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kykrilka [37]
3 years ago
6

Consider the structure of the amino acid alanine. indicate the hybridization about each interior atom.

Chemistry
2 answers:
OLEGan [10]3 years ago
5 0
The structure of Alanine is shown below,
Except the carbon atom of carbonyl group which is Sp² Hybridized, all remaining atoms are Sp³ Hybridized. The hybridization of each element is depicted in a box below,

Brilliant_brown [7]3 years ago
5 0

Answer:

Alanine is composed of three carbon atoms , two oxygen atoms ,one nitrogen atom and seven hydrogen atoms with chemical formula of   C_3H_7NO_2 , with structure shown in image attached.

C-1 atom is sp^3 hybridized as it can form  four single bonds to give four hybridized orbitals.

C-2 atom is sp^3 hybridized as it can form  four single bonds to give four hybridized orbitals.

C-3 atom is sp^2 hybridized as it can form two single bonds to give three hybridized orbitals .

N atom is sp^3 hybridized as it can form three single bonds to give four hybridized orbitals out of which one hybridized orbitals is occupied by the lone pair of the nitrogen atom.

O-1 atom is sp^2 hybridized which means there are three hybridized orbitals out of which two of the hybridized orbitals are occupied by the two lone pairs of the oxygen atom.

O-2 atom is sp^3 hybridized as it form two single bonds to give four hybridized orbitals out of which two of the hybridized orbitals are occupied by the two lone pairs of the oxygen atom .

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Explanation:

( a )

<u>The four types of  spread spectrum techniques are as follows -</u>

1. Direct sequence spread spectrum .

2. frequency hopping spread spectrum .

3. chirp spread spectrum .

4. time hopping spread spectrum .

( b )

<u>The Direct sequence spread spectrum was devised for eavesdropping in the military .</u>

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3 years ago
Given : 4.50 molte
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Which physical method can be used for obtaining a sample of salt from a sm
lana [24]

The physical method that can be used for obtaining a sample of salt from a small beaker of salt and water would be evaporation.

<h3>Separation of salt and water</h3>

A mixture of salt and water can be separated by a method known as evaporation. This is based on the assumption that the salt in question is a water-soluble salt.

In order to separate the salt/water mixture:

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  • Boil the mixture until all the water evaporates.
  • The remaining residue would be the salt

Care should be taken not to overheat the residue in order to avoid melting. Evaporation is generally used to separate a mixture of water and soluble salt. If the salt is insoluble, filtration using a suitable filter paper will filter off the salt while the water is collected as the filtrate.

More on evaporation can be found here: brainly.com/question/1097783

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Which physical method can be used for obtaining a sample of salt from a small beaker of salt water?

6 0
1 year ago
Use the given data at 500 K to calculate ΔG°for the reaction
Anton [14]

Answer : The  value of \Delta G^o for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0 = standard enthalpy of formation

Now put all the given values in this expression, we get:

\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]

\Delta H^o=-1035.6kJ=-1035600J

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

Now put all the given values in this expression, we get:

\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]

\Delta S^o=-153J/K

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 500 K.

\Delta G^o=(-1035600J)-(500K\times -153J/K)

\Delta G^o=-959100J=-959.1kJ

Therefore, the value of \Delta G^o for the reaction is -959.1 kJ

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