Answer:
10
Explanation:
The unbalanced combustion reaction is shown below as:-
On the left hand side,
There are 3 hydrogen atoms and 1 nitrogen atom and 2 oxygen atoms
On the right hand side,
There are 1 nitrogen atom and 2 hydrogen atoms and 2 oxygen atoms
Thus,
Right side, 
must be multiplied by 6 to balance hydrogen.
Left side, 
must be multiplied by 4 to balance hydrogen.
Also, Right side, 
is multiplied by 4 so to balance nitrogen.
Left side, 
must be multiplied by 5 to balance the whole reaction.
Thus, the balanced reaction is:-
Sum of Coefficient of product - 4 + 6 = 10
The balanced chemical equation for the <span>combustion of ethane is
2C</span>₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(l)
The stoichiometric ratio between C₂H₆(g) and CO₂(g) is 1 : 2
Hence,
moles of CO₂(g) produced = moles of reacted C₂H₆(g) x 2
= 1.00 mol x 2
= 2.00 mol
Hence, the correct answer is "C".
The correct answer is:
a positron is emitted when proton converts to a neutron.
The reaction can be described as following:
₁¹p (proton) → ₀¹n (neutron) + ₁°e (positron or ₁⁰β)
Positron is an antiparticle of a β particle (₋₁°β), which means it has an oposite charge to it, but same mass.
Try looking through this, don't actually read it if you don't want to
but if you skim through it you might find answers. Good Luck! :) ..Reactions of Aldehydes, Ketones And Phenols
Objective:
1. To carry out some simple chemicals test in order to distinguish
between aldehydes, ketones and phenols
2. To study the properties of aldehydes, ketones and phenols.
3. To identify the unknowns A, B, C, D and E.
Introduction:
Part I : Reaction of Aldehydes and Ketones
The carbonyl group is C=O and any compound containing this group that
can be described as a carbonyl compound. Carbonyl compounds fall into
two main classes: aldehydes and ketones on the one hand and carboxylic
acids and their derivatives on the other hand. The characteristic
reactions of the aldehydes and ketones are addition and oxidation
reactions occurring at the unsaturated carbonyl group. With the same
reagent, aldehydes usually react faster than ketones, mainly because
there is lees crowding at the carbonyl carbon and the steric effect.
Aldehydes are also more easily oxidized than ketones. The carbonyl and
other compounds investigated in this experiment are tested in each of
the following ways:
A) Chromic Acid (H2CrO4)
Chromic acid is a strong oxidant. Aldehydes are oxidized to carboxylic
acids by chromic acid. The Cr6+ in the chromic acid which is orange,
then is reduced to Cr3+ which is green/blue. Ketones are not oxidized by
chromic acid.image B) Tollen’s Test (Ag(NH3)2+ / OH-
Tollen’s reagent (Ag(NH3)2+ / OH- is a weak oxidant. Aldehydes are
readily oxidized to carboxylic acids by Tollen’s reagent to produce a
silver mirror on the inside of a clean test tube. Ketones are not
oxidized by Tollen’s reagent.image
C) Fehling’s solution
Fehling’s solution is an oxidizing agent. It is prepared by mixing equal
part of Fehling’s solution I (copper(II) sulfate) and Fehling’s
solution II (sodium potassium tartate and sodium hydroxide). Aldehydes
are easily oxidized to carboxylic acid by Fehling’s solution and will
reduce the cupric ion which complexed with tartate ion to cuprous oxide.
A positive result is indicated by the formation of a brick red
precipitate. Ketones are not oxidized by Fehling’s solution.image
D) 2,4-dinitrophenylhydrazine (DNP Test or Brady’s Reagent)
2,4-dinitrophenylhydrazine (Brady’s reagent) is an important reagent
related to hydrazine. Most aldehydes and ketones very readily with this
reagent to give the yellow orange and red precipitates of
2,4-dinitrophenylhydrazones. Unconjugated aldehydes and ketones give
precipitates toward the yellow while conjugated compound tend to be
deeper colour of red. The conversion of aldehydes and ketones into
hydrazone is an example of the addition-elimination reaction occurring
at the unsaturated carbonyl group.image
E) Iodoform Test
Iodoform test can be used for the detection of acetalaldehyde and all
methyl ketone which have the formula:image
Iodoform, CHI3 is a yellow solid with a strong medicinal smell. Iodoform
will precipitate out of a mixture of methyl ketone, iodine and
base.image For acetaldehyde, the following reaction shows the formation
of iodoform:image
Compounds that are easily oxidized to acetaldehyde and methyl ketones
also give a positive iodoform test. Only ethanol can be oxidized to
acetaldehyde and secondary alcohol that have the general formula
CH3CH(OH)R can be oxidized to methyl ketones.
Part II: Reactions
