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2 years ago

How many ATP are produced from a fatty acid that is 14 carbons long?

Chemistry

1 answer:

fenix001 [56]2 years ago

3 0

Answer:

92 ATP

Explanation:

Fatty acid oxidation results in the formation of large number of ATP molecules. Three important process of fatty acid are activation of the fatty acid, beta oxidation and entry of acetyl CoA in Krebs cycle.

14 carbon fatty acid is Miristic acid. The complete oxidation of Miristic acid results in the formation of 7 acetyl CoA + 6NADH and $6FADH_2$

1 Acetyl CoA gives 10 molecules of ATP then 7 acetyl CoA gives 70 molecules of ATP.

1NADH = 2.5 ATP, 6NADH = 15 ATP.

$1FADH_2$ = 1.5 ATP, $6FADH_2$ = 9 ATP.

2 ATP has been consumed in the activation of fatty acid.

Total ATP = 70+15+9-2

=92 ATP.

Thus, the total ATP generated from the oxidation of 14 carbon fatty acid is 92.

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Abdul swabs one of his sister’s arms with water. He swabs her other arm with rubbing alcohol.

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Answer:

B. The amounts of water and alcohol are not the same

Explanation:

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A 8.5-liter sample of a gas at 2.0 atm and 300.0 K has 1.2 moles of the gas. If 0.65 mole of the gas is added to the sample at t

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An atom of helium has a radius = and an average speed in the gas phase at of . . Suppose the speed of a helium atom at has been

Hunter-Best [27]

Answer:

$1.2\times 10^3 rNe$

Explanation:

Given that

Speed of neon = 350 m/s

Un-certainity in speed= (0.01 ÷ 100) × 350

= 0.035 m/s

As per heisenberg uncertainty principle

$\triangle X\times m \triangle \ v\geq \frac{h}{4\pi }$ ....... (i)

substituting the values in equation (i)

$\triangle X = 4.49 \times 10^-^8 m$

In terms of rNe i.e 38 pm = $38\times 10^-^1^2$

$\triangle X = \frac{4.49\times 10^-^8}{38\times 10^-^1^2}$

$= 0.118 \times 10^4 \times (rNe)$

$= 1.18\times 10^3 rN$

$= 1.2 \times 10^3 rNe$

Therefore the smallest possible length of the box inside in which the atom could be known for locating with certainty is $1.2\times 10^3 rNe$

6 0

2 years ago

The combustion of ethyne, shown below unbalance, produces heat which can be used to weld metals:

Andreyy89

Answer:

3.69 g

Explanation:

Given that:

The mass m = 325 g

The change in temperature ΔT = ( 1540 - 165)° C

= 1375 ° C

Heat capacity $c_p$ = 0.490 J/g°C

The amount of heat required:

q = mcΔT

q = 325 × 0.490 × 1375

q = 218968.75 J

q = 218.97 kJ

The equation for the reaction is expressed as:

$C_2H_{2(g)} + 5O_{2(g)} \to 2CO_{2(g)} + H_2O_{(g)} \ \ \ \ \ \Delta H^o_{reaction} = -1544 \ kJ$

Then,

1 mole of the ethyne is equal to 26 g of ethyne required for 1544 kJ heat.

Thus, for 218.97 kJ, the amount of ethyne gas required will be:

$= \dfrac{26 \ g}{1544 \ kJ} \times 218.97 \ kJ$

= 3.69 g

3 0

2 years ago

You want to determine ΔH o for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) To do so, you first determine the heat capacity

Assoli18 [71]

Answer:

(A) The heat capacity of the calorimeter is therefore = −2.1428KJ÷13.5°C

= −0.1587KJ/°C

(B) ΔHo for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) = –15.42KJ

Explanation:

Solution

Calculate the heat actually evolved.

q = mcΔt

Finding the mass of the reactants in grams we have.

Use density. (50 mL + 50 mL ) = 100 mL of solution.

100 mL X 1.04g/mL = 104 grams of solution. (mass = Volume X Density)

Find the temperature change.

Δt =tfinal - tinitial = 30.4°C – 16.9°C = 13.5°C

q = mcΔt

= 104grams × 3.93J/g°C × 13.5°C = 5.51772×103J

= 5.51772 × 103 J

This is the heat lost in the reaction between HCl and NaOH, therefore q = -5.52 × 103 J.

this is an exothermic heat producing reaction.

To calculate the total heat of the reaction or heat per mole we have

50.0 mL of HCl X 2.00 mol HCl /(1000 mL HCl ) = 0.100 mol HCl

The same quantity of base, 0.100 mole NaOH, was used.

The energy per unit mole is given by

i.e. molar enthalpy = J/mol = -5.52 × 103J / 0.100 mol

= -5.52 × 104 J/mol

= -55177.2 J/mol

= -55.177 kJ/mol

Therefore, the enthalpy change for the neutralization of HCl and NaOH, that is the enthalpy, heat, of reaction is ΔH = -55.177 kJ/mol

Heat absorbed by the calorimeter = −57.32kJ − 55.177 kJ = −2.1428KJ

The heat capacity of the calorimeter is therefore = −2.1428KJ÷13.5°C

= −0.1587KJ/°C

(B) For the ZnCl we have

Calculate the heat actually evolved.

q = mcΔt

Finding the mass of the reactants in grams we have.

Use density. 100 mL of solution of HCl

100 mL X 1.015g/mL = 101.5 grams of solution. (mass = Volume X Density)

Find the temperature change.

Δt =tfinal - tinitial = 20.5°C – 16.8°C = 3.7 °C

q = mcΔt

= 101.5grams × 3.95J/g°C × 3.7°C = 1483.422×103J

= -1483.422×103J

This is the heat lost in the reaction between HCl and NaOH, therefore q = -1.483 × 103 J.

this is an exothermic heat producing reaction.

To calculate the total heat of the reaction or heat per mole we have

100.0 mL of HCl X 1.00 mol HCl /(1000 mL HCl ) = 0.100 mol HCl

The energy per unit mole is given by

i.e. molar enthalpy = J/mol = -1.483 × 103J / 0.100 mol

= -1.483 × 104 J/mol

= -14834.22 J/mol

= -14.834 kJ/mol

Therefore, the enthalpy change for the neutralization of HCl and NaOH, that is the enthalpy, heat, of reaction is ΔH = -14.834 kJ/mol

ΔHo for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

= -14.834 kJ –(0.1587KJ/°C×3.7°C) = -15.42KJ

ΔHo for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) = –15.42KJ

5 0

2 years ago