At the start of beta-oxidation, fatty acids are converted into ___ at the expense of ___ in the ____

Answers

Answer 1

As beta-oxidation starts, fatty acids are first converted into acyl-CoA in the cytosol at the expense of ATP. Acyl-CoA is then transported into the mitochondria for additional oxidation after this procedure.

The oxidation of fatty acids is a metabolic process that involves the breakdown of long-chain fatty acids into acetyl-CoA, which is then used for energy production. The oxidation process involves a series of enzymatic reactions that result in the release of energy in the form of ATP.

However, during the oxidation process, fatty acids are susceptible to oxidation themselves, which can lead to the formation of harmful reactive oxygen species (ROS). To prevent this, cells have evolved various mechanisms to regulate fatty acid oxidation and limit ROS formation. At the start of beta-oxidation, fatty acids are converted into acyl-CoA at the expense of ATP (adenosine triphosphate) in the oxidation process.

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Related Questions

Calcium is an element. What is the smallest particle of calcium that has the same chemical properties of calcium?.

Answers

The tiniest calcium particle with calcium, Chemical properties is called an atom. The smallest component of an element, an atom has exactly the same chemical makeup as the element as a whole.

First off, that is inaccurate, and your justification is plagiarism. The smallest calcium particle that possesses calcium's chemical properties is an atom. Any attribute that can only be generated by altering a substance's chemical identity is referred to as a chemical property. Chemical properties include any of a material's properties that emerge during or after a chemical reaction. Simply put, it is impossible to detect a substance's chemical qualities by looking at it or touching it; instead, a significant amount of the substance's internal structure must be altered in order to explore its chemical properties.

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what is the molarity of a solution made by dissolving 45.6 g glucose, c6h12o6, in enough water to make 350.0 ml of solution?

Answers

The molarity of a solution made by dissolving 45.6 g glucose, C₆H₁₂O₆, in enough water to make 350.0 ml of solution is 0.722 M.

given that :

volume = 350 mL = 0.350 L

mass of glucose = 45.6 g

molar mass of glucose = 180.15 g/mol

the number of moles of glucose = mass / molar mass

                                                      = 45.6 / 180.15

                                                      = 0.253 mol

the molarity expression is given as follows :

Molarity = moles / volumes in L

              = 0.253 / 0.350

              = 0.722 M                  

Thus, the molarity of the solution to make 350.0 mL of solution is 0.722 M.          

                       

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The element titanium-44 decays into other elements with a half-life of 63 years. Suppose a 14 kilogram deposit of titanium-44 was discovered. What is a function that represents the amount, A, in kilograms of the titanium-44 deposit t years after the discovery?

\(A(t)=63(\frac{1}{2} )^\frac{t}{14}\)

\(A(t)=2(14)^\frac{t}{63}\)

\(A(t)=14(\frac{1}{63} )^2^t\)

\(A(t)=14(\frac{1}{2})^\frac{t}{63}\)

Answers

The function that represents the amount of titanium-44 deposit in kilograms t years after discovery is: A(t) = 14 * e^(-t * ln(2)/63)

What is the decay function?

The decay of titanium-44 can be modeled by an exponential decay function, where the amount of titanium-44 remaining after t years is given by:

A(t) = A₀ * e^(-kt)

where;

A₀ is the initial amount of titanium-44, k is the decay constant, and e is the mathematical constant approximately equal to 2.71828.

To find k, we use the half-life of titanium-44, which is 63 years. The half-life is the time it takes for half of the initial amount of titanium-44 to decay. Therefore, we have:

0.5 A₀ = A₀ * e^(-k*63)

Dividing both sides by A₀, we get:

0.5 = e^(-k*63)

Taking the natural logarithm of both sides, we get:

ln(0.5) = -k*63

Solving for k, we get:

k = ln(2)/63

Substituting k into the equation for A(t), we get:

A(t) = A₀ * e^(-t * ln(2)/63)

Now we need to find A₀. We know that the deposit initially weighed 14 kilograms, so we have:

A₀ = 14 kg

Putting it all together, the function that represents the amount of titanium-44 deposit in kilograms t years after discovery is:

A(t) = 14 * e^(-t * ln(2)/63)

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An called “G” has a half life of 4 years. Only 60g of the sample are left after 12 years. How much did you start with?

