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wariber [46]
3 years ago
6

By what mechanism does cyclohexanol react when treated in sulfuric acid and what compound results?A) E 1; methoxycyclohexane B)

E2: methoxycyclohexane C) SN 1; methoxycycloheXafle D) E2; cyclohexene E) E 1: cyclohexene
Chemistry
1 answer:
tensa zangetsu [6.8K]3 years ago
8 0

Answer:

E 1: cyclohexene

Explanation:

This reaction is an example of the dehydration of cyclic alcohols. The reaction proceeds in the following steps;

1) The first step of the process is the protonation of the cyclohexanol by the acid. This now yields H2O^+ attached to the cyclohexane ring.

2) the water molecule, which a good leaving group now leaves yielding a carbocation. This now leaves a cyclohexane carbocation which is highly reactive.

3) A water molecule now abstracts a proton from the carbon adjacent to the carbocation leading to the formation of cyclohexene and the regeneration of the acid catalyst. This is an E1 mechanism because it proceeds via a carbocation intermediate and not a concerted transition state, hence the answer.

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A chemistry graduate student is given of a chlorous acid solution. Chlorous acid is a weak acid with . What mass of should the s
DerKrebs [107]

Answer:

11.31g NaClO₂

Explanation:

<em> Is given 250mL of a 1.60M chlorous acid HClO2 solution. Ka is 1.110x10⁻². What mass of NaClO₂ should the student dissolve in the HClO2 solution to turn it into a buffer with pH =1.45? </em>

It is possible to answer this question using Henderson-Hasselbalch equation:

pH = pKa + log₁₀ [A⁻] / [HA]

<em>Where pKa is -log Ka = 1.9547; [A⁻] is the concentration of the conjugate base (NaClO₂), [HA] the concentration of the weak acid</em>

You can change the concentration of the substance if you write the moles of the substances:

[Moles HClO₂] = 250mL = 0.25L×(1.60mol /L) = <em>0.40 moles HClO₂</em>

Replacing in H-H expression, as the pH you want is 1.45:

1.45 = 1.9547 + log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]

-0.5047 = log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]

<em>0.3128 = </em>[Moles NaClO₂] / [0.40 moles HClO₂]

0.1251 = Moles NaClO₂

As molar mass of NaClO₂ is 90.44g/mol, mass of 0.1251 moles of NaClO₂ is:

0.1251 moles NaClO₂ ₓ (90.44g / mol) =

<h3>11.31g NaClO₂</h3>
5 0
3 years ago
A student decomposed 3.67g of copper (ii) hydroxide into copper (ii) oxide. how many ml of 3m h2so4 is need to react with all th
vredina [299]
In the presence of heat, copper (II) hydroxide decomposes in to copper (II) oxide. 
Cu(OH)₂ (s) ----> CuO (s) + H₂O (l)
upon decomposition, water is removed from Cu(OH)₂
the amount of Cu(OH)₂ decomposed - 3.67 g
number of moles of Cu(OH)₂  - 3.67 g / 97.5 g/mol = 0.038 mol
stoichiometry of Cu(OH)₂ to CuO is 1:1
therefore number of CuO moles formed are - 0.038 mol
CuO reacts with sulfuric acid to form CuSO₄ 
CuO + H₂SO₄ ---> CuSO₄ + H₂O
stoichiometry of CuO to H₂SO₄ is 1:1
therefore number of H₂SO₄ moles that should react is 0.038 mol
the molarity of H₂SO₄ is 3M
this means that in 1000 ml - 3 mol of H₂SO₄ present 
so if 3 mol are present in 1000 ml 
then volume for 0.038 mol = 1000/3 * 0.038 
                                           = 12.67 ml
6 0
3 years ago
Caclulate the pH for a solution with an [H+] = 9.75 x10-4
vovangra [49]

Answer:

pH=3.01

Explanation:

pH=-log[H+]

=-log(9.74x10e-4)

=3.01

3 0
3 years ago
Unit: Stoichiometry
Reika [66]

Answer:

1. 2.41 × 1023 formula units

2. 122 L

3. 7.81 L

Explanation:

1. Equation of the reaction: 2 Na(NO3) + Ca(CO3) ---> Na2(CO3) + Ca(NO3)2

Mole ratio of NaNO3 to CaCO3 = 2 : 1

Moles of CaCO3 = mass/molar mass

Mass of CaCO3 = 20 g; molar mass of CaCO3 = 100 g

Moles of CaCO3 = 20 g/100 g/mol = 0.2 moles

Moles of NaNO3 = 2 × 0.2 moles = 0.4 moles

1 Mole of NaNO3 = 6.02 × 10²³ formula units

0.4 moles of NaNO3 = 0.4 × 6.02 × 10²³ = 2.41 × 1023 formula units

2. Equation of reaction : 2 H2O ----> 2 H2 + O2

Mole ratio of oxygen to water = 1 : 2

At STP contains 6.02 × 10²³ molecules = 1 mole of water

6.58 × 10²⁴ molecules = 6.58 × 10²⁴ molecules × 1 mole of water/ 6.02 × 10²³ molecules = 10.93 moles of water

Moles of oxygen gas produced = 10.93÷2 = 5.465 moles of oxygen gas

At STP, 1 mole of oxygen gas = 22.4 L

5.465 moles of oxygen gas = 5.465 moles × 22.4 L/1 mole = 122 L

3.Equation of reaction: 6 K + N2 ----> 2 K3N

Mole ratio of Nitrogen gas and potassium = 6 : 1

Moles potassium = mass/ molar mass

Mass of potassium = 90.0 g, molar mass of potassium = 39.0 g/mol

Moles of potassium = 90.0 g / 39.0 g/mol = 2.3077moles

Moles of Nitrogen gas = 2.3077 moles / 6 = 0.3846 moles

At STP, 1 mole of nitrogen gas = 22.4 L

0.3486 moles of oxygen gas = 0.3486 moles × 22.4 L/1 mole = 7.81 L

7 0
3 years ago
Plz help me it a timed test
Shtirlitz [24]

Answer:

B

Explanation:

6 0
2 years ago
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