What is a insect and name one.
2 answers:
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Answer:
1. pH = 1.23.
2.
Explanation:
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1. In this case, for the ionization of H2C2O4, we can write:
It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:
Whereas the pKa is:
The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:
2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:
It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:
Which is also shown in net ionic notation.
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Answer:
Aluminum metal
Explanation:
In order to properly answer this or a similar question, we need to know some basic rules about galvanic cells and standard reduction potentials.
First of all, your strategy would be to find a trusted source or the table of standard reduction potentials. You would then need to find the half-equations for aluminum and gold reduction:
Since we have a galvanic cell, the overall reaction is spontaneous. A spontaneous reaction indicates that the overall cell potential should be positive.
Since one half-equation should be an oxidation reaction (oxidation is loss of electrons) and one should be a reduction reaction (reduction is gain of electrons), one of these should be reversed.
Thinking simply, if the overall cell potential would be obtained by adding the two potentials, in order to acquite a positive number in the sum of potentials, we may only reverse the half-equation of aluminum (this would change the sign of E to positive):
Notice that the overall cell potential upon summing is:
Meaning we obey the law of galvanic cells.
Since oxidation is loss of electrons, notice that the loss of electrons takes place in the half-equation of aluminum: solid aluminum electrode loses 3 electrons to become aluminum cation.