All categories

3 years ago

What is the hybridization around the O atom?

1 answer:

7 0

Your answer is......... with little bit explains....and read correctly.......The nitrogen is sp3 hybridized which means that it has four sp3 hybrid orbitals. Two of the sp3 hybridized orbitals overlap with s orbitals from hydrogens to form the two N-H sigma bonds. One of the sp3 hybridized orbitals overlap with an sp3 hybridized orbital from carbon to form the C-N sigma bond. The lone pair electrons on the nitrogen are contained in the last sp3 hybridized orbital. Due to the sp3 hybridization the nitrogen has a tetrahedral geometry. However, the H-N-H and H-N-C bonds angles are less than the typical 109.5o due to compression by the lone pair electrons.

You might be interested in

Which diagram represents an object at static equilibrium?

LuckyWell [14K]

Answer:

it is not D!! i took the test and it was wrong on plato

Explanation:

6 0

3 years ago

Under what conditions can a non-spontaneous reaction be made spontaneous?

Kitty [74]

It can be c I think tho

7 0

3 years ago

Read 2 more answers

How many carbons are present in the hydrocarbon 3-ethyl-2,3,4-trimethylpentane

fiasKO [112]

I would say there is 10 (2 in the ethyl group, 3 in the three methyl groups and 5 in the pent chain)

4 0

3 years ago

The molar solubilities of the following compounds (in mol/L) are:

IceJOKER [234]

Answer: The decreasing order of $K_{sp}$ is $AgSCN>AgBr>AgCN$

Explanation:

For AgBr:

The balanced equilibrium reaction for the ionization of silver bromide follows:

$AgBr\rightleftharpoons Ag^{+}+Br^-$

s s

The expression for solubility constant for this reaction will be:

$K_{sp}=[Ag^{+}][Br^-]$

We are given:

$s=7.3\times 10^{-7}mol/L$

Putting values in above equation, we get:

$K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(7.3\times 10{-7})^2=5.33\times 10^{-13}$

Solubility product of AgBr = $5.33\times 10^{-13}$

For AgCN:

The balanced equilibrium reaction for the ionization of silver cyanide follows:

$AgCN\rightleftharpoons Ag^{+}+CN^-$

s s

The expression for solubility constant for this reaction will be:

$K_{sp}=[Ag^{+}][CN^-]$

We are given:

$s=7.7\times 10^{-9}mol/L$

Putting values in above equation, we get:

$K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(7.7\times 10{-9})^2=5.93\times 10^{-17}$

Solubility product of AgCN = $5.33\times 10^{-17}$

For AgSCN:

The balanced equilibrium reaction for the ionization of silver thiocyanate follows:

$AgSCN\rightleftharpoons Ag^{+}+SCN^-$

s s

The expression for solubility constant for this reaction will be:

$K_{sp}=[Ag^{+}][SCN^-]$

We are given:

$s=1.0\times 10^{-6}mol/L$

Putting values in above equation, we get:

$K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(1.0\times 10{-6})^2=1.0\times 10^{-12}$

Solubility product of AgSCN = $1.0\times 10^{-12}$

The decreasing order of $K_{sp}$ follows:

$AgSCN>AgBr>AgCN$

4 0

4 years ago

What new scientific evidence led to changes in Dalton's atomic theory?

Masja [62]

Drawbacks of Dalton's Atomic Theory. The indivisibility of an atom was proved wrong: an atom can be further subdivided into protons, neutrons and electrons. However an atom is the smallest particle that takes part in chemical reactions. According to Dalton, the atoms of same element are similar in all respects.

6 0

3 years ago