The standard free energy change, ΔG°', for this reaction is +6.7 kJ/mol. However, the observed free energy change (ΔG) for this reaction in pig heart mitochondria is +0.8 kJ/mol. What is the ratio of [isocitrate]/[citrate] in these mitochondria at 25.0 °C?

Answers

Answer 1
Mannnn son it’s 0.52 backwards baybeeee

Related Questions

The solubility of cadmium oxalate, , in 0.150 M ammonia is mol/L. What is the oxalate ion concentration in the saturated solution? If the solubility product constant for cadmium oxalate is , what must be the cadmium ion concentration in the solution? Now, calculate the formation constant for the complex ion

Answers

Answer:

[Cd²⁺] = 2.459x10⁻⁶M

Kf = 9.96x10⁶

Explanation:

Solubility of CdC₂O₄ is 6.1x10⁻³M and ksp is 1.5x10⁻⁸

The ksp of CdC₂O₄ is:

CdC₂O₄(s) ⇄ Cd²⁺(aq) + C₂O₄²⁻(aq)

ksp = [Cd²⁺] [C₂O₄²⁻] = 1.5x10⁻⁸

As solubility is 6.1x10⁻³M, concentration of C₂O₄²⁻ ions is 6.1x10⁻³M. Replacing:

[Cd²⁺] = 1.5x10⁻⁸ / [6.1x10⁻³M]

[Cd²⁺] = 2.459x10⁻⁶M

All Cd²⁺ in solution is 6.1x10⁻³M and exist as Cd²⁺ and as Cd(NH₃)₄²⁺. That means concentration of Cd(NH₃)₄²⁺ is:

[Cd(NH₃)₄²⁺] + [Cd²⁺] = 6.1x10⁻³M

[Cd(NH₃)₄²⁺] = 6.1x10⁻³M - 2.459x10⁻⁶M = 6.098x10⁻³M

[Cd(NH₃)₄²⁺] = 6.098x10⁻³M

In the same way, the whole concentration of NH₃ in solution is 0.150M, as you have 4ₓ6.098x10⁻³M = 0.024M of NH₃ producing the complex, the concentration of the free NH₃ is:

[0.150M] = [NH₃] + 0.024M

0.1256M = [NH₃]

The equilibrium of the complex formation is:

Cd²⁺ + 4 NH₃ → Cd(NH₃)₄²⁺

The kf, formation constant, is defined as:

Kf = [Cd(NH₃)₄²⁺] / [Cd²⁺] [NH₃]⁴

Replacing:

Kf = [6.098x10⁻³M] / [2.459x10⁻⁶M] [0.1256M]⁴

Kf = 9.96x10⁶

If you have 101 g of hydrogen gas (H2) and excess amount of nitrogen gas (N2), how many grams of ammonia gas (NH3) can you make?

Answers

Answer:

572. 3 g of NH3

Explanation:

Equation of the reaction: 3H2 + N2 ----> 2NH3

From the equation of reaction, 3 moles of H2 reacts with 1 mole of N2 to produce 2 moles of NH3.

Since N2 is in excess in the given reaction, H2 is the limiting reactant.

Molar mass of H2 = 2 g/mol

Molar mass of NH3 = 17 g/mol

Therefore 3 * 2 g of H2 reacts to produce 2 * 17 g of NH3

6 g of H2 produces 34 g of NH3

101 g of H2 will produce (34 * 101)/6 g of NH3 = 572.3 g of NH3

Therefore, 572.3 g of NH3 are produced

Answer:

572.33g of NH3.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

N2 + 3H2 —> 2NH3

Next, we shall determine the mass of the H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:

Molar Mass of H2 = 2x1 = 2g/mol

Mass of H2 from the balanced equation = 3 x 2 = 6g

Molar Mass of NH3 = 14 + (3x1) = 17g/mol

Mass of NH3 from the balanced equation = 2 x 17 = 34g.

From the balanced equation above,

6g of H2 reacted to produce 34g of NH3.

Finally, we can determine the mass of ammonia (NH3) produced by reacting 101g of H2 as follow:

From the balanced equation above,

6g of H2 reacted to produce 34g of NH3.

