An acid with molar mass 84.48 g/mol is titrated with 0.650 M KOH. What volume of KOH solution is needed to titrate 1.70 grams of
1 answer:
Answer:
Explanation:
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In this case, since the acid is monoprotic and the KOH has one hydroxyl ion only, we can see that at the equivalence point the moles of both of them are the same:
Thus, since we are given 1.70 g of the acid, we compute the moles of acid that were titrated:
Which equal the moles of KOH. In such a way, since the molarity is defined as moles over liters (M=n/V), the liters are moles over molarity (V=n/M), thus, the resulting volume is:
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I used the genetic code table. mRNA codon ===> amino acid
1st base 2nd base 3rd base
A U U ===> Isoleucine
A U C ===> Isoleucine
The point mutation of codon AUU to AUC is a neutral mutation because it neither benefits nor deter the ability of the organism to survive and reproduce.
As you can see, Both codons result to the Isoleucine amino acid.
Another codon that will still result to the Isoleucin amino acid is AUA.
1st base 2nd base 3rd base
A U U ===> Isoleucine
A U C ===> Isoleucine
The point mutation of codon AUU to AUC is a neutral mutation because it neither benefits nor deter the ability of the organism to survive and reproduce.
As you can see, Both codons result to the Isoleucine amino acid.
Another codon that will still result to the Isoleucin amino acid is AUA.