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3 years ago

15

Which statement is false?

1 answer:

baherus [9]3 years ago

6 0

Answer:

"The PDH complex is considered to be a reaction of the citric acid cycle" is the false statement

Explanation:

The PDH complex is involved in the oxidation of piruvate, producing acetate and free CO2 (besides ATP and NADH). Concomitantly, in presence of coenzyme A, it binds the acetate molecule to form Acetyl-CoA.

Acetyl-CoA, later, will enter the Krebs cycle reacting with oxalacetate to form citric acid.

The other statements options are true

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Why are coastal zones productive?

Anit [1.1K]

The surface waters in regions of coastal up-welling are cold and nutrient-rich, promoting robust growth of plants and the animals that feed on them. ... When too many nutrients are released into seas, bays, and estuaries, it can create an overabundance of decaying plants and animals, depleting oxygen from the water.

7 0

3 years ago

Salt breaks into a chlorine ion and sodium ion when dissolved in water. The chlorine ion has a negative charge. Which end of the

Monica [59]

The positive, hydrogen end.<span />

3 0

4 years ago

Someone who’s good with science this is for you btw I’ll brainlist and rate 5 star due today.

alekssr [168]

Answer:

Solid Osmium transition metal reacts with Oxygen gas to produce solid Osmium tetroxide.

Os(s) + 2O₂(g) -> OsO₄(s)

Explanation:

Osmium tetroxide is another way of writing Osmium (VIII) oxide.

Leaving powdered osmium exposed to air in a room will slowly create osmium tetroxide at room temperature.

Similarly, osmium tetroxide vapor will readily be released from a liquid solution at room temperature.

7 0

3 years ago

Butane C4H10 (g),(Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Delta.Hf = –393.5 kJ/mol), and H2O(g) (Delta

shusha [124]

Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ

Explanation:

The balanced chemical reaction is,

$2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The expression for enthalpy change is,

$\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]$

Putting the values we get :

$\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]$

$\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]$

$\Delta H=-5314.8kJ$

2 moles of butane releases heat = 5314.8 kJ

1 mole of butane release heat = $\frac{5314.8}{2}\times 1=2657.4kJ$

Thus enthalpy of combustion per mole of butane is -2657.4 kJ

3 0

3 years ago

A sample of calcium phosphate was found to have a mass of 125.3 g. How many molecules were contained in the sample?

Viktor [21]

The answer for the following problem is mentioned below.

<u><em>Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules.</em></u>

Explanation:

Given:

mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = 125.3 grams

We know;

molar mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = (40×3) + 3 (31 +(4×16))

molar mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = 120 + 3(95)

molar mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = 120 +285 = 405 grams

<em>We also know;</em>

No of molecules at STP conditions( $N_{A}$ ) = 6.023 × 10^23 molecules

To solve:

no of molecules present in the sample(N)

We know;

$\frac{m}{M} =\frac{N} }{}$ N÷ $N_{A}$

$\frac{405}{125.3}$ = $\frac{N}{6.023*10^23}$

N =(405×6.023 × 10^23) ÷ 125.3

N = 19.3 × 10^23 molecules

<u><em>Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules</em></u>

3 0

3 years ago