Therefore, 101g of H2 will react to produce = ( 101 x 34) / 6 = 572.33g of NH3.

Therefore, 572.33g of NH3 is produced from the reaction.

need help and quick answer as fast as possible

Answers

yes. arthropod are animals such as insects, crabs, lobsters etc

A chemist working as a safety inspector finds an unmarked bottle in a lab cabinet. A note on the door of the cabinet says the cabinet is used to store bottles of diethylamine, tetrahydrofuran, chloroform, ethanolamine, and acetone. First, from her collection of Material Safety Data Sheets (MSOS), the chemist finds the following information:
liquid density
diethylamine 1.1 gcm-3
tetrahydrofuran 0.7 9gcm-3
chloroform 0.71 gcm-3
ethanolamine 0.89 gcm-3
acetone 1.6 gcm-3
Next, the chemist measures the volume of the unknown liquid as 0.767 L and the mass of the unknown liquid as 682 g.
1. Calculate the density of the liquid.
2. Given the data above, is it possible to identify the liquid?
3. If it is possible to identify the liquid, do so.
a. dimethyl sulfoxide.
b. acetone.
c. diethylamine.
d. tetrahydrofuran .
e. carbon tetrachloride

Answers

Answer:

1. density = 0.89 g/cm3

2. Yes is possible to identify the liquid

3. ethanolamine

Explanation:

Data:

mass = 682 g

volume = 0.767 L = 767 mL or cm3

1.

To calculate the density of the liquid it is necessary to know that the density formula is:

[tex]density=\frac{mass(g)}{volume(cm^{3}) }[/tex]

The data obtained is replaced in the formula:

[tex]density=\frac{682g)}{767(cm^{3}) }=0.89\frac{g}{cm^{3} }[/tex]

2.

With the given data it is possible to identify the liquid, this because the density value is a basic property of each liquid.

3.

It is possible to determine what liquid it is, since when comparing the value obtained with those reported in the collection of Material Safety Data Sheets (MSOS), the value that agrees is that of ethanolamine.

What is a good title for this chart?

Answers

Answer:

pH of the acid

Explanation:

Nitroglycerin, an explosive, decomposes according to the following equation 4C3H5(NO3)3(s) → 12CO2(g) + 10H2O(g) + 6N2(g) + O2(g) Calculate the total volume of gases produced when collected at 1.45 atm, and 18.0°C from 2.70 × 102 g of nitroglycerin.

Answers

Answer:

6.65dm³

Explanation:

Equation of reaction,

4C3H5(NO3)3(s) → 12CO2(g) + 10H2O(g) + 6N2(g) + O2(g)

From the equation of reaction, 4 moles of Nitroglycerin gave 29 moles of various gases.

Molar mass of nitroglycerin C₃H₅(NO₃)₃ = 908g

Since all the product of the reaction are in gaseous phase, let's assume that law of conservation of matter is held hence there's no loss in mass.

908g of C₃H₅(NO₃)₃ = 908g of products

2.70×10²g of C₃H₅(NO₃)₃ = 2.70×10²g of products

Number of moles = mass / molar mass

Molar mass of C₃H₅(NO₃)₃ = 908g/mol

Number of moles = 2.70×10² / 908

Number of moles = 0.297 moles

But 1 mole = 22.4dm³

0.297mole = x dm³

x = (0.297 × 22.4) / 1

x = 6.65dm³

The volume of gas that'll be produced when 2.70×10²g of C₃H₅(NO₃)₃ would be 6.65dm³

please help!!!! Chem question

Answers

Answer : The net ionic equation will be,

[tex]Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)[/tex]

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The given balanced ionic equation will be,

[tex]Ba(OH)_2(aq)+H_2SO_4(aq)\rightarrow 2H_2O(aq)+BaSO_4(s)[/tex]

The ionic equation in separated aqueous solution will be,

[tex]Ba^{2+}(aq)+2OH^-(aq)+2H^{+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)+2H^+(aq)+2OH^{-}(aq)[/tex]

In this equation, [tex]H^+\text{ and }OH^-[/tex] are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

[tex]Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)[/tex]

According to the ideal gas law, what happens to the volume of a gas when the
temperature doubles (all else held constant)?
A. The volume stays constant.
B. The volume doubles.
OOO
C. It cannot be determined
D. The volume is halved

Answers

According to the ideal gas law, when the temperature of a gas doubles, its volume doubles as well (Option B).