Answers

The amount (in grams) of the sample you started with given that only 60 grams remains after 12 years is 480 g

How do I determine the amount (in grams) I started with?

We'll begin our calculation by obtaining the number of half lives that has elapsed after 12 years. This is shown below:

Half-life (t½) = 4 yearsTime (t) = 12 yearsNumber of half-lives (n) =?

n = t / t½

n = 12 / 4

n = 3

Thus, 3 half-lives has elapsed!

Finally, we shall determine the original amount you started with. Details below:

Amount remaining (N) = 60 gNumber of half-lives (n) = 3Original amount (N₀) = ?

N = N₀ / 2ⁿ

60 = N₀ / 2³

60 = N₀ / 8

Cross multiply

N₀ = 60 × 8

N₀ = 480 g

Thus, we can conclude that the original amount you started with is 480 grams

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Glucose and fructose are simple carbohydrates with the molecular formula c6h12o6. Sucrose , or table sugar, is a complex carbohydrate with molecular formula c12h22o11 that consists of a glucose unit covalently linked to a fructose unit (a water molecule is released as a result of the reaction between glucose and fructose to form sucrose). Estimate the standard enthalpy of combustion of sucrose (in kj/mol) from the standard enthalpies of formation of the reactants and products.

Answers

The standard enthalpy of combustion is ΔHcombustion = ΔHf(products) - ΔHf(reactants) = -6353.8 kJ/mol - (-2218.2 kJ/mol) = -4135.6 kJ/mol.

How to estimate the standard enthalpy of combution of sucrose?To estimate the standard enthalpy of combustion of sucrose, we can use the standard enthalpies of formation of the reactants and products. Here are the steps to follow:Write out the balanced equation for the combustion of sucrose: C12H22O11 + 12O2 --> 12CO2 + 11H2OLook up the standard enthalpies of formation for each of the reactants and products. These can be found in tables of thermodynamic data.Use the equation ΔH combustion = ΔHf(products) - ΔHf(reactants) to calculate the standard enthalpy of combustion.The standard enthalpies of formation of reactants are C12H22O11 (sucrose) + 12O2(g) = -2218.2 kJ/mol + 0 kJ/mol = -2218.2 kJ/molThe standard enthalpies of formation of products are 12CO2(g) + 11H2O(l) = -3935.4 kJ/mol + -2418.4 kJ/mol = -6353.8 kJ/molTherefore the standard enthalpy of combustion is ΔHcombustion = ΔHf(products) - ΔHf(reactants) = -6353.8 kJ/mol - (-2218.2 kJ/mol) = -4135.6 kJ/mol.

