If the half-cell containing 1.00 M Fe(NO3)2(aq) in galvanic cells Y and Z is replaced with a half-cell containing 5.00 M Fe(NO3)
1 answer:
The details of the galvanic cells are missing so i have attached it.
Answer:
The voltage will increase in cell Y and decrease in cell Z.
Explanation:
Looking at the reaction, in half cell 3, we see that reduction takes place in galvanic cell Y while oxidation takes place in galvanic cell Z.
In galvanic cell Y, aluminum gains electrons which is reduction while in galvanic cell Z, Copper loses electrons which is oxidation.
Now, electricity involves movement of electrons and where there is a loss of electrons, it means a decrease in electricity and when there is a gain in electrons, it means there is an increase in electricity.
Now, in electricity, Voltage is directly proportional to electric power.
Thus, the higher the voltage, the higher the electric power.
Thus, in galvanic cell Y that undergoes gain in electrons, there will be an increase in voltage while in galvanic cell Z that undergoes loss of electrons, there will be a decrease in voltage.
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Answer:
The ionization equation is
⇄
(1)
Explanation:
The ionization equation is
⇄
(1)
As the Bronsted definition sais, an acid is a substance with the ability to give protons thus, H2PO4 is the acid and HPO42- is the conjugate base.
The Ka expression is the ratio between the concentration of products and reactants of the equilibrium reaction so,
The pKa is
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In the body H2CO3 is a more optimal buffer for regulating pH due to the combination of the two acid-base equilibriums and the two pKa.
If the urine is acidified, according to Le Chatlier's Principle the equilibrium (1) moves to the left neutralizing the excess proton concentration.