What does the ideal law state?

The ideal gas law relates the pressure, volume, number of moles and temperature of an ideal gas.

Let's consider the equation of the ideal gas law.

P . V = n . R .T

V = n . R . T / P

As we can see, there is a direct relationship between the volume and the temperature. Thus, if the temperature doubles, the volume will double as well.

According to the ideal gas law, when the temperature of a gas doubles, its volume doubles as well (Option B).

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Describe why some acids are strong while other acids are weak

Answers

Answer:

I hope this help you. Mark me as brainliest and rate please

Explanation:

the terms strong and weak as applied to acids. As a part of this it defines and explains what is meant by pH, Ka and pKa.

It is important that you don't confuse the words strong and weak with the terms concentrated and dilute.

As you will see below, the strength of an acid is related to the proportion of it which has reacted with water to produce ions. The concentration tells you about how much of the original acid is dissolved in the solution.

It is perfectly possible to have a concentrated solution of a weak acid, or a dilute solution of a strong acid.

8) What is the molarity (M) of an aqueous solution containing 22.5 g of sucrose (C12H22011) in 35.5 mL of solution?
A) 3.52 M
B) 1.85 x 10-2M
C) 0.104 M
D) 0.0657 M
E) 1.85 M

Answers

Answer:

E) 1.85 M

Explanation:

M(C12H22O11) = 342.3 g/mol

22.5 g * 1mol/342.3 g = 0.0657 mol

35.5 mL = 0.0355 L

Molarity = mol solute/L solution = 0.0657 mol/0.0355L =1.85 mol/L = 1.85 M

The molarity of the aqueous solution is 1.85 M. The correct option is E) 1.85 M

From the question,

We are to determine the molarity (that is, concentration) of the given sucrose solution

First, we will determine the number of moles present in the given mass of sucrose

Mass of sucrose = 22.5 g

Using the formula

[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]

Molar mass of sucrose = 342.2965 g/mol

∴ Number of moles of sucrose present = [tex]\frac{22.5}{342.2965}[/tex]

Number of moles of sucrose present = 0.0657325 moles

Now, for the molarity (concentration) of the sucrose solution

From the formula

Number of moles = Concentration × Volume

Then,

[tex]Concentration = \frac{Number\ of\ moles}{Volume}[/tex]

From the question,

Volume = 35.5 mL = 0.0355 L

∴ [tex]Concentration = \frac{0.0657325}{0.0355}[/tex]

Concentration = 1.85 M

Hence, the molarity of the aqueous solution is 1.85 M. The correct option is E) 1.85 M

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tertbutylamine and ammonia. Which is more basic

Answers

Answer:

ammonia

Explanation:

Click on the Delta H changes sign whan a process is reversed button within the activity and analyze the relationship between the two reactions that are displayed. The reaction that was on the screen when you started and its derivative demonstrate that the reaction enthalpy, ΔH, changes sign when a process is reversed. Consider the reaction H2O(l)→H2O(g), ΔH =44.0kJ What will ΔH be for the reaction if it is reversed?

Answers

Answer:

ΔH = - 44.0kJ

Explanation:

H2O(l)→H2O(g), ΔH =44.0kJ

In the reaction above, liquid water changes to gaseous water. This occurs through a process known as boiling. This process requires heat, hence the ΔH  is positive.

If he reaction is reversed, we have;

H2O(g)→H2O(l)

In this reaction, gaseous water changes to liquid water. This process is known as condensation. The water vapor loses heat in this reaction. Hence ΔH would be negative but still  have the same value.

When you turn on the air conditioner during a hot summer day the cooler air will sink to the floor, while warmer air rises to the
ceiling
Which type of heat transfer is this an example of?
(A) conduction
(B) convection
(C) radiation
(D)
kinetic

Answers

It’s B convection it makes air rise and sink

What type of bond will be formed for atoms that have a +1 or -1 charge?

