answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Sladkaya [172]
2 years ago
9

For each of the following substituents, indicate whether it withdraws electrons inductively, donates electrons by hyperconjugati

on, withdraws electrons by resonance, or donates electrons by resonance. (Effects should be compared with that of a hydrogen; remember that many substituents can be characterized in more than one way.)
a. Br
b. CH2CH3
c. NHCH3
d. OCH3
Chemistry
1 answer:
AveGali [126]2 years ago
3 0

Answer:

a. withdraws electrons inductively

b. donates electrons by hyperconjugation

c. donates electrons by resonance

d.  withdraws electrons inductively

Explanation:

a.  The bromide ion is a highly electronegative ion (in the halide series). Electronegative substituents on acids increase the acidity by inductive electron withdrawal method. The higher the electronegativity of a substance, the greater the acidity. The halogens have this order of electronegativity:

F > Cl > Br>I

b.  The carboxyl groups have a stabilization of the sigma and pi bonds. This is achieved through a special delocalization of electrons.  Because of the delocalization, hyperconjugation is the result effect.

c. The NHCH₃ group has a highly electonegative nitrogen atom that pulls the electron cloud towards itself. In this case, it withdraws electrons inductively. As a result, it donates electrons by resonance.

d. The OCH₃ group has a highly electonegative oxygen atom. This oxygen atom withdraws electron cloud towards itself. As a result, it withdraws electrons inductively.

You might be interested in
"The compound K2O2 also exists. A chemist can determine the mass of K in a sample of known mass that consists of either pure K2O
o-na [289]

Answer:

Yes, the chemist can determine which compound is in the sample.

Explanation:

In 1 mole of K₂O, the mass of K is 2 × 39.1 g = 78.2 g and the mass of K₂O is 94.2 g. The mass ratio of K to K₂O is 78.2 g / 94.2 g = 0.830.

In 1 mole of K₂O₂, the mass of K is 2 × 39.1 g = 78.2 g and the mass of K₂O₂ is 110.2 g. The mass ratio of K to K₂O₂ is 78.2 g / 110.2 g = 0.710.

If the chemist knows the mass of K and the mass of the sample, he or she must calculate the mass ratio of K to the sample.

  • If the ratio is 0.830, the compound is pure K₂O.
  • If the ratio is 0.710, the compound is pure K₂O₂.
  • If the ratio is not 0.830 or 0.710, the sample is a mixture.
6 0
2 years ago
1) Aluminum sulphate can be made by the following reaction: 2AlCl3(aq) + 3H2SO4(aq) Al2(SO4)3(aq) + 6 HCl(aq) It is quite solubl
kolezko [41]

Answer:

88.9%

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2AlCl3(aq) + 3H2SO4(aq) —> Al2(SO4)3(aq) + 6HCl(aq)

Step 2:

Determination of the masses of AlCl3 and H2SO4 that reacted and the mass of Al2(SO4)3 produced from the balanced equation.

Molar mass of AlCl3 = 27 + (35.5x3) = 133.5g/mol

Mass of AlCl3 from the balanced equation = 2 x 133.5 = 267g

Molar mass of H2SO4 = (2x1) + 32 + (16x4) = 98g/mol

Mass of H2SO4 from the balanced equation = 3 x 98 = 294g

Molar mass of Al2(SO4)3 = (27x2) + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96] = 342g/mol

Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g

Summary:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4 to produce 342g of Al2(SO4)3.

Step 3:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4.

Therefore, 25g of AlCl3 will react with = (25 x 294)/267 = 27.53g of H2SO4.

From the calculations made above, we see that only 27.53g out 30g of H2SO4 given were needed to react completely with 25g of AlCl3.

Therefore, AlCl3 is the limiting reactant and H2SO4 is the excess.

Step 4:

Determination of the theoretical yield of Al2(SO4)3.

In this case we shall be using the limiting reactant because it will produce the maximum yield of Al2(SO4)3 since all of it is used up in the reaction.

The limiting reactant is AlCl3 and the theoretical yield of Al2(SO4)3 can be obtained as follow:

From the balanced equation above,

267g of AlCl3 reacted to produce 342g of Al2(SO4)3.

