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adell [148]
3 years ago
7

I have these orgo problems for homework and I have no idea where to even begin. We have to find missing reagents and products ba

sed on regioselectivity and stereoselectivity. Any help or explanations would be greatly appreciated!!

Chemistry
1 answer:
almond37 [142]3 years ago
5 0

Answer:

Honestly makes no sense sorry :(

Explanation:

I can try though.. There are three types of selectivity possible for any synthesis: (i) Chemoselectivity is deciding which group reacts. (ii) Regioselectivity is where the reaction takes place in that group. (iii) Stereoselectivity is how the group reacts with respect to the stereochemistry of the product.

A stereospecific mechanism specifies the stereochemical outcome of a given reactant, whereas a stereoselective reaction selects products from those made available by the same, non-specific mechanism acting on a given reactant. Of stereoisomeric reactants, each behaves in its own specific way.

I tried to explain it the best I could.

Hopefully this helps you :)

Feel free to correct me If it was wrong

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What causes hot air to rise, and cool air to sink?
nataly862011 [7]

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8 0
3 years ago
What is the energy change associated with 1.5 mole of D being formed? 
AleksandrR [38]

Answer:

–36 KJ.

Explanation:

The equation for the reaction is given below:

2B + C —› D + E. ΔH = – 24 KJ

From the equation above,

1 mole of D required – 24 KJ of energy.

Now, we shall determine the energy change associated with 1.5 moles of D.

This can be obtained as illustrated below:

From the equation above,

1 mole of D required – 24 KJ of energy

Therefore,

1.5 moles of D will require = 1.5 × – 24 = –36 KJ.

Therefore, –36 KJ of energy is associated with 1.5 moles of D.

3 0
3 years ago
The K a of propanoic acid ( C 2 H 5 COOH ) is 1.34 × 10 − 5 . Calculate the pH of the solution and the concentrations of C 2 H 5
Zigmanuir [339]

Answer:

2.62.

Explanation:

Okay let us first write the parameters in the question in question above out. We are given the ka value of propanoic acid, C2H5COOH to be equals to 1.34 × 10^- 5. Also, we are given the value for the initial concentration of propanoic acid to be 0.441 M.

So, let us delve right into the solution to the question and we will be starting by writting the equation below;

C2H5COOH <--------> H^+ + C2H5COO^-.

Please note that this Reaction is a reversible Reaction.

Therefore, the basic things about acid is its great tendency to release Hydrogen ion in an aqeous solution.

So, we will be taken equation above and correspond it with the time and Concentration.

C2H5COOH <----> H^+ C2H5COO^-.

Initial concentration of the C2H5COOH = 0.441 M and the initial concentration of H^+ and C2H5COO^- are both zero.

So, after a time, t, concentration of C2H5COOH= 0.441 - x and at that time the concentration of H^+ and C2H5COO^- are both x and x respectively.

Hence, Ka = [C2H5COO^-] [H^+]/ C2H5COOH. -----------------------(**).

Therefore, slotting in the values from above into equation (**), we have;

1.34 × 10^-5 = [x] [x]/ [0.441 - x].

1.34 × 10^-5= x^2/ [0.441 - x].

x^2 = 1.34 × 10^-5(0.441) - 1.34 × 10^-5x.

x^2 + 1.34 × 10^-5x - 5.91× 10^-6.

x = 2.4×10^-3.

Hence, the concentration of the propanoic acid at time, t= 0.441 - 2.4 ×10^-3.

==> 0.44 M.

pH = -log [H^+].

Then, we have; pH= - log[2.4× 10^-3].

pH= 2.62.

4 0
3 years ago
Which type of climate does Florida have due to its latitude?
nevsk [136]
The land and the way the heat changes around there
5 0
3 years ago
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