The Michaelis-Menten equation is an expression of the relationship between the initial velocity,V0, of an enzymatic reaction and substrate concentration, [S]. There are three conditions that are useful for simplifying the Michaelis-Menten equation to an expression from which the effect of [S] on the rate can be more readily determined. Match the condition (e.g. [S] = Km) with the statement(s) that describe it:

1. Doubling [S] will almost double the rate.
2. Half of the active sites are occupied by substrate.
3. About 90% of the active sites are occupied by substrate.
4. Doubling [S] will have little effect on the rate.
5. Less than 10% of the active sites are occupied by substrate.
6. This condition will result in the highest rate.

Answer:

No HBr will be left over by the chemical reaction.

Explanation:

The reaction of hydrobromic acid with sodium hydroxide is:

HBr + NaOH → NaBr + H₂O

80,0g of HBr are:

80,0g×(1mol/80,91g) = 0,989 moles

55,0g of NaOH are:

55,0g×(1mol/40g) = 1,375 moles

That means excess moles of NaOH are:

1,375-0,989 = 0,286 moles of NaOH

Thus, HBr is the limiting reactant and will be used completely. As a result, no HBr will be left over by the chemical reaction.

I hope it helps!

Answer:

To produce 0.9888 moles of NaBr and 0.9888 moles of H2O, there will remain 0.3862 moles of NaOH. All the HBr will be consumed, no HBr will be left over by the chemical reaction

Explanation:

Step 1: Data given

Mass of hydrobromic acid = 80.00 grams

Mass of sodium hydroxide = 55.00 grams

Molar mass HBr = 80.91 g/mol

Molar mass NaOH = 40 g/mol

Step 2: The balanced equation

HBr(aq) + NaOH(s) → NaBr(aq) + H₂O(l)

Step 3: Calculate moles of HBr

Moles HBr = Mass HBr / molar mass HBr

Moles HBr = 80.00 grams / 80.91 g/mol

Moles HBr = 0.9888 moles

Step 4: Calculate moles NaOH

Moles NaOH = 55.00 grams / 40.00 g/mol

Moles NaOH = 1.375 moles

Step 5: Calculate limiting reactant

For 1 mol of HBr we need 1 mol of NaOH to produce 1 mol of NaBr and 1 mol of H2O

The limiting reactant is HBr. It will completely be consumed. (0.9888 moles). NaOH is in excess. There will react 0.9888 moles. There will remain 1.375 - 0.9888 = 0.3862 moles

This means to produce 0.9888 moles of NaBr and 0.9888 moles of H2O, there will remain 0.3862 moles of NaOH. All the HBr will be consumed, no HBr will be left over by the chemical reaction

Explanation:

In a hexagonal-close-pack (HCP) unit cell, the ratio of lattice points to octahedral holes to tetrahedral holes =  1 : 1 : 2

let the :

Number of lattice point = 1x.

Number of octahedral points = 1x

Number of tetrahedral  points = 2x

If anions occupy the HCP lattice points and cations occupy half of the octahedral holes.

Number of anions occupying the HCP lattice points, A= 1x

Number of cations occupying the octahedral points, B =  1x

The formula of the compound will be = [tex]A_{1x}B_{1x}=AB[/tex]

If anions occupy the HCP lattice points and cations occupy all of  the octahedral and the tetrahedral holes.

Number of anions occupying the HCP lattice points, A= 1x

Number of cations occupying the octahedral points, B =  x

Number of cations occupying the tetrahedral points, B =  2x

total number of cations = x + 2x = 3x

The formula of the compound will be = [tex]A_{1x}B_{3x}=AB_3[/tex]

Answer: [tex]KOCH_2CH_3[/tex], [tex]CH_3CH_2COONa[/tex], NaOH are all ionic compounds.

Explanation:

It is known that ionic compounds are the compounds in which one atom transfer its valence electrons to another atom. Hence, during this transfer partial opposite charges tend to develop on the combing atoms due to which strong force of attraction exists between the atoms.

An ionic bond is always formed between a metal and a non-metal.

For example, [tex]KOCH_2CH_3[/tex], [tex]CH_3CH_2COONa[/tex], NaOH are all ionic compounds.

On the other hand, a compound formed due to sharing of electrons between the combining atoms is known as a covalent compound. Generally, a non-metal with same or different non-metal tends to form a covalent bond.

For example, [tex]POCl_{3}[/tex], [tex]SOCl_{2}[/tex] etc are all covalent compounds.

Thus, we can conclude that [tex]KOCH_2CH_3[/tex], [tex]CH_3CH_2COONa, NaOH are all ionic compounds.

Answer:

A) - 0.26 V

Explanation:

Here Ni undergoes oxidation by loss of electrons, thus act as anode. Standard hydrogen electrode undergoes reduction by gain of electrons and thus act as cathode.

[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]

Where both [tex]E^0[/tex] are standard reduction potentials.

