Suppose the gas is ideal gas. According to the ideal gas equation of state, pV = nRT, the pressure is unchanged. So Vi/Ti = nR/p = Vn/Tn. (i stands for initial and n stands for new) So the new volume Vn = 3.67 L
Answer:
the weakest acid is B) HIO
Explanation:
pKa = - log Ka
the higher the value of pKa, the lower the dissociation, therefore, an acid will be stronger the lower its pKa.
a) HC2H3O2; Ka = 1.8 E-5
⇒ pKa1 = - Log (1.8 E-5) = 4.745
b) HIO; Ka = 23 E-11
⇒ pKa2 = - Log ( 23 E-11 ) = 9.638
c) HBrO; Ka = 23 E-9
⇒ pKa3 = - Log ( 23 E-9 ) = 7.638
d) HClO; Ka = 2.9 E-8
⇒ pKa4 = - Log ( 2.9 E-8 ) = 7.537
e) HCO2H; Ka = 63 E-5
⇒ pKa5 = - Log ( 63 E-5 ) = 3.200
from the values pKa, we places the acids from the weakest to the least weak:
1) pKa2; HIO (weakest)
2) pKa3
3) pKa4
4) pKa1
5) pKa5
There are one antibonding molecular orbitals present in molecular orbital model of c.
The cyclobutadiene has a pi system comprised of four individual atomic p - orbital and thus should have a four pi molecular orbitals. The compound is the prototypical antiaromatic hydrocarbon with 4 
- electrons . Its rectangular structure is the result of jahn teller reaction which disorder the molecule and lowers its symmetry , converting the triplet to a singlet ground state. It is a small annulene . The delocalisation energy of the electrons of the cyclobutene is predicted to be zero .
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