Answers

covalent bonding. example lithium bond with fluorine since lithium has a valence charge of +1 and fluorine has a valence charge of +7. they will bond together to give u a stable full electron

What can be known about the salt sample that Gerry is looking at?

Answers

Answer:

That its small pointed. Pink(Himalayan salt)or white(normal salt)

Explanation:

Summa dees questions are so stupid, deys makin me salty.

What is the rate of a reaction if the value of kis 0.1, [A] is 1 M, and [B] is 2 M?
Rate = K[A]2[B]2
A. 1.6 (mol/L)/s
B. 0.8 (mol/L)/S
C. 0.2 (mol/L)/S
D. 0.4 (mol/L)/S

Answers

Answer:

D.  0.4 (mol/L)/S

Explanation:

You simply have to plug in the given values into the rate law.

Rate = k[A][B]

Rate = (0.1)(1)²(2)²

Rate = (0.1)(1)²(4)²

Rate = 0.4

According to the following reaction, what amount of Al2S3 remains when 20.00 g of Al2S3 and 2.00 g of H2O are reacted? A few of the molar masses are as follows: Al2S3 = 150.17 g/mol, H2O = 18.02 g/mol.

Answers

Answer:

Mass of Al2S3 remaining is 17.212 g

Explanation:

Equation of the reaction is given below:

Al2S3 + 6H2O -----> 2Al(OH)3 + 3H2S

From the balanced equation above

6 mole of H20 reacts with 1 mole of Al2S3

i.e. 6 * 18.02 g of H2O reacts with 1 * 150.71 g of Al2S3

= 108.12 g of H2O reacts with 150.71 g of Al2S3

Therefore 2.0 g of water will react with 2.0 * (150.71/108.12) g of Al2S3

= 2.788 g of Al2S3

Mass of Al2S3 remaining = 20.0 g - 2.788 g = 17.212 g

According to the properly balanced chemical equation, the amount (mass) of [tex]AL_2S_3[/tex] that remains after the chemical reaction is 17.22 grams.

Given the following data:

Mass of [tex]AL_2S_3[/tex] = 20.00 gramsMass of [tex]H_2O[/tex] = 2.00 gramsMolar mass of [tex]AL_2S_3[/tex] = 150.17 g/molMolar mass of [tex]H_2O[/tex] = 18.02 g/mol.

To calculate the amount (mass) of [tex]AL_2S_3[/tex] that remains after the chemical reaction:

First of all, we would write a properly balanced chemical equation for this chemical reaction.

                        [tex]Al_2S_3 + 6H_2O ---> 2Al(OH)_3 + 3H_2S[/tex]

By stoichiometry:

1 mole of [tex]AL_2S_3[/tex] reacts with 6 moles of [tex]H_2O[/tex]

Next, we would calculate the mass of each compound.

For [tex]AL_2S_3[/tex]:

[tex]Mass = Number\;of\;moles \times molar\;mass\\\\Mass = 1 \times 150.17[/tex]

Mass = 150.17 grams

For [tex]H_2O[/tex]:

[tex]Mass = Number\;of\;moles \times molar\;mass\\\\Mass = 6 \times 18.02[/tex]

Mass = 108.12 grams

108.12 grams of [tex]H_2O[/tex] = 150.17 grams of [tex]AL_2S_3[/tex]

2.00 grams of [tex]H_2O[/tex] = X grams of [tex]AL_2S_3[/tex]

Cross-multiplying, we have:

[tex]108.12 \times X = 150.17 \times 2\\\\108.12X = 300.34\\\\X = \frac{300.34}{108.12}[/tex]

X = 2.78 grams of [tex]AL_2S_3[/tex]

Remaining mass = [tex]20.00 - 2.78[/tex]

Remaining mass = 17.22 grams of [tex]AL_2S_3[/tex]

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1. ______The force that keeps the nucleons bound inside the nucleus of an atom
A. Strong electrostatic force
B. Strong nuclear force
C. Strong centripetal force
D. Gravitational attraction

