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kolbaska11 [484]
2 years ago
10

Phenol, c6h5oh, is a stronger acid than methanol, ch3oh, even though both contain an o]h bond. Draw the structures of the anions

resulting from loss of h1 from phenol and methanol, and use resonance structures to explain the difference in acidity.

Chemistry
1 answer:
timurjin [86]2 years ago
8 0

In resonance structures, the chemical connectivity in the molecule is same but the distribution of electrons are different around the structure.  They are created by moving electrons in double or triple bonds, and not atoms.

Phenol, C_6H_5OH and methanol, CH_3OH both are alcohols  that contain an -OH group attached to carbon atom.

Due to loss of 1 H from phenol, it forms phenoxide anion and due to presence of double bond in the benzene ring the negative charge on the oxygen atom (which represents electrons) will resonate with double bonds of benzene ring as shown in the image. The resonance-stabilized phenoxide ion is more stable. Whereas when methanol lose 1 H it forms methoxide anion and there are no such electrons present in the structure of methoxide that will result in the movement of electron. Since, due to resonance-stabilized phenoxide ion is more stable than methoxide ion, so it is a stronger acid.

The structures of the anions resulting from loss of 1 H from phenol and methanol is shown in the image.

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How many grams of NH3 can be prepared from 85.5 grams of N2 and 17.3 grams of H2 ?
Tcecarenko [31]
N2 + 3H2 ---> 2NH3

mass of N2 = 28g
mass of H2 = 2g
mass of NH3 = 17g

according to the reaction:
28g N2----------------- 3*2g H2
85,5g N2-------------------- x
x = 18,32g H2 >>>  so, nitrogen is excess

according to the reaction:
2*3g H2---------------------- 2*17g NH3
17,3g H2 ------------------------- x
x = 98,03g NH3

<u>answer: 98,03g of NH3</u>
4 0
2 years ago
How many pints of a 30% sugar solution must be added to a 5 pint of a 5% sugar solution to obtain a 20% sugar solution?
ser-zykov [4K]

You must add 7.5 pt of the 30 % sugar to the 5 % sugar to get a 20 % solution.

You can use a modified dilution formula to calculate the volume of 30 % sugar.

<em>V</em>_1×<em>C</em>_1 + <em>V</em>_2×<em>C</em>_2 = <em>V</em>_3×<em>C</em>_3

Let the volume of 30 % sugar = <em>x</em> pt. Then the volume of the final 20 % sugar = (5 + <em>x</em> ) pt

(<em>x</em> pt×30 % sugar) + (5 pt×5 % sugar) = (<em>x</em> + 5) pt × 20 % sugar

30<em>x</em> + 25 = 20x + 100

10<em>x</em> = 75

<em>x</em> = 75/10 = 7.5

5 0
2 years ago
Brian's aunt has cats. When Brian recently visited her, he started sneezing badly and believes that it was because
kifflom [539]

Answer:

By visiting other households with cats.

Explanation:

This will give Brian a variety of other houses and determine if it is truly cats or just alleries from other items. This is the most direct way to get Brian the answer he is looking for.

4 0
2 years ago
Determine ΔH for the reaction CaCO3 → CaO + CO2 given these data: 2 Ca + 2 C + 3 O2 → 2 CaCO3 ΔH = −2,414 kJ C + O2 → CO2 ΔH = −
kicyunya [14]

Answer:

The ΔH for the reaction is -456.5 KJ

Explanation:

Here we want to determine ΔH for the reaction;

Mathematically;

ΔH = ΔH(product) - ΔH(reactant)

In the case of the first reaction;

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)  ...........................(*)

From the other reactions, we can get the respective ΔH for the individual molecule in the reaction

In second reaction;

Kindly note that for elements, molecule of gases, ΔH = 0

What this means is that throughout the solution;

ΔH(Ca)  = 0 KJ

ΔH(O2) = 0 KJ

ΔH(C) = 0 KJ

Thus, in writing the equation for the subsequent chemical reactions, we shall need to write and equate the overall ΔH for the reaction to that of the product alone

So in the second reaction

ΔH = 2ΔH(CaCO3)

Thus;

-2414/2 = ΔH(CaCO3)

ΔH(CaCO3) = -1,207  KJ

Moving to the third reaction, we have;

ΔH = ΔH(CO2)

Hence ΔH(CO2) = -393.5 KJ

For the last reaction;

ΔH = ΔH(CaO)

Hence ΔH(CaO) = -1270 KJ

Going back to equation *

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)

Using the values of the ΔH  of the respective molecules given above,

ΔH  = -1270 + (-393.5) - (-1207)

ΔH  = -456.5 KJ

8 0
2 years ago
You wish to make a 0.339 m h2so4 solution from a stock solution of 12.0 m h2so4. how much concentrated acid must you add to obta
Alex_Xolod [135]
When preparing diluted solutions from concentrated solutions , we can use the following equation;
c1v1 =c2v2
Where c1 and v1 are the concentration and volume of the concentrated solution
c2 is the concentration of the diluted solution to be prepared
v2 is the volume of the diluted solution
Substituting the values;
12.0 M x v1 = 0.339 M x 100 mL
v1 = 2.825 mL needs to be taken from the stock solution
8 0
2 years ago
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