Answers
Answer:
Explanation:
In the picture you have the answer.
Now, let's analize the structure, so you can know why the structure in the picture is the correct structure.
The aniline is the name that receives the benzene with a NH2 group as one of it's substituent. Now, This group is a really strong activating group and in the nomenclature priority, it has more order priority than any halide.
Now, it says that the chloro it's on the para position. The "para" position in a aromatic ring, in this case, the benzene, refers to the position of this substituent to the first substitued position. In this case, the NH2 it's on the position 1 or carbon 1, the para position, means that it's on position 4 of the ring. The ortho position is carbon 2, and meta position is carbon 3 of the benzene. So, according to this, the p-chloroaniline it's on picture attached.
Related Questions
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
Answers
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.
Indicate whether the substance exists in aqueous solution (a) entirely in molecular form, (b) entirely as ions, (c) or as a mixture of molecules and ions. HF, CH3CN, NaClO4, Ba(OH)2
Answers
Answer:
In aqueous solution, HF is an acid, exist as a mixture of molecules and ions
CH3CN is none of the above, entirely in molecular form
NaCIO4 is a salt, entirely as ions
Ba(OH)2 is a base, entirely as ions
The substance exists in aqueous solution entirely in molecular form is Methyl cyanide, entirely as ions is NaClO₄ & Ba(OH)₂ and HF as a mixture of molecules and ions.
What is aqueous solution?Aqueous solutions are those solution in which water is present as a solvent and we know that nature of water is polar in water means it consist positive or negative charges.
HF is a weak acid and it shows partial dissociation only, so it is present in the mixture form of molecules and ions.Methyl cyanide is a covalent molecule not dissolve in aqueous solution and present in the molecular form.NaClO₄ is a salt and it is completely dissolve in aqueous solution and present in the ions form.Ba(OH)₂ is a strong base and it is also completely dissociates into their ions.Hence NaClO₄ & Ba(OH)₂ is completely dissociates into ions.
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Which functional group does not contain an oxygen atom? Group of answer choices amine amide aldehyde ketone ether ester alcohol carboxylic acid
Answers
Answer: Amine does not contain an oxygen atom
Explanation:
amine is represented by -NH2
amide is represented by -CONH2
aldehyde is represented by -CHO
ketone is represented by -CO
ether is represented by -CO
ester is a represented by -COOR
alcohol is represented by -OH
carboxylic acid is represented by -COOH
You will observe that the only functional group without an oxygen atom is amine with -NH2
Amines are the functional group that does not contain an oxygen atom, distinguishing them from aldehydes, ketones, carboxylic acids, esters, ethers, alcohols, and amides. Option A is correct.
The functional group that does not contain an oxygen atom among the listed options is the amine group. Functional groups like aldehydes, ketones, carboxylic acids, esters, ethers, alcohol, and amide are characterized by the presence of oxygen in their structures.
Notably, the carbonyl group common to aldehydes, ketones, carboxylic acids, and esters features a carbon-oxygen double bond. Amines, on the other hand, are compounds containing nitrogen atoms bonded to alkyl or aryl groups, and they do not incorporate oxygen within their functional group.
Hence, A. is the correct option.
The complete question is:
Which functional group does not contain an oxygen atom? Group of answer choices
A) amine
B) amide
C) aldehyde
D) ketone
E) ether
F) ester
G) alcohol
H) carboxylic acid
(a) Write the rate law for the reaction 2A + B → C if the reaction (1) is second order in B and overall third order, –rA = ______ (2) is zero order in A and first order in B, –rA = ______ (3) is zero order in both A and B, –rA = ______ (4) is first order in A and overall zero order. –rA = ______ (b) Find and write the rate laws for the following reactions (1) H2 + Br2 → 2HBr (2) H2 + I2 → 2HI
Answers
Answer:
A. Write the rate law for this reaction
2A + B -----> C
Rate law is expressed as
-rA = kaCa^αCb^β
n = α + β
Where, ka is the rate constant;
Ca is the concentration of reactant A, Cb is the conversation of reactant B, α is the rate order for reactant A, β is the rate order for reactant B, n is the overall order.
1. β = 2, n = 3; α = 3-2 = 1
Rate law is;
-rA = ka[A][B]^2
2. α = 0, β = 1
-rA = ka[A]^0[B]
-rA = ka[B]
3. α=β=0
-rA = ka[A]^0[B]^0
-rA = ka
4. α=1, n = 0
The reaction is zero order as it is independent of the reactants.
-rA = ka
B. Rate law for the following;
a. H2+ Br2 ------>2HBr
-rA1 = ka[H2] . [Br2]
-rA2 = kb[HBr]^2
Comparing both rate, rA1= rA2
ka.[H2][Br2] = kb[HBr]^2
ka/kb = K = [HBr]^2 / [H2] [Br2]
b. H2 + I2 ------>2HI
K = [HI]^2 / [H2] [I2]
The rate laws for different reaction scenarios and specific reactions.
Explanation:(a) The rate law for the reaction 2A + B → C can be determined by examining the reaction orders for each reactant. For the given scenarios:
When the reaction is second order in B and overall third order, the rate law is -rA = k[A]2[B].When the reaction is zero order in A and first order in B, the rate law is -rA = k[B].When the reaction is zero order in both A and B, the rate law is -rA = k.When the reaction is first order in A and overall zero order, the rate law is -rA = k[A].(b) The rate laws for the given reactions are:
The rate law for the reaction H2 + Br2 → 2HBr is -rH2 = k[H2][Br2].The rate law for the reaction H2 + I2 → 2HI is -rH2 = k[H2][I2].Learn more about Rate laws here:https://brainly.com/question/35884538
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Calculate the concentration of hydronium in a solution that contains 5.5x10-5 M OH- at 25C. Indentify the solution as acidic, basic, or neutral.
Answers
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
What does ""electron density in a particular tiny volume of space"" mean?
Answers
Explanation:
Electron Density is nothing but finding of some electron in a tiny space of per unit volume. It is rather a probability of finding electron in per unit volume, it should be called electron probability density. And there is function Ψ^2 that represent probability of finding electron in some tiny volume of of the atom.
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.
Answers
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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