Therefore, 25g of AlCl3 will react to produce = (25 x 342) /267 = 32.02g of Al2(SO4)3.

Therefore, the theoretical yield of Al2(SO4)3 is 32.02g

Step 5:

Determination of the percentage yield of Al2(SO4)3.

This can be obtained as follow:

Actual yield of Al2(SO4)3 = 28.46g

Theoretical yield of Al2(SO4)3 = 32.02g

Percentage yield of Al2(SO4)3 =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 28.46/32.02 x 100

Percentage yield = 88.9%

Therefore, the percentage yield of Al2(SO4)3 is 88.9%

3 0
2 years ago
At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2 + O2(g) → 2SO3(g) At equilibrium, the partia
lapo4ka [179]

Answer:

The partial pressure of SO₃ is 82.0 atm

Explanation:

The equilibrium constant Kp is equal to <em>the equilibrium pressure of the gaseous products raised to the power of their stoichiometric coefficients divided by the equilibrium pressure of the gaseous reactants raised to the power of their stoichiometric coefficients</em>.

For the reaction,

2 SO₂(g) + O₂(g) → 2 SO₃(g)

Kp = 0.345 = \frac{(pSO_{3})^{2} }{(pSO_{2})^{2} \times pO_{2} }\\pSO_{3} = \sqrt[]{0.345 \times (pSO_{2})^{2} \times pO_{2} } \\pSO_{3} = \sqrt[]{0.345 \times (35.0)^{2} \times 15.9 } \\pSO_{3} = 82.0 atm

4 0
2 years ago
A mixture of carbon dioxide and an unknown gas was allowed to effuse from a container. The carbon dioxide took 1.25 times as lon
Aleks04 [339]

Answer:

CO

Explanation:

From Graham's law, time taken to diffuse is directly proportional to the molecular mass of the gases. For two different gases.

t1/t2=√m1/m2

Since gas 1 diffuse 1.25 times as slowly as gas 2 and gas 1 is CO2 with m as 44g

1.25/1=√44/m2

Therefore m2=28g CO

7 0
2 years ago
In order to reduce the exposure to organic solvents like turpentine, some art instructors recommend the students clean brushes a
kykrilka [37]
Generally speaking, organic molecules tend to dissolve in solvents that have similar physical properties. A good rule of thumb is that "like dissolves like". Meaning, polar compounds can dissolve polar compounds and nonpolar compounds can dissolve nonpolar compounds.

To apply this to the current problem, we are told that the brushes are being cleaned with vegetable oil or mineral oil. In this case, the oils are used as solvents. In order for these solvents to be effective, the compounds they are trying to dissolve must be similar in structure and properties to other oils. Therefore, vegetable oil or mineral oil will be most effective in removing oil-based paints, as these will have the similar properties needed to dissolve in the oil solvents.
6 0
2 years ago
Other questions:
  • Mr. martin's spanish class is 45 minutes long. if it starts at 3:30, what time does it end?
    5·1 answer
  • An iron atom has a mass of ×9.2710−23g and a cooking pot has a mass of 0.500kg
    12·2 answers
  • A student titrates 10.00 milliliters of hydrochloric acid of unknown molarity with 1.000 m naoh. it takes 21.17 milliliters of b
    11·1 answer
  • Benzene, C6H6, reacts with oxygen, O2, to form CO2 and H2O. How much O2 is required for the complete combustion of 1.0 mol C6H6?
    6·2 answers
  • Tamara is doing a scientific investigation to answer the question, "Do corn seeds germinate faster at warmer temperatures than t
    12·1 answer
  • a) (1 point) Build anthracene, optimize its geometry and examine its structure. Describe its shape. b) (1 point) Measure the C-C
    13·1 answer
  • Kaia, a chemical engineering graduate, has documented all titration procedures in her project report. She refers to this report
    12·1 answer
  • The particles of a substances are arranged in a definite geometric pattern and are constantly vibrating. This substances can be
    10·1 answer
  • In one compound of lead and sulfur, there are 6.46 g of lead
    13·1 answer
  • You have 3.00 L of a 3.12 M solution of NaCl(aq) called solution A. You also have 2.00 L of a 2.00 M solution of AgNO3(aq) calle
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!