Nitrogen ion (Nitride) is ... Cation/Anion/Neither? # of protons? # of electrons? Charge (1-, 2-, 3-, 1+, 2+, 3+, or 0) Number o
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The question is incomplete, the complete question is:
Nitrogen and fluorine react to form nitrogen fluoride according to the chemical equation:
A sample contains 19.3 g of is reacted with 19.3 g of
. Now we need to find the amount of
that can be formed by the complete reactions of each of the reactants.
If all of the was used up in the reaction, how many moles of
would be produced?
<u>Answer:</u> 1.378 moles of are produced in the reaction.
<u>Explanation:</u>
The number of moles is defined as the ratio of the mass of a substance to its molar mass.
......(1)
Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.
Excess reagent is defined as the reagent which is left behind after the completion of the reaction.
In the given chemical reaction, is considered as a limiting reagent because it limits the formation of the product and it was completely consumed in the reaction.
We are given:
Mass of = 19.3 g
Molar mass of = 28.02 g/mol
Putting values in equation 1:
For the given chemical reaction:
By the stoichiometry of the reaction:
1 mole of produces 2 moles of
So, 0.689 moles of will produce =
of
Hence, 1.378 moles of are produced in the reaction.
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
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 hybridized.
So, Oxygen is also 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 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 hybridized.
As, 2 + 2 = 4 which means it is hybridized.
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
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
So, Oxygen is also
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
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
As, 2 + 2 = 4 which means it is