There are 2 peaks are in the proton spin decoupled 13C NMR spectrum of 1,3,5-trinitrobenzene .
The 13C NMR spectrum give the peak which is directly about the carbon skeleton not just the proton attached to it . The number of signals tell us how many different carbons or set of equivalent carbons . The splitting of a signal tells us how many hydrogens are attached to each carbon.
In 1,3,5-trinitrobenzene molecule , there are 2 peaks are in the proton spin decoupled 13C NMR spectrum three carbon give one signal and another three carbon give another one signal .
To learn more about NMR spectrum
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Answer:
- Sn²⁺ ⇒ 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰
- Ti⁺ ⇒ 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 4f¹⁴ 6s² 5d¹⁰
- As⁺³ ⇒ 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
Explanation:
The <em>electron configuration</em> indicates the way the electrons of an atom or ion are structured.<u> In the case of cations</u>, by knowing the electronic configuration of the atom (which is neutral), we can find out the cations' configuration by substracting <em>n</em> outermost electrons, where <em>n</em> is the charge of the cation.
Mg⁰ ⇒ [Ne] 3s² = 1s² 2s² 2p⁶ 3s². Thus
Mg⁺² ⇒ [Ne] = 1s² 2s² 2p⁶.
In a similar fashion, the answers are:
Sn²⁺ ⇒ 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰
K⁺ ⇒ 1s² 2s² 2p⁶ 3s² 3p⁶
Al³⁺ ⇒ 1s² 2s² 2p⁶
Ti⁺ ⇒ 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 4f¹⁴ 6s² 5d¹⁰
As⁺³ ⇒ 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
Correct answer is 
.
Phosphoric acid is a polyprotic acid having 3 acidic hyrdogen therefore it will have 3 pka values.
The equations for the release of acidic hydrogen can be written as:
From the pka values we can judge the idea of pH as using Henderson-Hasselbalch Equation, we get the relation between pH and pka.
Using the following equation, relation of pH and pka is
![pH=pka+log\frac{[A^-]}{HA}](https://tex.z-dn.net/?f=pH%3Dpka%2Blog%5Cfrac%7B%5BA%5E-%5D%7D%7BHA%7D)
Using this equation, we can find that the equation having pka= 2. 14 is closest to the pH=3.2 so the ionic form in this equation will be dominant at the same pH.
Therefore at pH=3.2 the ionic form of 
is dominant.
Answer:
(a) 1s2 2s1
Explanation:
Electron configurations of atoms are in their ground state when the electrons completely fill each orbital before starting to fill the next orbital.
<h3><u>
Understanding the notation</u></h3>
It's important to know how to read and interpret the notation.
For example, the first part of option (a) says "1s2"
- The "1" means the first level or shell
- The "s" means in an s-orbital
- The "2" means there are 2 electrons in that orbital
<h3><u>
</u></h3><h3><u>
Other things to know about electron orbitals</u></h3>
It important to know which orbitals are in each shell:
- In level 1, there is only an s-orbital
- In level 2, there is an s-orbital and a p-orbital
- in level 3, there is an s-orbital, a p-orbital, and a d-orbital <em>(things get a little tricky when the d-orbitals get involved, but this problem is checking on the basic concept -- not the higher level trickery)</em>
So, it's also important to know how many electrons can be in each orbital in order to know if they are full or not. The electrons should fill up these orbitals for each level, in this order:
- s-orbitals can hold 2
- p-orbitals can hold 6
- d-orbitals can hold 10 <em>(but again, that's beyond the scope of this problem)</em>
<h3><u>
Examining how the electrons are filling the orbitals</u></h3>
<u>For option (a):</u>
- the 1s orbital is filled with 2, and
- the 2s orbital has a single electron in it with no other orbitals involved.
This is in it's ground state.
<u>For option (b):</u>
- the 1s orbital is filled with 2,
- the 2s orbital is filled with 2,
- the 2p orbital has 5 (short of a full 6), and
- the 3s orbital has a single electron in it.
Because the 3s orbital has an electron, but the lower 2p before it isn't full. This is NOT in it's ground state.
<u>For option (c):</u>
- the 1s orbital is filled with 2,
- the 2s orbital has 1 (short of a full 2), and
- the 2p orbital is filled with 6
Although the 2p orbital is full, since the 2s orbital before it was not yet full, this is NOT in it's ground state.
<u>For option (d):</u>
- the 1s orbital has 1 (short of a full 2), and
- the 2s orbital is filled with 2
Again, despite that the final orbital (in this case, the 2s orbital), is full, since the 1s orbital before it was not yet full, this is NOT in it's ground state.
80% carbon and 20.% is hydrogen
