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3 years ago

Which of the following additions to alkenes occur(s) specifically in an syn fashion?

Chemistry

1 answer:

BlackZzzverrR [31]3 years ago

8 0

Answer:

E) A, B, and C

Explanation:

Syn addition refers to the addition of two substituents on the same face or side of a double bond. This differed from anti addition which a occurs across opposite face of the double bond.

Hydrogenation, hydroboration and dihydroxylation all involve syn addition to the double bond, hence the answer chosen above.

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Ranging from 0 to 14, a pH value indicates how acidic or basic a solution is. Which of these pH values would be a strong acid? *

Eduardwww [97]

Answer:

13 aaaàaàaaaaaaaaaßaaaaasssss

4 0

3 years ago

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The reaction of sodium peroxide and water produces sodium hydroxide and oxygen gas. The following balanced chemical equation rep

a_sh-v [17]

Answer:

3.925 mol.

Explanation:

From the balanced equation:

2 Na₂O₂(s) + 2 H₂O(l) → 4 NaOH(s) + O₂(g) ,

It is clear that 2 moles of Na₂O₂ react with 2 moles of H₂O to produce 4 moles of NaOH and 1 mole of O₂ .

Using cross multiplication:

4 moles of NaOH produced with → 1 mole of O₂ .

15.7 moles of NaOH produced with → ??? mole of O₂ .

∴ The no. of moles of O₂ made = (1 mole)(15.7 mole)/(4 mole) = 3.925 mol.

8 0

3 years ago

An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni

Svetllana [295]

Answer: The pH of acid solution is 4.58

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For KOH:

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

$1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol$

For propanoic acid:

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

$0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol$

The chemical reaction for propanoic acid and KOH follows the equation:

$C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O$

Initial: 0.1372 0.04514

Final: 0.09206 - 0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})$

$pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of propanoic acid = 4.89

$[C_2H_5COOK]=\frac{0.04514}{0.26594}$

$[C_2H_5COOH]=\frac{0.09206}{0.26594}$

pH = ?

Putting values in above equation, we get:

$pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58$

Hence, the pH of acid solution is 4.58

7 0

3 years ago

The specific heats at constant pressure of some common gases are provided as a thirdorder polynomial: �;<<< = � + �� +

3241004551 [841]

Answer:

1.991 kJ

Explanation:

Calculate the amount of heat ( J )

CH4 ;

coefficients are : a = 19.89 , b = 5.02 * 10^-2 , c = 1.269 * 10^-5 , d = -11.01 * 10^-9

attached below is the detailed solution

3 0

3 years ago

Are your knees the elbows of your legs or are your elbows the knees of your

Nimfa-mama [501]

Answer:

Explanation:

it is confusing lol

but i think that it is C.both

hope it helps

have a nice day

5 0

3 years ago

Read 2 more answers