<span>Response:
A 1.00 L solution that includes 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.
Using the formula Cu(en)2^2+
0.000300 moles of Cu(NO3)2 reacts with double that amount of en = 0.000600 mol of en.
Thus, 0.00240 moles of ethylenediamine - 0.000600 mol of en reacted leaves 0.00180 mol en unreacted.
According to the formula Cu(en)2^2+
0.000300 moles of Cu(NO3)2 reacts to yield an equivalent of 0.000300 moles of Cu(en)2^2+
The formation constant Kf for Cu(en)2^2+ is 1x10^20.
Therefore,
1 Cu+2 and 2 en --> Cu(en)2^2+
Kf = [Cu(en)2^2+] / [Cu+2] [en]^2
1x10^20 = [0.000300] / [Cu+2] [0.00180 ]^2
Solving for [Cu+2] gives [Cu+2] = [0.000300] / (1x10^20) (3.24 e-6)
Thus, Cu+2 = 9.26 e-19 Molar.
Since Kf only has 1 significant figure, round that to 9 X 10^-19 Molar Cu+2.</span>
The aqueous solution of Na3PO4 is described as "strongly basic."
The atomic number corresponds to the number of protons
Protons are denoted as P and Electrons as E P = E
The atomic mass equals the sum of Neutrons and Protons
Atomic number = atomic mass = neutrons
P = E
Atomic mass - atomic number = Neutrons
Example:
Calcium consists of 20 Protons 20P = 20E
Atomic mass - atomic number = neutron count:)
Hi,
Due to calcium hydroxide being a strong base, its full dissociation will yield both calcium and hydroxyl ions:
Thus, the concentration of hydroxyl ions mirrors that of the calcium hydroxide, allowing for the calculation of pOH as demonstrated below:
Now, pH relates to pOH as:
Consequently, the final pH is achieved.
Best regards.
Response:
Clarification:
Hello,
In this scenario, since a single drop equates to 0.05 mL of the solution provided, with a concentration of 0.02 g/mL, the mass of oleic acid in one drop calculates to:
Best wishes.