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(a) Using the half-equivalence point data from the experiment, what is the experimental pKa for acetic acid in this reaction?
(b) Using this experimental pKa value, what is the experimental Ka for acetic acid? (Must show all work to receive credit)
(c) Look up the accepted "actual" Ka value for acetic acid. How does the experimental value compare? Calculate the percent error for the experimental value. (Must show all work to receive credit.)
Table 1: Hot Sauce and Ketchup Titration Data
Hot Sauce (Trial 1)
Hot Sauce (Trial 2)
Ketchup (Trial 1)
Ketchup (Trial 2)
Mass of sauce (g)
1.5 g
1.5 g
1.5 g
1.5 g
Concentration of NaOH used (M)
0.1
0.1
0.1
0.1
(mL) NaOH needed to reach equivalence point
2 ml
10 ml
4 ml
3.5 ml
(mol) NaOH needed to reach equivalence point [show work below]
.0002 mol
.001 mol
.0004 mol
.00035 mol
Concentration of C2H4O2 (mol/g of sauce) [show work below]
1.33 x mol/g
6.67 x mol/g
2.67x 10-4 mol/g
2.33 x 10-4 mol/g
Average Concentration of C2H4O2 (mol/g of sauce)
4.0 x mol/g
2.5 x mol/g
pH of solution at equivalence point
3.98
7.53
7.97
6.23
NaOH needed to reach half-equivalence point (mL)
1 ml
5 ml
2 ml
1.75 ml
pH of solution at half-equivalence point
3.76
4.92
4.95
4.85
Concentration Calculations:
Mol NaOH needed (Hot Sauce Trial 1):
Mol of NaOH needed to reach equivalence point = Molarity of NaOH × equivalence point volume in liter
Equivalence point volume in liter = 2 ml = 0.002 L
Mol of NaOH needed to reach equivalence point = 0.1 mol/L x .002 L = .0002 mol
Concentration of C2H4O2 (Hot Sauce Trial 1):
At equivalence point, Mol of NaOH = mole of CH3COOH = 0.0002 mol
Conc. of C2H4O2 (mol/g) = .0002 mol / 1.5 g = 1.33 x 10-4 mol/g
Mol NaOH needed (Hot Sauce Trial 2):
Equivalence point volume in liter = 10 ml = 0.01 L
Mol of NaOH needed to reach equivalence point = 0.1 mol/L × 0.01 L = 0.001 mol
Concentration of C2H4O2 (Hot Sauce Trial 2):
At equivalence point, Mol of NaOH = mole of CH3COOH = 0.001 mol
Conc. of C2H4O2 (mol/g) = 0.001 mol / 1.5 g = 6.67 × 10-4 mol/g
Mol NaOH needed (Ketchup Trial 1):
Equivalence point volume in liter = 4 ml = .004 L
Mol of NaOH needed to reach equivalence point = 0.1 mol/L x .004 L = .0004 mol
Concentration of C2H4O2 (Ketchup Trial 1):
At equivalence point, Mol of NaOH = mole of CH3COOH = 0.0004 mol
Conc. of C2H4O2 (mol/g) = 0.0004 mol / 1.5 g = 2.67x 10-4
Mol NaOH needed (Ketchup Trial 2):
Equivalence point volume in liter = 3.5 ml = .0035 L
Mol of NaOH needed to reach equivalence point = 0.1 mol/L x .0035 L = .00035 mol
Concentration of C2H4O2 (Ketchup Trial 2):
At equivalence point, Mol of NaOH = mole of CH3COOH = 0.00035 mol
Conc. of C2H4O2 (mol/g) = 0.00035 mol / 1.5 g = 2.33 x 10-4

Answers

The experimental values have varying degrees of error when compared to the accepted actual Ka value.

(a) Using the half-equivalence point data from the experiment, the experimental pKa for acetic acid is as follows:

Hot sauce (Trial 1): The pH of solution at the half-equivalence point is 3.76.

Using the formula, pH = pKa + log([A-]/[HA]), the calculation is as follows: pKa = pH - log([A-]/[HA]) = 3.76 - log(0.5) = 3.26

Hot sauce (Trial 2): The pH of solution at the half-equivalence point is 4.92.

Using the formula, pH = pKa + log([A-]/[HA]), the calculation is as follows: pKa = pH - log([A-]/[HA]) = 4.92 - log(0.5) = 4.42

Ketchup (Trial 1): The pH of solution at the half-equivalence point is 4.95.

Using the formula, pH = pKa + log([A-]/[HA]), the calculation is as follows: pKa = pH - log([A-]/[HA]) = 4.95 - log(0.5) = 4.45

Ketchup (Trial 2): The pH of solution at the half-equivalence point is 4.85.

Using the formula, pH = pKa + log([A-]/[HA]), the calculation is as follows: pKa = pH - log([A-]/[HA]) = 4.85 - log(0.5) = 4.35

Therefore, the experimental pKa for acetic acid in this reaction is as follows:

Hot sauce (Trial 1) pKa = 3.26

Hot sauce (Trial 2) pKa = 4.42

Ketchup (Trial 1) pKa = 4.45

Ketchup (Trial 2) pKa = 4.35

(b) Using this experimental pKa value, the experimental Ka for acetic acid is calculated as follows:

Hot sauce (Trial 1): Ka = \(10^{-pKa} = 10^{-(3.26)} = 5.12 * 10^{-4}\)

Hot sauce (Trial 2): Ka = \(10^{-pKa} = 10^{-(4.42)} = 2.51 * 10^{-5}\)

Ketchup (Trial 1): Ka = \(10^{-pKa} = 10^{-(4.45)} = 2.23 * 10^{-5}\)