Given that:- [tex]E^0=0.26\ V[/tex]

Also, [tex]E^0_{[H^{+}/H_2]}=0\ V[/tex]

So,

[tex]E^0=E^0_{[H^{+}/H_2]}-E^0_{[Ni^{2+}/Ni]}[/tex]

[tex]0.26\ V=0\ V- E^0_{[Ni^{2+}/Ni]}[/tex]

[tex]E^0_{[Ni^{2+}/Ni]}=-0.26\ V[/tex]

The standard reduction potential for the Ni2+/Ni half-cell is - 0.26 V.

The standard hydrogen electrode is a reference electrode that was arbitrarily assigned an electrode potential of 0.00V. We know that metals that are above hydrogen in the electrochemical series will always have a negative electrode potential.

As such, the standard reduction potential for the Ni2+/Ni half-cell is - 0.26 V.

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Answer:

There are formed 98.05 g of MnCl₂

Explanation:

The reaction is this one:

MnO₂ + 4 HCl   →  MnCl₂ +  2 H₂O + Cl₂

First of all, determinate moles. Divide mass /molar mass

150 g / 36.45 g/m = 4.11 moles of HCl

Ratio between HCl and MnCl₂ is 4:1

4 moles of HCl produce 1 mol of Chloride

4.11 moles of HCl  'll produce (4.11 . 1)/ 4 =1.03 moles of chloride

Molar mass . Moles = Mass

Molar Mass MnCl₂ = 95.2 g/m

95.2 g/m  . 1.03 moles = 98.05 grams

Answer:

There are 129.4 grams of MnCl2 formed

Explanation:

Step 1: Data given

Mass of HCl = 150.0 grams

MnO2 = excess

Molar mass of HCl = 36.46 g/mol

Step 2: The balanced equation

MnO2+4HCl → MnCl2+2H2O+Cl2

Step 3: Calculate moles of HCl

Moles HCl = Mass HCl / molar mass HCl

Moles HCl = 150.0 grams / 36.46 g/mol

Moles HCl = 4.114 moles

Step 4: Calculate Moles of MnCl2

For 1 mol of MnO2 we need 4 moles of HCl to produce 1 mol of MnCl2, 2 moles of H2O and 1 mol Cl2

For 4.114 moles of HCl we'll have 4.114/4 = 1.0285 moles of MnCl2

Step 5: Calculate mass of MnCl2

Mass MnCl2 =moles MnCl2 * molar mass MnCl2

Mass MnCl2 = 1.0285 * 125.84 g/mol

Mass MnCl2 = 129.4 grams

There are 129.4 grams of MnCl2 formed

Answer:

[tex]MF_2[/tex]

Explanation:

In a simple cubic lattice lattice, the atoms are present at the eight corners of the cibe.

Since it is mentioned that the fluorine is present at the corners and also 1 corners are shared by 8 unit cells. So, share of atom in one unit cell is:- [tex]8\times \frac{1}{8}=1[/tex]

Also, the metal, M occupy half of the body centre. A cube has only one body centre. So, share of M in each unit cell:- [tex]\frac{1}{2}[/tex]

Thus, the formula is:-

[tex]M_{\frac{1}{2}}F[/tex] Or simplifying [tex]MF_2[/tex]

Answer:

Explanation:

given that fluoride ions occupy simple cubic lattice and metal ions occupy the body center of half the cube

therefore,

The number of [tex]F^-[/tex] ions per unit cell = [tex]\frac{1}{8}*1[/tex]

[tex]= \frac{1}{8}[/tex]

There is only one body center per unit cell

The total number of metal ions per unit cell = [tex]\frac{1}{2}*1[/tex]

[tex]= \frac{1}{2}[/tex]

Therefore the formula for metal fluoride

[tex]= MF_2[/tex]

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Answer:B

Explanation:

The solubility product is the product of the concentration of all the ions present as found in the rate equation. The solubility product is the equilibrium expression that shows the extent to which a substance dissolves in water. The full solution is shown in the image attached.

Answer:

4.0E^-18

which makes the answer A

Answer:

A

Explanation:

As electrolysis progresses, the iodide ion migrates towards the anode (positive electrode) and is discharged as iodine gas while the metal migrates towards the cathode (negative electrode). The evolution of gas at the positive electrode signifies the reaction:

2I-=I2 +2e

The brown color formed near an electrode in a KI solution during electrolysis indicates the positive terminal due to oxidation of iodide ions. However, the provided choices have an incongruity as the brown color appears near the positive electrode, not the negative one, as option (C) incorrectly states.

To identify the polarities of the power supply terminals using a 0.1-molar KI solution with platinum electrodes, we need to understand the processes occurring at each electrode during electrolysis. When direct current is applied, oxidation occurs at the anode (positive electrode) and reduction occurs at the cathode (negative electrode).

For a solution containing potassium iodide (KI), oxidation at the anode results in the liberation of iodine, which can cause a brown coloration in the solution near the positive electrode. At the cathode, reduction occurs, which could lead to the evolution of hydrogen gas. Hence, the brown color appearing near the anode indicates that it is the positive terminal.

