1V1 = nRT1
P2V2 = nRT2
Divide one by the other:
P1V1/P2V2 = nRT1/nRT2
From which:
P1V1/P2V2 = T1/T2
(Or P1V1 = P2V2 under isothermal conditions)
Inverting and isolating T2 (final temp)
(P2V2/P1V1)T1 = T2 (Temp in K).
Now P1/P2 = 1
V1/V2 = 1/2
T1 = 273 K, the initial temp.
Therefore, inserting these values into above:
2 x 273 K = T2 = 546 K, or 273 C.
Thus, increasing the temperature to 273 C from 0C doubles its volume, assuming ideal gas behaviour. This result could have been inferred from the fact that the the volume vs temperature line above the boiling temperature of the gas would theoretically have passed through the origin (0 K) which means that a doubling of temperature at any temperature above the bp of the gas, doubles the volume.
From the ideal gas equation:
V = nRT/P or at constant pressure:
V = kT where the constant k = nR/P. Therefore, theoretically, at 0 K the volume is zero. Of course, in practice that would not happen since a very small percentage of the volume would be taken up by the solidified gas.