2._____The amount of energy needed to split the nucleus into individual protons and neutrons
A. Nuclide transfer energy
B. Nuclear binding energy
C. Mass energy equivalence
D. Nuclear energy
3._______ The difference between the mass of the nucleons and the mass of an Atom
A. Mass of nucleus
B. Mass defect
C. Atomic mass
D. Isotopic mass

Answers

Answer:

1). strong nuclear force 2). nuclear binding energy 3), mass defect

Explanation:

Right on Edge

1. Strong nuclear force the force that keeps the nucleons bound inside the nucleus of an atom.

2. Nuclear binding energy the amount of energy needed to split the nucleus into individual protons and neutrons.

3. Mass defect the difference between the mass of the nucleons and the mass of an Atom.

What is strong nuclear force ?

The term strong nuclear force is defined as the force that binds protons and neutrons together. It also binds them all together in a nucleus and is responsible for the energy released in nuclear reactions.

The examples of strong nuclear force are the force that hold protons and neutrons in nuclei of atoms. The elements' greater than the hydrogen atom. The fusion of hydrogen into helium in the sun's core.

Thus, 1. option B, 2. option B and 3. option B is correct.

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iron oxide + oxygen equals to ?

Answers

Answer:

It's ferric oxide Fe2O3

Explanation:

I don't say u must have to mark my ans as brainliest but if it has really helped u plz don't forget to thank me plz...

Iron+ oxygen= Fe+ 3O2 hopefully this will help!

Consider the following reaction. I– 2 H2O2 (l) 2 H2O (l) + O2 (g) A solution contains 15 mL 0.1 M KI, 15 mL of DI water and 5 mL of 3% H2O2. After the decomposition of H2O2 is complete, you titrate the solution with 0.1 M AgNO3. If the catalyst, I–, is not consumed in the reaction and is completely recovered, what volume of the 0.1 M AgNO3 is required to reach the end point?

Answers

Answer:

Explanation:

The given chemical reaction is:

[tex]2H_2O_{(l)} \to^{I^-}} 2H_2O_{(l)}+O_2_{(g)}[/tex]

From above equation  [tex]I^-[/tex] serves as catalyst which is not consumed by the reaction and also it is completely recovered; as a result to that , the full volume of KI will definitely react with AgNO₃.

Given that :

the volume of potassium iodide [tex]V_{KI} = 15 \ ml[/tex]

the molarity of potassium [tex]M_{KI} = 0.1 \ M[/tex]

the volume of distilled water [tex]V_W = 15 \ mL[/tex]

The volume of 3% [tex]H_2O_2 \ \ V_{H_2O_2} = 5 \ mL[/tex]

Molarity of AgNO₃ [tex]M_{AgNO_3} = 0.1 \ M[/tex]

Let take an integral look with the reaction between KI and AgNO₃; we have

[tex]KI + AgNO_3 \to KNO_3 + AgI[/tex]

At the end point; the moles of KI will definitely be equal to the moles of AgNO₃

So;

[tex]M_{KI}V_{KI}= M_{AgNO_3}V_{AgNO_3} \\ \\ V_{AgNO_3} = \dfrac{M_{KI}V_{KI}}{M_{AgNO_3}} \\ \\ \\ V_{AgNO_3} = \dfrac{ 0.1*15}{0.1}[/tex]

[tex]V_{AgNO_3} = 15 \ ml[/tex]

Thus; the volume of 0.1 M AgNO₃  needed to reach the end point is 15 mL

The three‑dimensional structure of a generic molecule is given. Identify the axial and equatorial atoms in the three‑dimensional structure. What is the shape of this molecule?

Answers

Answer:

Explanation:

CHECK THE ATTACHMENT FOR THE COMPLETE QUESTION AND THE DETAILED EXPLANATION

NOTE:

Equatorial atoms are referred to atoms that are attached to carbons in the cyclohexane ring which is found at the equator of the ring.

Axial atoms are atoms that exist in a bond which is parallel to the axis of the ring in cyclohexane

Give the IUPAC name for the following compound

Answers

Answer:

3–bromo–5–chloro–4–methylhexane.

Explanation:

To name the compound given in the question, the following must be observed:

1. Locate the longest continuous carbon chain. This gives the parent name of the compound. In this case, the longest chain is carbon 6 i.e Hexane.