Ketchup (Trial 2): Ka = \(10^{-pKa} = 10^{-(4.35)} = 2.81 * 10^{-5}\)

Therefore, the experimental Ka for acetic acid is as follows:

Hot sauce (Trial 1) Ka = \(5.12 * 10^{-4}\)

Hot sauce (Trial 2) Ka =  \(2.51 * 10^{-5}\)

Ketchup (Trial 1) Ka =  \(2.23 * 10^{-5}\)

Ketchup (Trial 2) Ka =  \(2.81 * 10^{-5}\)

(c) The accepted actual Ka value for acetic acid is \(1.8 * 10^{-5} M\).

The percent error for the experimental value is calculated as follows:

Hot sauce (Trial 1): Percent error = \(((5.12 * 10^{-4} - 1.8 * 10^{-5}) / (1.8 * 10^{-5})) * 100% = 27%\)

Hot sauce (Trial 2): Percent error = \(((2.51 * 10^{-5} - 1.8 * 10^{-5}) / (1.8 * 10^{-5})) * 100% = 40%\)

Ketchup (Trial 1): Percent error = \(((2.23 * 10^{-5} - 1.8 * 10^{-5}) / (1.8 * 10^{-5})) * 100% = 24%\)

Ketchup (Trial 2): Percent error = \(((2.81 * 10^{-5} - 1.8 * 10^{-5}) / (1.8 * 10^{-5})) * 100% = 56%\)

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g. What is crystallization?Name any two go mixtures that can be separated by this method:name any two mixtures that can be separated by ​

Answers

crystallization is a process which helps to separate a pure solid from a solution in its crystal form. This is the in use to purify solid. For an example the salt we get from seawater can have many impurities in it. Hence, the process of crystallization is in use to remove these impurities.

Sugar and salt are examples of products where crystallization does not only serve as separation/purification technique, but where it is also responsible for getting crystals with the right size (and shape) for further application of the products.

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Answer:

Crystallization is the process by which a solid forms, where the atoms or molecules are highly organized into a structure known as a crystal. Some of the ways by which crystals form are precipitating from a solution, freezing, or more rarely deposition directly from a gas.

Sugar and salt are two mixtures that can be seperated by crystallization

Explanation:

The average atomic mass of nickel is 58.69 amu. It has 28 protons. How many electrons does nickel have

Answers

Answer:

Nickel has a total of 28 electrons.

Explanation:

Question 7
(01.03 LC)
Which of the following happens to a molecule of an object when the object is cooled?
a) Its temperature increases,

b) It loses kinetic energy.

c) Its size increases.

d) It moves faster.

Answers

Answer:

D

Explanation:

Answer:

D

Explanation:

:)

Could a juman stay alive if one body system is not working? could the body maintain homeostasis?

Answers

Answer:

no i think

Explanation:

As a result of this process, the proportions of oxygen and carbon dioxide in

air breathed in and air breathed out change.

Which one of the statements is true? Tick the correct box. [1]


- Air breathed out has less carbon dioxide and more oxygen than air breathed in.

- Air breathed out has less carbon dioxide and less oxygen than air breathed in.

- Air breathed out has more carbon dioxide and less oxygen than air breathed in.

- Air breathed out has more carbon dioxide and more oxygen than air breathed in.

Answers

Answer:

the third one

Explanation:

When you breathe in, you inhale oxygen and exhale carbon dioxide

Explain the connection between atoms and electricity

Answers

Answer:

The protons and electrons of an atom are attracted to each other. They both carry an electrical charge. Protons have a positive charge (+) and electrons have a negative charge (-). The positive charge of the protons is equal to the negative charge of the electrons.

Explanation:

What are the basic forces that drive the water cycle? Explain what they are.

Answers

Answer:

kcgiobb jkffiob hffyio hiiddhoo hhioiv

Answer: Have a blessed day

Explanation:

Water Cycle: Component # 1. Evaporation:

Evaporation is the process where a liquid changes from its liquid state to a gaseous state. Liquid water becomes water vapour. Temperature is the primary factor for evaporation.

