Answer:
pH = 9.74
The solution is basic
Explanation:
To find the pH of the solution, we need to find the pOH of the solution.
From the question, the concentration of OH^- = 5.5x10^-5 M
pOH = - Log[OH^-]
pOH = - Log 5.5x10^-5
pOH = 4.26
Recall,
pH + pOH = 14
pH = 14 — pOH
pH = 14 — 4.26
pH = 9.74
Since the pH is above 7, the solution is alkaline ie basic
Which element in each of the following sets would you expect to have the highest IE₂?
(a) Na, Mg, Al (b) Na, K, Fe (c) Sc, Be, Mg
Explanation:
Ionization energy is defined as the energy required to remove the most loosely bound electron from a neutral gaseous atom.
With increase in atomic size of the atom, there will be less force of attraction between the nucleus and the valence electrons of the atom. Hence, with lesser amount of energy the valence electrons can be removed easily.
Since, Na, Mg and Al are all period 3 elements. And, when we move across a period from left to right then there occurs a decrease in atomic size of the atoms. Hence, smaller is the size of an atom more energy is required to remove an electron.
Therefore, out of Na, Mg, the highest [tex]IE_{2}[/tex] will be that of Na. This is because when sodium will lose one electron then it forms [tex]Na^{+}[/tex] ion which is stable in nature.
Hence, in order to remove another electron from [tex]Na^{+}[/tex] will be difficult. Therefore, it will have high [tex]IE_{2}[/tex].
Similarly, Na will have highest [tex]IE_{2}[/tex] as compared to K and Fe. Also because sodium is smaller in size than K.
Since, beryllium is smallest in size as compared to Mg and Sc. Hence, Be will have the highest [tex]IE_{2}[/tex].
Final answer:
The elements expected to have the highest second ionization energy (IE₂) from the given sets are Mg for set (a), Fe for set (b), and Be for set (c), based on their electronic configurations and positions on the periodic table.
Explanation:
The student is asking about the second ionization energy (IE₂) for various sets of elements. Ionization energy is the energy required to remove an electron from an atom or ion. The second ionization energy specifically refers to the energy required to remove a second electron after one has already been removed. Generally, this energy is greater than the first ionization energy because the remaining electrons feel a greater effective nuclear charge.
For the sets given:
(a) Na, Mg, Al: Mg (Magnesium) expected to have the highest IE₂ because it will be removing an electron from a full s-orbital, which requires more energy.
(b) Na, K, Fe: Fe (Iron) is likely to have the highest IE₂ as it is a transition metal with more protons in the nucleus, resulting in a stronger attraction to the remaining electrons.
(c) Sc, Be, Mg: Be (Beryllium) should have the highest IE₂ because removing the second electron will remove a completely filled s-orbital, which is a stable configuration requiring more energy to disrupt.
Plasmid DNA and a gene of interest are cut with the enzyme PpuMI. Write a possible sequence of bases for the sticky end of the gene in the 5' to 3' direction.
The possible 5' to 3' sequence of bases for the sticky end of the gene cut by PpuMI is: 5'-ACGGA-3'.
The enzyme PpuMI cuts DNA at recognition sequences. In the 5' to 3' direction, "ACGGA." might be the sticky end of the gene snipped by PpuMI. This indicates that one DNA strand reads 5'-ACGGA-3' and the other 3'-TGCCT-5'.
These overhanging single-stranded ends are called "sticky ends" because they can base-pair with complementary sequences in another DNA fragment cut with the same enzyme. In genetic engineering, like cloning, DNA fragments with matching sticky ends can be joined to form recombinant DNA. PpuMI-generated sticky ends with the sequence "ACGGA" allow scientists to insert genes of interest into plasmids or other DNA molecules for study and practical applications.
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The possible sequence of bases for the sticky end of the gene in the 5' to 3' direction after being cut with PpuMI is 5'-GAAC (gene) TTGC-3'.
Explanation:The sequence of bases for the sticky end of the gene in the 5' to 3' direction after being cut with the enzyme PpuMI can be determined from the given information. The restriction enzyme PpuMI leaves a 2- to 4-nucleotide single-stranded overhang on each strand of the DNA after cutting. The sequence that is recognized by PpuMI is a palindrome, meaning it reads the same forward and backward. Therefore, the possible sequence of bases for the sticky end of the gene in the 5' to 3' direction is: 5'-GAAC (gene) TTGC-3'
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