Color they look the same hope this helps.
Answer:
They are eukaryotic, which means they have a nucleus. Most have mitochondria
Answer:The product formed on reaction with hydroxide ion as nucleophile is 2R-hexane-2-ol.
The product formed on reaction with water would be a 50:50 mixture of
2S-hexane-2-ol. and 2R-hexane-2-ol.
Explanation:
2S-iodohexane on reactiong with hydroxide ion would undergo SN² substitution reaction that is substitution bimolecular. Hydroxide ion has a negative charge and hence it is a quite good nucleophile .
The rate of a SN² reaction depends on both the substrate and nucleophile . Here the substrate is a secondary carbon center having Iodine as a leaving group.SN² reaction takes place here as hydroxide ion is a good nucleophile and it can attack the secondary carbon center from the back side leading to the formation of 2R-hexane-2-ol.
In a SN² reaction since the the nucleophile attacks from the back-side so the product formation takes place with the inversion of configuration.
When the same substrate S-2-iodohexane undergoes a substitution reaction with water as a nucleophile then the reaction occurs through (SN¹) substitution nucleophilic unimolecular mechanism .
The rate of a SN¹ reaction depends only on the nature of substrate and is independent of the nature of nucleophile.
The SN¹ reaction is a 2 step reaction , in the first step leaving group leaves leading to the formation of a carbocation and once the carbocation is formed then any weaker nucleophile or even solvent molecules can attack leading the formation of products.
In this case a secondary carbocation would be generated in the first step and then water will attack this carbocation to form the product in the second step.
The product formed on using water as a nucleophile would be a racemic mixture of R and S isomers of hexane -2-ol in 50:50 ratio. The two products formed would be 2R-hexane-2-ol and 2S-hexane-2-ol.
Kindly refer the attachment for reaction mechanism and structure of products.
Answer:
0.73L
Explanation:
The following data were obtained from the question :
V1 = 0.65 L
P1 = 3.4 atm
T1 = 19°C = 19 + 273 = 292K
V2 =?
P2 = 3.2 atm
T2 = 36°C = 36 + 273 = 309K
The bubble's volume near the top can be obtain as follows:
P1V1 /T1 = P2V2 /T2
3.4 x 0.65/292 = 3.2 x V2 /309
Cross multiply to express in linear form as shown below:
292 x 3.2 x V2 = 3.4 x 0.65 x 309
Divide both side by 292 x 3.2
V2 = (3.4 x 0.65 x 309) /(292 x 3.2)
V2 = 0.73L
Therefore, the bubble's volume near the top is 0.73L
Answer:
% = 5.69%
Explanation:
To do this, we need to write the equations taking place here. First, this is a REDOX reaction where the hypoclorite and thiosulfate solution reacts. The balanced equations are:
ClO⁻ + 2I⁻ + 2H⁺ -------> Cl⁻ + I₂ + H₂O
I₂ + 2S₂O₃²⁻ -----------> 2I⁻ + S₄O₆²⁻
We already have the required volume and concentration of the thiosulfate solution, so we can calculate the moles of thiosulfate. With this moles, we can calculate the moles of hypochlorite, then the mass and finally the %.
The moles of thiosulfate would be:
moles S₂O₃²⁻ = V * M
moles S₂O₃²⁻ = 0.01324 * 0.0732 = 9.69x10⁻⁴ moles
Now according to the above reactions, we can see that
moles I₂ = moles ClO⁻
and
moles I₂ / moles S₂O₃²⁻ = 1/2
Therefore, let's calculate the moles of ClO⁻:
moles ClO⁻ = 9.69x10⁻⁴ / 2 = 4.845x10⁻⁴ moles
Now, we can calculate the mass of these moles, using the molar mass of sodium hypochlorite which is 74.44 g/mol:
m = 74.44 * 4.845x10⁻⁴
m = 0.036 g
Finally the % of this, in the bleach sample would be:
% = 0.036 / 0.634 * 100
<h2>
% = 5.69%</h2>