<span>Answer:
A 1.00 L solution containing 3.00x10^-4 M Cu(NO3)2 and 2.40x10^-3 M ethylenediamine (en).
contains
0.000300 moles of Cu(NO3)2 and 0.00240 moles of ethylenediamine
by the formula Cu(en)2^2+
0.000300 moles of Cu(NO3)2 reacts with twice as many moles of en = 0.000600 mol of en
so, 0.00240 moles of ethylenediamine - 0.000600 mol of en reacted = 0.00180 mol en remains
by the formula Cu(en)2^2+
0.000300 moles of Cu(NO3)2 reacts to form an equal 0.000300 moles of Cu(en)2^2+
Kf for Cu(en)2^2+ is 1x10^20.
so
1 Cu+2 & 2 en --> Cu(en)2^2+
Kf = [Cu(en)2^2+] / [Cu+2] [en]^2
1x10^20. = [0.000300] / [Cu+2] [0.00180 ]^2
[Cu+2] = [0.000300] / (1x10^20) (3.24 e-6)
Cu+2 = 9.26 e-19 Molar
since your Kf has only 1 sig fig, you might be expected to round that off to 9 X 10^-19 Molar Cu+2</span>
Answer:
pH = 2.56
Explanation:
The Henderson-Hasselbalch equation relates the pH to the Ka and ratio of the conjugate acid-base pair as follows:
pH = pKa + log([A⁻]/[HA]) = -log(Ka) + log([A⁻]/[HA])
Substituting in the value gives:
pH = -log(1.77 x 10⁻⁴) + log((0.0065M) / (0.10M))
pH = 2.56
1. In first reaction reactant a is the electron donor, while b is the electron acceptor,
the oxidized product is c while the reduced product is d
2. in the second equation e is the electron donor, f is the electron acceptor
g is the oxidized product while h is the reduced product.
3. In the third reaction i is the electron donor, j is the electron acceptor , k is the oxidized product while l is the reduced product.
I think it is aaaaaaaa I think it's aaa
The electronic geometry for the carbonate ion, in CO₃²⁻ is trigonal planar, and molecular geometry will be trigonal planar.
The electronic geometry = Total number of atoms + lone pair around the central atom
= 3 atoms + 0 lone pair
= sp₂ (trigonal planar)
The molecular geometry = As there is no lone pair its geometry will be trigonal planar.
