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2 years ago

Which of the following statements about anabolic pathways is true?

Chemistry

2 answers:

nadezda [96]2 years ago

8 0

Answer:

They consume energy to build up polymers from monomers.

Explanation:

Anabolic reactions are the metabolic reactions performed in the body which synthesizes the large molecules from the small molecules or synthesizes polymers from the monomers.

The synthesis of polymers requires energy which is provided during the reaction by ATP, NADH or NADPH molecules.

The examples of anabolism include the synthesis of biomolecules like the nucleic acid from the nucleotides, the synthesis of proteins from the amino acids, the synthesis of sugars from the carbon dioxide and water and many others.

Thus, the selected option is the correct answer.

Sedbober [7]2 years ago

6 0

Answer:

They consume energy to build up polymers from monomers.

Explanation:

Metabolism may be defined as all the biochemical reactions that are present in the living organisms. These reactions are important for the sustenance of the life.

Two main type of metabolic reactions are catabolic reaction and anabolic reactions. The anabolic reactions includes the formation of the large product from the small molecules. The large compounds are synthesized in the cells in the anabolic reactions.

Thus, the correct answer is option (2).

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There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.

Similarly, for the second compound

$m(\text{O}) = m(\text{loss}) = 0.016 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012 \; g$

$n = m/M$ ; $0.4172 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol$
$n(\text{Hg atoms}) = 0.4012 \; g / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol$

$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

$\text{Hg}_{2} \text{O}$ .

8 0

3 years ago

Which of the following molecules has a bent shape?

poizon [28]

A, because the shape of it forms and arc much like a H2O molecule.

4 0

2 years ago

How many grams are in 9.05 x 1023 atoms of silicon

UkoKoshka [18]

Answer:

42.2075 grams

Explanation:

4 0

2 years ago

What is the (OH-) in a solution with a pOH of 6.48

inn [45]

Through manipulation of equations, we are able to obtain the equation:

$-pOH= log [ OH^{-}]$

Then we can transform the equation into:

$[ OH^{-}]= 10^{-pOH}$

Then we are able to plug in the pOH and directly get [OH-]:

$[ OH^{-}] = 10^{-6.48}$

$[ OH^{-}]=3.31* 10^{-7} M$

3 0

3 years ago

What is the empirical formula of a compound containing 90 grams carbon, 11 grams hydrogen, and 35 grams nitrogen? (5 points)

konstantin123 [22]

C3 H4 N1 ~~that’s what I think

8 0

3 years ago