During the water cycle the water of the oceans and freshwater bodies, such as lakes and rivers, is warmed by the sun and evaporates. During the process of evaporation, impurities in the water are left behind. As a result, the water that goes into the atmosphere is cleaner than it was on Earth.

Water Cycle: Component # 2. Condensation:

Condensation is the opposite of evaporation. Condensation is the process where a gas is changed into a liquid. Condensation occurs when the temperature of the vapour decreases.

The water droplets formed from condensation are very small and they remain suspended in the atmosphere. There millions of droplets of suspended water form clouds in the sky or fog at ground level. Water condenses into droplets only when there are small dusts particles present around.

Water Cycle: Component # 3. Precipitation:

When the temperature and atmospheric pressure are right, the small droplets of water in clouds form larger droplets or raindrops and it is called precipitation. The raindrops fall to earth.

As a result of evaporation, condensation and precipitation, water travels from the surface of the Earth goes into the atmosphere, and returns to Earth again.

Water Cycle: Component # 4. Surface Runoff:

Much of the water that returns to Earth as precipitation runs off the surface of the land and flows down-hill into streams, rivers, ponds and lakes. Small streams flow into the ocean. Surface runoff is an important part of the water cycle because, through surface runoff, much of the water returns again to the oceans, where a great deal of evaporation occurs.

Water Cycle: Component # 5. Infiltration:

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Infiltration is an important process where rain water soaks into the ground, through the soil and underlying rock layers. Some of this water ultimately returns to the surface at springs or in low spots downhill. Some of the water remains underground and is called groundwater.

As the water infiltrates through the soil and rock layers, many of the impurities in the water are filtered out. This filtering process helps in cleaning of the water.

Water Cycle: Component # 6. Transpiration or Evapotranspiration:

This is performed by plants. As plants absorb water from the soil, the water moves from the roots through the stems to the leaves. Once the water reaches the leaves, some of it evaporates from the leaves, adding to the amount of water vapour in the air. This process of evaporation through plant leaves is called transpiration. In large forests, an enormous amount of water will transpire through leaves.

is the dissolution reaction of urea enthalpically favored?

Answers

No, since the biochemical system gains energy, it does not have enthalpy favored.

Is the breakdown of urea driven by entropy or by enthalpy?

Urea dissolves quickly in liquid, making it a good choice for enthalpy change measurements during dissolution. Entropy is the process' driving force since it is endothermic and spontaneous.

How do you determine whether a reaction was enthalpically advantageous?

If stronger bonds are created than are broken, if a stronger acid or base is generated in the result, or if more molecules are produced, processes will be thermodynamically beneficial. Other examples include when more molecules are produced.

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Question
Drag each label to the correct location in the table
Match each career to the correct career pathway.
Science and Mathematics
Engineering and Technology
anthropologist
fuel ce technician
Civil drafter
computer programmer
meteorologs
electrice
Submit

Answers

Answer: Science and Mathematics- Anthropologist, Meteorologist

Engineering and technology- Fuel Cell Technician, Computer programmer, electrical Drafter, Civil Drafter

Answer:

i have the same question  

Explanation:math and science is the meteorologs and the  anthropologist

What is the morality of a solution that contains 80.0 G Al2 (SO4)3 (aluminum sulfate) in 625 g H2O?

Answers

Answers

Explanations

Can anyone help me on this please !

Can anyone help me on this please !

Answers

Answer:

16.....................

HELP ASAP!!
A student mixes baking soda and vinegar in a glass. The results are shown at left. Do you think any new substances are being created in this mixture? If so, how do you know?

Answers

Yes I do belive that new substances are being formed because there is a chemical reaction between the baking soda and vinegar turning it into a bubbly substances instead of a powder and liquid.


Hope this helps pls mark me brainliest

Which biochemical pathways add or subtract carbons with the end product regenerating the molecule needed at the start?.

Answers

The biochemical pathways which add or subtract carbons with the end product regenerating the molecule needed at the start are called cyclic pathways.

A biochemical pathway is a chain of chemical reactions taking place within a cell. These reactions lead to the synthesis or breakdown of molecules which are important for various cellular processes such as growth, reproduction, and metabolism. Biochemical pathways are also called metabolic pathways.

