How are radar and an approaching siren different? How are they similar?

Methionine, ch3sch2ch2ch(nh2)co2h, is an amino acid found in proteins. what is the hybridization type of each carbon, oxygen, th

Lady_Fox [76]

Answer 1) To find the hybridization of carbon we need to draw the lewis structure first; which is attached in the answer.

So the formula for finding hybridization is
Lone pairs + Bonding pairs = hybridisation.

Here, in the molecule of methionine Carbon is bonded to all 4 atoms and has got no lone pairs so we will have;

0 + 4 = 4 which means it is $sp^{3}$ hybridized.

Answer 2) For oxygen we can see in the lewis structure that it is having 2 lone pairs and double bonded with carbon atom, also there is another oxygen atom with two lone pair but bonded with carbon and hydrogen at the end.

So, in both the cases we can calculate;

where 2 lone pairs are present
2 + 2 = 4 so, it will be $sp^{3}$ hybridized.

So, Oxygen is also $sp^{3}$ hybridized.

Answer 3) Nitrogen present in the methionine is bonded to carbon atom and hydrogen atoms and has only one lone pair on it.

So when we are substituting this in the formula we get 1 + 3 = 4;

which means 1 lone pair and 3 bond pairs so, it is also $sp^{3}$ hybridized.

Answer 4) Sulfur atom in methionine is bonded to two carbon atoms forming a linear structure with 2 lone pairs on it.

So, it has 2 lone pairs and 2 bond pairs,

hence the hybridization will be   $sp^{3}$ hybridized.

As, 2 + 2 = 4 which means it is   $sp^{3}$ hybridized.

5 0

3 years ago

Aqueous hydrochloric acid HCl reacts with solid sodium hydroxide NaOH to produce aqueous sodium chloride NaCl and liquid water H

allochka39001 [22]

Answer : The theoretical yield of water formed from the reaction is 0.54 grams.

Solution : Given,

Mass of HCl = 1.1 g

Mass of NaOH = 2.1 g

Molar mass of HCl = 36.5 g/mole

Molar mass of NaOH = 40 g/mole

Molar mass of $H_2O$ = 18 g/mole

First we have to calculate the moles of HCl and NaOH.

$\text{ Moles of }HCl=\frac{\text{ Mass of }HCl}{\text{ Molar mass of }HCl}=\frac{1.1g}{36.5g/mole}=0.030moles$

$\text{ Moles of }NaOH=\frac{\text{ Mass of }NaOH}{\text{ Molar mass of }NaOH}=\frac{2.1g}{40g/mole}=0.525moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that

As, 1 mole of HCl react with 1 mole of NaOH

So, 0.030 mole of HCl react with 0.030 mole of NaOH

From this we conclude that, $NaOH$ is an excess reagent because the given moles are greater than the required moles and $HCl$ is a limiting reagent and it limits the formation of product.