2. Identify the substituents attached. In this case the substituents attached are:

a. Chloro i.e Cl.

b. Bromo ie Br.

c. Methyl i.e CH3.

3. Give the substituents the lowest possible count alphabetically. Bromo comes before Chloro alphabetically, so we shall consider bromo first. Their positions are given below:

Bromo i.e Br at carbon 3

Chloro i.e Cl is at carbon 5

Methyl i.e CH3 is at carbon 4

4. Combine the above to get the name of the compound.

Therefore, the name of the compound is:

3–bromo–5–chloro–4–methylhexane.

A base has a molarity of 1.5 M with respect to the hydroxyl ion (OH-) concentration. If 7.35 cm³ of this base is taken and diluted to 147 cm³, then what is the concentration of the hydroxyl ion. How many moles of hydroxyl ion are there in the 7.35 cm³? In the 147 cm³?

Answers

Answer:

0.077M is the concentration of the hydroxyl ion

Explanation:

If 7.35 cm3 of this base is take and diluted to 147 cm3, then what is the concentration of the hydroxyl ion?

Use the dilution equation:

M1V1 = M2V2

M1 * 147cm³ = 1.5 M * 7.35 cm³

M1 = 1.5 M * 7.35 cm³ / 147 cm³

M1 = 0.077 M

0.077M is the concentration of the hydroxyl ion

How many moles of hydroxyl ion are there in the 7.35 cm3?

1000 cm³ contains 1.5 mol OH- ions

7.35 cm³ contains : 7.35 cm³ / 1000 cm³ *1.5 mol

= 0.011025 mol

Answer correct to 2 significant digits = 0.011 mol OH- ions.

What is the systematic name of the following compound?
Mn3(PO4)2
The polyatomic ion phosphate has the formula PO

Answers

Answer:

Manganese(II) phosphate | Mn3(PO4)2 - PubChem

Answer:

Magnese(ll) posphate M23 (p042) Molecular weight.

That is what the leters stand for!

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Consider 10.0 g of helium gas (He) in a rigid steel container. If you add 10.0 g of neon gas (Ne) to this container, which of the following best describes what happens? (Assume the temperature is constant.)
a) The pressure in the container doubles.
b) The pressure in the container more than doubles.
c) The volume of the container doubles.
d) The volume of the container more than doubles.
e) The pressure in the container increases but does not double.

Answers

Answer: (e) The pressure in the container increases but does not double.

Explanation:

To solve this, we need to first remember our gas law, Boyle's law states that the pressure and volume of a gas have an inverse relationship. That is, If volume increases, then pressure decreases and vice versa, when temperature is held constant. Therefore, increasing the volume in this case does not double the pressure owning to out gas law, but an increase in pressure would be noticed if temperature is constant

The pressure in the container increases but does not double.

At constant temperature and volume, the pressure of a given mass of gas is directly proportional to the number of moles of gas present.

Number of moles of He = 10 g/4 g/mol = 2.5 moles

Number of moles of Ne = 10 g/20 g/mol  = 0.5 moles

We can see that the number of moles only increases by 1/5 of its initial value therefore, the pressure in the container increases but does not double.

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A mixture of compounds containing diethylamine, phenol, ammonia, and acetic acid is separated using liquid-liquid extraction as follows: Step 1: Concentrated HCl is added followed by draining the aqueous layer. Step 2: Dilute NaOH is added to the organic layer followed by draining the aqueous layer. Step 3: Concentrated NaOH is added to the organic layer followed by draining the aqueous layer. Which compound would you expect to be extracted into the aqueous layer after the addition of dilute HCl, step 1? Group of answer choices

Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is  ammonia

Explanation:

The mixture contains two base compound which are

           ammonia,

and     diethylamine

Now the addition of HCl which is  a strong acid in step 1  will cause the protonation of  the  two base compound , which makes the soluble hence resulting in them being extracted to the aqueous layer as represented in below

       [tex]NH_3 + HCl\to NH_4 ^{+} + Cl^-[/tex]

and

     [tex](CH 3CH 2) 2NH + HCl \to (CH 3CH 2) 2NH_2^{+} + Cl[/tex]

       

I WILL GIVE BRAINLIEST

Answers

Molarity= no. of molecules of solute /1 liter
one moles of sodium hydroxide =49 gm of sodium hydroxide
So we can say that if we want to prepare 1 molar NaOH solution then we need 40 gm NaOH dissolve in one liter of water so it can become one 1 molar NaOH solution.

Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produced 39.61 g CO2 and 9.01 g H2O. The molar mass of equilin is 268.34 g/mol. Find its molecular formula.

Answers

Answer: The molecular formula for the given organic compound is [tex]C_{18}H_{20}O_2[/tex]

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of [tex]CO_2=39.61g[/tex]

Mass of [tex]H_2O=9.01g[/tex]

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 39.61 g of carbon dioxide, [tex]\frac{12}{44}\times 39.61=10.80g[/tex] of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 9.01 g of water, [tex]\frac{2}{18}\times 9.01=1.00g[/tex] of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.80 + 1.00) = 1.62 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = [tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.80g}{12g/mole}=0.9moles[/tex]

Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1g}{1g/mole}=1moles[/tex]

Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.62g}{16g/mole}=0.10moles[/tex]

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.10 moles.

For Carbon = [tex]\frac{0.9}{0.10}=9[/tex]

For Hydrogen = [tex]\frac{1}{0.10}=10[/tex]

For Oxygen = [tex]\frac{0.10}{0.10}=1[/tex]

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9 : 10 : 1

Hence, the empirical formula for the given compound is [tex]C_9H_{10}O[/tex]

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

[tex]n=\frac{\text{Molecular mass}}{\text{Empirical mass}}[/tex]

We are given:

Mass of molecular formula = 268.34 g/mol

Mass of empirical formula = 134 g/mol

Putting values in above equation, we get:

[tex]n=\frac{268.34g/mol}{134g/mol}=2[/tex]

Multiplying this valency by the subscript of every element of empirical formula, we get:

[tex]C_{(9\times 2)}H_{(10\times 2)}O_{(1\times 2)}=C_{18}H_{20}O_2[/tex]

Thus, the molecular formula for the given organic compound is [tex]C_{18}H_{20}O_2[/tex].

Glycine, C2H5O2N, is important for biological energy. The combustion reaction of glycine is given by the equation 4C2H5O2N(s) + 9O2(g) → 8CO2(g) + 10H2O(l) + 2N2(g) ΔH°rxn = –3857 kJ/mol Given that ΔH°f[CO2(g)] = –393.5 kJ/mol and ΔH°f[H2O(l)] = –285.8 kJ/mol, calculate the enthalpy of formation of glycine.

Answers

Answer:

ΔH°f C₂H₅O₂N(s)  = -537.2kJ

Explanation:

Based on the reaction:

4 C₂H₅O₂N(s) + 9O₂(g) → 8CO₂(g) + 10H₂O(l) + 2N₂(g)

ΔHrxn = ΔH°f products - ΔH°f reactants.

As:

ΔH°fO₂(g) = 0

ΔH°fCO₂(g) = -393.5kJ/mol

ΔH°fH₂O(l) = -285.8kJ/mol

ΔH°fN₂(g) = 0

The ΔHrxn is:

ΔHrxn = (8×-393.5kJ/mol + 10×-285.8kJ/mol) - (4×ΔH°fC₂H₅O₂N(s)) = -3857kJ/mol

-6006kJ/mol - (4×ΔH°fC₂H₅O₂N(s)) = -3857kJ/mol

-4×ΔH°fC₂H₅O₂N(s) = 2149kJ/mol

ΔH°fC₂H₅O₂N(s) = 2149kJ/mol / -4

ΔH°f C₂H₅O₂N(s)  = -537.2kJ

Trans-4-hexen-3-ol can be synthesized starting from acetaldehyde. One of the key reagents is ethyl grignard.
1. Synthesize ethyl grignard from acetaldehyde in the steps below using the reagents provided.
2. Synthesize (trans)-4-hexen-3-ol from acetaldehyde.

Answers

find the given attachment

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