Cyclic pathways are metabolic pathways in which the end product of the pathway regenerates the molecule required at the beginning of the pathway. These pathways are also called circular pathways or recycling pathways. Cyclic pathways play a crucial role in energy production and metabolism in living cells.Cyclic pathways involve the addition or subtraction of carbons in their reaction sequence. Some of the examples of cyclic pathways are the citric acid cycle, the Calvin cycle, and the urea cycle.

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Which of the following equation is balanced?

Which of the following equation is balanced?

Answers

Answer:

2fe2o3+3c=4fe+3co2

Explanation:

because 2×2 =4 fe &2×3=6 so we put it like 3co2

A 1.0 L balloon has a pressure of 2 atm. When the pressure increases to 1,000 kPa, what is the volume

Answers

When the pressure increases to 1,000 kPa, the volume of the balloon will be 0.203 L.

For calculating the final volume, we can use Boyle's Law. Boyle's Law states that the pressure and volume of a gas are inversely proportional to each other at a constant temperature. It can be mathematically represented as

P1V1 = P2V2

Where, P1 and V1 are the initial pressure and volume of the gas, and P2 and V2 are the final pressure and volume of the gas.

So, the equation becomes

2 atm × 1 L = 1,000 kPa × V2

Here, we need to convert 1,000 kPa to atm.

To convert kPa to atm, we divide kPa by 101.325 kPa/atm.

1,000 kPa ÷ 101.325 kPa/atm ≈ 9.87 atm

Now, substituting these values, we get

2 atm × 1 L = 9.87 atm × V2

Simplifying the above equation, we get

V2 = (2 atm × 1 L) ÷ 9.87 atm

V2 = 0.203 L

Therefore, the volume of the balloon when the pressure increases from 2 atm to 1,000 kPa is 0.203 L.

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Identify each structure based on its label. compound a is a amine. compound b is a amine. compound c is a amine. the common name for the amine d is .

Answers

Compound a is a primary amine compound b is a secondary amine compound c is a tertiary amine the common name for the amine d is ammonia

Amines are the organic compound that contain nitrogen atom with a lone pair and they are derived from ammonia means NH₃ in which one or more hydrogen atom is replaced by an alkyl or aryl group and they are known as alkylamines and arylamines

Primary amines are formed when one hydrogen atom in ammonia is substituted by an alkyl or aromatic group and secondary amine have two carbon bonded to the nitrogen and tertiary amines have the three carbon bonded to the nitrogen and ammonia is the gas made up of nitrogen and hydrogen

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2Na+S→Na2S

What is the total number of moles of S that react when 4.0 moles of Na were completely consumed?​

Answers

Answer:

2.0 moles S

Explanation:

To find the number of moles of S, you need to convert the moles Na to moles S via the mole-to-mole ratio. This ratio is represented by the coefficients in the balanced equation. Because you wish to find moles S, you want to put this number in the numerator. Because you want to eliminate the moles Na, this number should be in the denominator.

2 Na + 1 S ---> Na₂S

4.0 moles Na             1 mole S
----------------------  x  ---------------------  =  2.0 moles S
                                 2 moles Na

Those solids which can be described as possessing a highly regular arrangement of their components are said to be:______.

Answers

Those solids which can be described as possessing a highly regular arrangement of their components are said to be crystalline.

Crystalline materials don't have their constituent parts arranged in regular arrays like those found in crystalline solids, which are held together by homogeneous intermolecular forces.

Any non-crystalline substance whose atoms and molecules lack a clearly defined lattice pattern is said to be amorphous. These solids include gel, glass, as well as plastic.

Therefore, Those solids which can be described as possessing a highly regular arrangement of their components are said to be crystalline.

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3. When 100 cm³ of 1.00 mol dm sodium hydroxide solution, NaOH(aq), is added to 100 cm³ of 1.00 mol
dm hydrochloric acid, HCl(aq), the temperature increases from 19.3 °C to 26.1 °C.
Determine the enthalpy change of neutralization for the reaction.

Answers

From the calculation, the enthalpy of the reaction is  -56.8 kJ/mol.

What is the enthalpy of the reaction?

