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Travka [436]
2 years ago
11

Give three properties of organic compounds​

Chemistry
2 answers:
Jlenok [28]2 years ago
4 0

Answer:

<em>Properties. The physical properties of organic compounds typically of interest include both quantitative and qualitative features. Quantitative information includes a melting point, boiling point, and index of refraction. Qualitative properties include odor, consistency, solubility, and color.</em>

Paraphin [41]2 years ago
3 0

The three poporites of organic compounds are

1) Melting point

2) Boiling point

3)Index of refraction

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Acetylene gas (C2H2) burns completely in the presence of oxygen gas (O2) to yield carbon dioxide
Luda [366]

A. The balanced equation for the reaction is

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

B. The volume of the oxygen gas required to burn 0.700 L of acetylene gas is 1.75 L

C. The volume of carbon dioxide gas produced is 1.4 L

D. The volume of water vapor produced is 0.7 L

<h3>A. Balanced equation </h3>

The balanced equation for the reaction between acetylene gas (C₂H₂) and oxygen gas (O₂) is given below

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

<h3>B. How to determine the volume of oxygen </h3>

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

Since the reaction occurred at standard temperature and pressure, we can thus say that:

From the balanced equation above,

2 L of C₂H₂ reacted with 5 L O₂.

Therefore,

0.7 L of C₂H₂ will react with = (0.7 × 5) / 2 = 1.75 L of O₂

Thus, 1.75 L of O₂ is needed for the reaction

<h3>C. How to determine the volume of carbon dioxide </h3>

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

From the balanced equation above,

2 L of C₂H₂ reacted to produce 4 L of CO₂

Therefore,

0.7 L of C₂H₂ will react to produce = (0.7 × 4) / 2 = 1.4 L of CO₂

Thus, 1.4 L of CO₂ were produced

<h3>D. How to determine the volume of water. </h3>

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

From the balanced equation above,

2 L of C₂H₂ reacted to produce 2 L of H₂O

Therefore,

0.7 L of C₂H₂ will also react to produce 0.7 L of H₂O

Thus, 0.7 L of H₂O was produced

Learn more about stoichiometry:

brainly.com/question/14735801

8 0
3 years ago
What happens to the ph of the reaction mixture as the oxidation reaction proceeds in an oxidation of an aromatic side chain?
andrew11 [14]

Answer:

The pH decreases.

Explanation:

Hello,

In organic chemistry, oxidation accounts for either the increasing of C-O bonds or the increasing in the oxygen atoms into the molecule. Thus, if we consider the oxidation from benzyl alcohol to benzoic acid, there will be a carboxyl functional group instead of a hydroxile one. Now, the presence of the polar -O-H bonds that are ionizable, there will be a H^{+1} releasing causing the pH to decrease (increase acidity).

Regards.

4 0
3 years ago
HELP!!!<br> IF YOU DONT KNOW THE ANSWER THEN PLEASE DONT ANSWER!!
Simora [160]

Answer:

1. melting

2. freezing

5. condensation

6. evaporation

4 0
3 years ago
Which of the following is not included in John Dalton's modern theory of the atom?
Akimi4 [234]

The answer for the following question is option "C".

Option C is not included in the John Dalton's modern theory of an atom.

  • "It states atoms of different elements combine to form new compound" but not new elements

Explanation:

According to John's Dalton's modern theory of an atom:

1. All matter is composed of atoms.

2. Atoms cannot be created,destroyed or subdivided in the ordinary chemical reactions.

3. Atoms of one element differ in the properties from atoms of an another element.

(i.e.)Each and every atom of the element has its own unique properties of their own.

4. Atoms of one  element combine with the atoms of another element to <u>form new compound.</u>

5. Atoms that make up an element are identical to each other.

7 0
3 years ago
The enzyme, phosphoglucomutase, catalyzes the interconversion
Fittoniya [83]

Answer:

K_{eq = 19

ΔG° of the reaction forming glucose 6-phosphate =  -7295.06 J

ΔG° of the reaction  under cellular conditions = 10817.46 J

Explanation:

Glucose 1-phosphate     ⇄     Glucose 6-phosphate

Given that: at equilibrium, 95% glucose 6-phospate is  present, that implies that we 5% for glucose 1-phosphate

So, the equilibrium constant K_{eq can be calculated as:

K_{eq = \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}

K_{eq= \frac{0.95}{0.05}

K_{eq = 19

The formula for calculating ΔG° is shown below as:

ΔG° = - RTinK

ΔG° = - (8.314 Jmol⁻¹ k⁻¹ × 298 k ×  1n(19))

ΔG° = 7295.05957 J

ΔG°≅ - 7295.06 J

b)

Given that; the concentration  for  glucose 1-phosphate = 1.090 x 10⁻² M

the concentration of glucose 6-phosphate is 1.395 x 10⁻⁴ M

Equilibrium constant  K_{eq can be calculated as:

K_{eq = \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}

K_{eq}= \frac{1.395*10^{-4}}{1.090*10^{-2}}

K_{eq} = 0.01279816514  M

K_{eq} = 0.0127 M

ΔG° = - RTinK

ΔG° = -(8.314*298*In(0.0127)

ΔG° = 10817.45913 J

ΔG° = 10817.46 J

5 0
3 years ago
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