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Lady bird [3.3K]
2 years ago
15

A critical review of analytical methods in pretreatment of lignocelluloses: Composition, imaging, and crystallin

Chemistry
1 answer:
Alla [95]2 years ago
4 0

A critical review of analytical methods in pretreatment of lignocelluloses: Composition, imaging, and crystallinity.

<h3>What is the abstract?</h3>

Lignocelluloses are being studied extensively as renewable substrates for the production of biofuels such as ethanol, methane, hydrogen, and butanol, as well as chemicals such as citric acid, lactic acid, and xanthan gum. However, because lignocelluloses have a recalcitrance structure that makes them resistant to microbial and enzymatic attacks, several physical, thermal, chemical, and biological pretreatment methods have been devised to open up their structure. These pretreatments' efficacy was investigated utilising a number of analytical approaches that addressed their image, composition, crystallinity, degree of polymerization, enzyme adsorption/desorption, and accessibility.

To learn more about polymerization visit:

brainly.com/question/3200802

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Using the mmoles of (35)-2,2,-dibromo-3,4-dimethylpentane calculated earlier and the molecular weight of the product (962 g/mol)
Sedbober [7]

Answer:

The yield of the product in gram is \mathsf{{w_P}=0.26 \ gram}

Explanation:

Given that:

the molecular mass weight of the product = 96.2 g/mol

the molecular mass of the reagent (3S)-2,2,-dibromo-3,4-dimethylpentane is 257.997 g

given that the millimoles of the reagent = 2,7 millimoles = 2.7 \times 10^{-3} \ moles

We know that:

Number of moles = mass/molar mass

Then:

2.7 \times 10^{-3} = \dfrac{ mass}{257.997}

mass = 2.7 \times 10^{-3} \times 257.997

mass = 0.697

Theoretical yield = (number of moles of the product/ number of moles of reactant) × 100

i.e

Theoretical yield = \dfrac{n_P}{n_R}\times 100\%

where;

n_P = \dfrac{w_P}{m_P}    and n_R = \dfrac{w_R}{m_R}

Theoretical yield = \dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%

Given that the theoretical yield = 100%

Then:

100\% =\dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%

\dfrac{w_P}{m_P}=\dfrac{w_R}{m_R}

{w_P}=\dfrac{w_R \times m_P}{m_R}

where,

w_P = derived weight of the product

m_P =the molecular mass of the derived product

m_R = the molecular mass of the reagent

w_R = weight in a gram of the reagent

{w_P}=\dfrac{w_R \times m_P}{m_R}

{w_P}=\dfrac{0.697 \times 96.2}{257.997}

\mathsf{{w_P}=0.26 \ gram}

8 0
3 years ago
How many of the following are oxidation-reduction reactions? I. reaction of a metal with a nonmetal II. synthesis III. combustio
nadezda [96]

Answer : The oxidation-reduction reactions are:

I. reaction of a metal with a nonmetal

II. synthesis

III. combustion

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

I. Reaction of a metal with a nonmetal :

When sodium react with chlorine gas then it react to give sodium chloride.

2Na+Cl_2\rightarrow 2NaCl

In this reaction, the oxidation state of sodium changes from (0) to (+1) and shows oxidation and the oxidation state of chlorine changes from (0) to (-1) and shows reduction. So, It is an oxidation-reduction reaction.

II. Synthesis reaction :

A chemical reaction where multiple substances or reactants combine to form a single product.

When hydrogen react with oxygen then it react to give water.

2H_2+O_2\rightarrow 2H_2O

In this reaction, the oxidation state of hydrogen changes from (0) to (+1) and shows oxidation and the oxidation state of oxygen changes from (0) to (-2) and shows reduction. So, It is an oxidation-reduction reaction.

III. Combustion reaction :

A chemical reaction in which a hydrocarbon reaction with the oxygen to give product as carbon dioxide and water.

When methane react with oxygen gas then it react to give carbon dioxide and water.

CH_4+2O_2\rightarrow CO_2+2H_2O

In this reaction, the carbon of methane gain oxygen and shows oxidation and the oxygen gas gain hydrogen and shows reduction. So, It is an oxidation-reduction reaction.

IV. Precipitation reaction :

It is defined as the reaction in which an insoluble salt formed when two aqueous solutions are combined.

The insoluble salt that settle down in the solution is known an precipitate.

It is a double displacement reaction. So, it is not an oxidation-reduction reaction.

V. Decomposition reaction :

A chemical reaction in which the the larger molecule decomposes to give two or more smaller molecules.

The oxidation state remains same on reactant and product side. So, it is not an oxidation-reduction reaction.

3 0
3 years ago
What do helium (He), neon (Ne), and argon (Ar) have in common?
IrinaK [193]

Answer:

The answer is: D

Explanation:

A. They have the same number of electron energy shells.  Is false, all the elements are in different periods so, they have different number of lectron energy shells.

B. They are all Halogens.  No, is wrong, halogens are F, Cl ,Br

C. They have the same number of electrons.  is wrong, if they had the same number of electrons they must be they same element and they aren't the same.

D. They are all Noble gases. Yes, it's true they are noble gases, they are the first group in the periodic table from the right.

6 0
2 years ago
Read 2 more answers
How many hours are in 188 years
AURORKA [14]

Answer: about 1,705,536 hrs

24 hrs = 1 day

189 hrs = 1 week

756 = 1 month

9072 = 1 yr

1,705,536 = 188 yrs

Explanation:

Trust Milky. Milky smart

7 0
3 years ago
What do elements in the same group in the periodic table have in common
Shalnov [3]

Answer:

The elements in each group have the same number of electrons in the outer orbital. Those outer electrons are also called valence electrons.

Explanation:

8 0
2 years ago
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