We know that the enthalpy has to do with the energy that may have been absorbed or evolved in a reaction. We know that in the case of the reaction that we have, the reaction occurs as follows; \(NaOH(aq) + HCl(aq) ---- > H_{2} O(l) + NaCl\)

The we have the following;

Number of moles acid = 1 M * 100/1000 = 0.1 moles

Number of moles of the base = 1 M * 100/1000 = 0.1 moles

In this case the reacting moles of the acid and the base are equal

Mass of the acid solution = 100 g

Mass of the base solution = 100 g

Total mass of the solution = 200 g

We then have the heat as;

H = mcdT

m = mass of the solution'

c = Heat capacity of the solution

dT = temperature change

Now;

H = 200  * 4.18 * (26.1 - 19.3)

H = 5.68 kJ

Then the enthalpy of the reaction is; -(5.68 kJ)/0.1 moles

= -56.8 kJ/mol

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Place these unknown pH test papers in order from most acidic to most alkaline.

Four pH indicator strips.

Answers

Answer:

In order of the colors of the pH test papers from most acidic solutions to most alkaline solutions, the colors of the pH test papers would be

               Red > Orange/Pink > Purple > Blue

(Red) Most acidic                                (Blue)Most alkaline

Explanation:

We can use the different colors of pH test papers to determine the acidity or alkalinity of a solution.

Definitions

A pH test paper is a strip of indicator paper which changes colour when placed in acidic or alkaline solutions.

A solution is said to be acidic if it turns the colour of the pH test paper RED

A solution is said to be alkaline/basic if it turns the colour of the pH test paper BLUE

Color changes

If the pH test paper turns red, then the solution is a very strongly acidic solution with pH less than 3 ( ie pH between 1-3)

If the pH test paper turns orange or pink, then the solution is a weak acidic solution with pH greater than 3 ut less than 7

If the pH test paper turns purple, then the solution is a weak alkaline solution with pH greater between 8-11

If the pH test paper turns blue, then the solution is a very strong alkaline solution with pH between 11 - 14

Summary

Therefore, in order of color changes from the most acidic solutions to most alkaline solutions, the colors of the pH test papers would be

Red > Orange/Pink > Purple > Blue

Most acidic                                Most alkaline

Answer:

d<c<a<b

Explanation:

any carbon substituent formed by removing one hydrogen from an alkane (for example, ch3ch2ch2-) is called a(n)

Answers

Any carbon substituent formed by removing one hydrogen from an alkane is called an alkyl group.

An alkane is a type of hydrocarbon that contains only single bonds and has the general formula \(CnH_{2}n+2\). When one hydrogen atom is removed from an alkane, the resulting carbon group is called an alkyl group. The name of the alkyl group depends on the number of carbon atoms in the group and the location of the removed hydrogen.

For example, \(CH_{3}CH_{2}CH_{2}\)- is a propyl group, while \(CH_{3}\)- is a methyl group. Alkyl groups are commonly used as substituents in organic chemistry and can greatly influence the chemical and physical properties of a molecule.

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Riders are associated with what divice
A.
a type of odometer
B.
Trundle wheel
C.
triple beam balance
D.
graduated cylinder

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D you welcome lovey.

at a time around 3.5 billion years ago, the primordial atmosphere that had a large fraction of carbon dioxide transformed into one that had a larger fraction of free oxygen. what cause this to occur?

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The primordial atmosphere of the Earth is the atmosphere that sustained Earth 5 million years ago when the sun and some other celestial bodies formed after an explosion of a supernova.

When Earth formed 4.6 billion years ago from a hot mix of gases and solids, it had almost no atmosphere. The surface was molten. As Earth cooled, an atmosphere formed mainly from gases spewed from volcanoes. It included hydrogen sulfide, methane, and ten to 200 times as much carbon dioxide as today's atmosphere.Free oxygen was present in the Earth's primordial atmosphere. Earth's primordial atmosphere most likely included: carbon dioxide, water vapor, sulfur dioxide, and methane.

Carbon dioxide gas was also removed form the early atmosphere by diffusing in the water of the oceans. volcanos also produced nitrogen gas, witch gradually built up in the earth atmosphere and there many have also been small proportions of methane and ammonia gases.

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can y'all help me :(

can y'all help me :(

Answers

it’s supposed to be compression, hope this helps

Answer:

compression

Explanation:

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