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Kisachek [45]
3 years ago
15

Complete the mechanism for the acid-catalyzed hydrolysis of the epoxide in alcohol by adding any missing atoms, bonds, charges,

nonbonding electrons, and curved arrows.Figure:Different bonds in the figure

Chemistry
1 answer:
Sauron [17]3 years ago
4 0

Answer:

See explanation below

Explanation:

You are missing the structure, therefore, I will do an example with one that I found on another place to try to explain.

This acid mechanism always involves carbocations, and positive charges, never negative because we are in acidic mediums.

In the first step, the lone pairs of the oxigen from the epoxide, substract one hydrogen of the reactant.

Second step, the lone pairs of the oxygen from the reactant, do a nucleophylic attack to the carbon of the epoxide. In this case, it will do it to the most substitued carbon.

Then, in the third step by acid base equilibrium, the hydrogen from the reactant that attacked, is substracted from the molecule by a molecule of water (We are in acid medium, therefore, there is traces of water) and the final structure is formed.

Check picture for mechanism:

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skad [1K]
Inches because that is a trait of measurement
5 0
3 years ago
Hydrazine (N2H4), a rocket fuel , reacts with oxygen to form nitrogen gas and water vapor. The reaction is represented with the
lilavasa [31]

Answer:

D)  8.40 L H₂O(g).

Explanation:

  • The balanced equation for the mentioned reaction is:

<em>2N₂H₄(l) + O₂(g) → N₂(g) + 2H₂O(g), </em>

It is clear that 2.0 moles of N₂H₄ react with 1.0 mole of O₂ to produce 2.0 moles of N₂ and 2.0 moles of H₂O.

  • At STP, 4.20L of O₂ reacts with N₂H₄:

It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.

<u>using cross multiplication: </u>

1.0 mol of O₂ represents → 22.4 L.

??? mol of O₂ represents → 4.2 L.

∴ 4.2 L of O₂ represents = (1.0 mol)(4.2 L)/(22.4 L) = 0.1875 mol.

  • To find the no. of moles of H₂O produced:

Using cross multiplication:

1.0 mol of O₂ produce → 2.0 mol of H₂O, from stichiometry.

0.1875 mol of O₂ produce → ??? mol of H₂O.

∴ The no. of moles of H₂O = (2.0 mol)(0.1875 mol)/(1.0 mol) = 3.75 mol.

  • Again, using cross multiplication:

1.0 mol of H₂O represents → 22.4 L, at STP.

3.75 mol of H₂O represents → ??? L.

<em>∴ The no. of liters of water vapor will be produced </em>= (3.75 mol)(22.4 L)/(1.0 mol) = <em>8.4 L.</em>

<em></em>

<em>So, the right choice is: D)  8.40 L H₂O(g).</em>

<em></em>

3 0
3 years ago
What is the oh- in a solution with a poh of 5.71
Rudik [331]

Answer:- The hydroxide ion concentration of the solution is 1.95*10^-^6 .

Solution:- The formula used to calculate pOH from hydroxide ion is:

pOH=-log[OH^-]

When pOH is given and we are asked to calculate hydroxide ion concentration then we multiply both sides by negative sign and take antilog and what we get on doing this is:

[OH^-]=10^-^p^O^H

pOH is given as 5.71 and we are asked to calculate hydrogen ion concentration. Let's plug in the given value in the formula:

[OH^-]=10^-^5^.^7^1

[OH^-] = 0.00000195 or 1.95*10^-^6

So, the hydroxide ion concentration of the solution is 1.95*10^-^6 .



3 0
3 years ago
What is the pressure of a sample of gas at a volume of .335 L if it occupies 1700 mL at 850 mm Hg?
dangina [55]

Answer:

Pressure = 4313.43mmHg

Explanation:

P1 = ?

V1 = 0.335L

V2 = 1700mL =1700*10^-3L = 1.7L

P2 = 850mmhg

From Boyle's law, the volume of a fixed mass of gas is inversely proportional to its pressure provided that temperature remains constant.

P = k / v

K = pv. P1V1 = P2V2 = P3V3 =........=PnVn

P1V1 = P2V2

Solve for P1,

P1 = (P2*V2) / V1

P1 = (850 * 1.7) / 0.335

P1 = 4313.43mmHg

The pressure of the gas was 4313.43mmHg

7 0
3 years ago
A 0.75M solution of CH3OH is prepared in 0.500 kg of water. How many moles of CH3OH are needed?
4vir4ik [10]

Answer:

We need 0.375 mol of CH3OH to prepare the solution

Explanation:

For the problem they give us the following data:

Solution concentration 0,75 M

Mass of Solvent is 0,5Kg

knowing that the density of water is 1g / mL,  we find the volume of water:

                           d = \frac{g}{mL} \\\\ V= \frac{g}{d}  = \frac{500g}{1 \frac{g}{mL} } = 500mL = 0,5 L

Now, find moles of CH_{3} OH are needed using the molarity equation:

                           M = \frac{ moles }{ V (L)} \\\\\\molesCH_{3}OH  = M . V(L) = 0,75 M . 0,5 L\\\\molesCH_{3}OH = 0,375 mol

therefore the solution is prepared using 0.5 L of H2O and 0.375 moles of CH3OH,  resulting in a concentration of 0,75M

5 0
4 years ago
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