H2(g) +C2H4(g)→C2H6(g)
H-H +H2C =CH2→H3C-Ch3
2C -H bonds and one C-C bond are formed while enthalpy change (dH) of the reaction,
H-H: 432kJ/mol
C=C: 614kJ/mol
C-C: 413 kJ/mol
C-C: 347 kJ/mol
dH is equal to sum of the energies released during the formation of new bonds or negative sign, and sum of energies required to break old bonds or positive sign.
The bond which breaks energy is positive.
432+614 =1046kJ/mol
Formation of bond energy is negative
2(413) + 347 = 1173 kJ/mol
dH reaction is -1173 + 1046 =-127kJ/mol
Carbon(C):
number of moles= mass/molar mass(Mr)
=65.5/12
=5.5 moles
Hydrogen(H):
number of moles=mass/molar mass (Mr)
=5.5/1
=5.5 moles
Oxygen (O):
number of moles = mass/molar mass (Mr)
=29.0/16
=1.8 moles
EF= lowest number of moles over each of the elements
So,
C= 5.5/1.8 = 3
H= 5.5/1.8 = 3
O= 1.8/1.8 = 1
Therefore Emperical formula= C3H3O
Answer:
15.4%
Explanation:
If Ka = 0.54 mM = 1.51x10⁻⁵
Then;
C₄H₈O₂ --------> C₄H₇O₂⁻ + H⁺
I 0.54x10⁻³ 0 0
E 0.54x10⁻³(1-x) 0.54x10⁻³x 0.54x10⁻³x
Recall that x is the percentage degree of dissociation
From the ICE table;
Ka = [C₄H₇O₂⁻] [ H⁺]/[C₄H₈O₂]
1.51x10⁻⁵=(0.54x10⁻³x) (0.54x10⁻³x)/ 0.54x10⁻³(1-x)
1.51x10⁻⁵ = 0.54x10⁻³x^2/1-x
1.51x10⁻⁵(1-x) = 0.54x10⁻³x^2
1.51x10⁻⁵ - 1.51x10⁻⁵x = 0.54x10⁻³x^2
Hence;
0.54x10⁻³x^2 + 1.51x10⁻⁵x - 1.51x10⁻⁵=0
x^2 + 0.028x - 0.028 = 0
Solving the quadratic equation here;
x = 0.154 or −0.182
Ignoring the negative result, x = 0.154
Hence, fraction of butanoic acid that is in the dissociated form in this solution = 15.4%
The molecular formula of quinine is C20H<span>24N2</span>O<span>2. For every 1 mole of quinine molecule, there are 20 moles of carbon. Simply multiplying 6.0 moles by 20, we get, 120 moles.
Therefore, there are 120 moles of carbon in 6.0 moles of quinine.</span>
