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almond37 [142]
3 years ago
5

During complete oxidation of the fatty acid CH3(CH2)14COOH, ________ molecules of acetyl-CoA are produced, and the fatty acid go

es through the β-oxidation cycle ________ times
Chemistry
1 answer:
zimovet [89]3 years ago
6 0

Answer:

During complete oxidation of the fatty acid CH3(CH2)14COOH, eight molecules of acetyl-CoA are produced, and the fatty acid goes through the β-oxidation cycle seven times.

Explanation:

In the β-oxidation of fatty acids, an acetyl-CoA molecule is removed from the fatty acid chain after every β-oxidation cycle that the fatty acid undergo, leaving behind a fatty acyl-CoA molecule shortened by two cabon atoms..

The removal of the acetyl-CoA molecule starts from the carboxyl end and shortens the fatty acid molecule by two carbon units. Successive β-oxidation cycles results in the complete oxidaton of the fatty acid molecle to acetyl -CoA molecules.

The compound CH3(CH2)14COOH, is a 16-carbon saturated fatty acid molecule known as palmitic acid.. It undergoes seven passes through the β-oxidation cycle to yield eight molecules of acetyl-CoA with each cycle yielding an acetyl-CoA molecule and a fatty acyl-CoA shortened by two carbon atoms. Finally, the seventh step yields two acetyl-CoA molecules.

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Write and balance the equation for the combination of ammonia gas with solid copper (II) oxide to produce copper metal, nitrogen
Katena32 [7]

Explanation :

The balanced chemical reaction will be,

3CuO(s)+2NH_3(g)\rightarrow 3Cu(s)+N_2(g)+3H_2O(l)

By the stiochiometry, 3 moles of solid copper(II)oxide react with 2 moles of ammonia gas to give 3 moles of copper metal, 1 mole of nitrogen gas and 3 moles of liquid water.

The states of matter of each elements and compound is,

Copper(II)oxide is in solid state

Ammonia is in gaseous state

Copper metal is in solid state

Nitrogen is in gaseous state

Water is in liquid state

7 0
3 years ago
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During the following chemical reaction, 46.3 grams of C3H6O react with 73.2 grams of O2
ra1l [238]

Answer:

a) O2 is the limiting reactant

b) 75.70 grams CO2 (theoretical yield)

c) There remains 12.81 grams of C3H6O

d) The actual yield CO2 is 34.29 grams

Explanation:

Step 1: Data given

Mass of C3H6O = 46.3 grams

Mass of O2 = 73.2 grams

Molar mass of C3H6O = 58.08 g/mol

Molar mass  of O2 = 32 g/mol

Step 2: The balanced equation

C3H6O + 4O2 → 3 CO2 + 3H2O

Step 3: Calculate moles C3H6O

Moles C3H6O = mass C3H6O / molar mass C3H6O

Moles C3H6O = 46.3 grams / 58.08 g/mol

Moles C3H6O = 0.793 moles

Step 4: Calculate moles O2

Moles O2 = 73.2 grams / 32 g/mol

Moles O2 = 2.29 moles

Step 5: Calculate limiting reactant

For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

O2 is the limiting reactant. It will completely be consumed. (2.29 moles).

C3H6O is in excess. There will react 2.29/4 = 0.5725 moles C3H6O

There will remain 0.793 - 0.5725 = 0.2205 moles C3H6O

This is 0.2205 moles * 58.08 g/mol =<u> 12.81 grams</u>

Step 6:  Calculate moles of CO2

For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

For 2.29 moles O2 we need 3/4 * 2.29 = 1.72 moles CO2

This is 1,72 moles * 44.01 g/mol = <u>75.70 grams CO2</u>

Step 7: Calculate actual yield

% yield = 45.3 % = 0.453 = (actual yield / theoretical yield)

actual yield = 0.453 * 75.70 = <u>34.29 grams</u>

3 0
3 years ago
I really need help plz I’ll give 40 points
polet [3.4K]
I’m pretty sure it’s b
4 0
3 years ago
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Whats the definition of hydrochloric acid​
svet-max [94.6K]

Answer:

Explanation:

a strongly acidic solution of the gas hydrogen chloride in water.

8 0
3 years ago
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What's the enthalpy of neutralization and explain briefly how it relates to esterification​
Liula [17]

Answer:

The enthalpy of neutralization and its relation to esterification​ is explained below in details.

Explanation:

The conventional enthalpy variation of neutralization is the enthalpy exchange when extracts of an acid and an alkali respond together under conventional circumstances to generate 1 mole of water. Remark that the enthalpy change of neutralization is always estimated per mole of water produced.

8 0
3 years ago
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