If the terminals are connected and the solution becomes brown near one of the electrodes, we can identify that electrode as the anode (positive). Therefore, the correct answer is:

(C) A brown color will appear in the solution near the negative electrode. However, this statement seems to contradict the previous explanation—it should state that the brown color appears near the positive electrode due to the oxidation of iodide ions. Hence, the correct phenomenon described does not match any of the given answer choices perfectly, and ambiguities in (C) suggest that a more detailed investigation may be required to ensure the most accurate determination of the terminal polarities.

Answer:

The molal freezing point depression constant for X is 4.12 °C/m

Explanation:

Apply the colligative property of freezing point depression:

ΔT = Kf . m . i

We assume the X lquid is non electrolytic, so i = 1

ΔT = T° freezing point of solvent pure - T° freezing point of solution

m = molality (moles of solute in 1kg of solvent)

Kf = Cryoscopic constant (the data we were asked for)

ΔT = 0.4° - (- 0.5°) = 0.9°

Let's calculate molality

Molar mass of urea = 60.06 g/m

Moles of urea = mass of urea / molar mass

5.90 g / 60.06 g/m = 0.098 moles

This moles are in 450 g of solvent, prepare a rule of three to find out the moles in 1000 g

450 g ____ 0.098 m

1000 g ____ ( 1000.  0.098)/450 = 0.218 m

0.9° = Kf . 0.218 m . 1

0.9° / 0.218 m = Kf → 4.12 °C/m

Answer:

Four products are possible.(in attachment)

2,3,4,5 carbons attached to the bromine.

Explanation:

NBS is N-bromosuccinamide which is used to brominate the allylic and benzylic positions of the compound.

Mainly used for the allylic bromination Here, four products are possible among them two are formed by the direct allylic bromination and remaining two products are formed by the rearrangement of the radical formation.

2,3,4,5 carbons of the compound are attached to the bromine.

Final answer:

When the alkene is brominated with NBS, bromine can become attached to multiple carbons in the structure.

Explanation:

When the alkene is brominated with NBS, the reaction can occur at multiple carbons. In this case, the alkene is a hexagonal structure of benzene with two hydrogen atoms replaced by bromine atoms. The bromine atoms can become attached to the carbons adjacent to each other, 1 carbon apart, or 2 carbons apart. So, the bromine can attach to all the numbered carbons in the structure, leading to different isomers of the product.

Hydrogen bonds in DNA are used instead of covalent bonds because they offer the right balance of strength and flexibility, allowing DNA strands to easily 'unzip' for replication and transcription while maintaining the double helix structure. They enable the precise pairing of complementary bases, crucial for genetic information integrity.

Hydrogen bonds, rather than covalent bonds, are responsible for holding the bases in DNA together. The reason for this lies in their strength and the functionality they provide to the DNA molecule. While covalent bonds are much stronger and hold the atoms of the individual DNA molecules together, hydrogen bonds are perfectly suited for their role in the double helix structure.

Hydrogen bonds are strong enough to keep the complementary bases paired but sufficiently weak to allow the strands to 'unzip' for essential biological processes such as replication and transcription.

The interplay between adenine and thymine, as well as guanine and cytosine, through specific hydrogen bonding arrangements is crucial for the stability and integrity of DNA. The cumulative effect of millions of these bonds provides stability for the double helix structure while still allowing it to separate when needed. The precise alignment of hydrogen bonds ensures the proper matching of bases, which is vital for preserving genetic information and ensuring accurate DNA copying in cells.

In summary, hydrogen bonds are essential because they provide a balance between stability and flexibility, which is paramount for the dynamic role DNA plays within living organisms.

Answer:

0.0269 M

Explanation:

There is some info missing. I think this is the original question.

Suppose 0.816 g of potassium nitrate is dissolved in 300 mL of a 14.0 mM aqueous solution of sodium chromate. Calculate the final molarity of potassium cation in the solution. You can assume the volume of the solution doesn't change when the potassium nitrate is dissolved in it. Be sure your answer has the correct number of significant digits.

The molecular equation corresponding to this reaction is:

2 KNO₃(aq) + Na₂CrO₄(aq) ⇄ K₂CrO₄(aq) + 2 NaNO₃(aq)

The full ionic equation is:

2 K⁺(aq) + 2 NO₃⁻(aq) + 2 Na⁺(aq) + CrO₄²⁻(aq) ⇄ 2 K⁺(aq) + CrO₄²⁻(aq) + 2 Na⁺(aq) + 2 NO₃⁻(aq)

As we can see, the moles of K⁺ are equal to the initial moles of KNO₃. The molar mass of KNO₃ is 101.10 g/mol. The moles of KNO₃ (and K⁺) are:

0.816 g × (1 mol/ 101.10 g) = 8.07 × 10⁻³ mol

The molarity of K⁺ is:

8.07 × 10⁻³ mol / 0.300 L = 0.0269 M

Answer:

Heat transfer = 3564 Jolues

The same value

Explanation:

The heat of combustion is the heat released per 1 mole of substunce experimenting the combustion at standard conditions of pressure and temperature ( 1 atm, 298 K):

Qtransfer = - mol x ΔHºc Qtransfer

So look up in appropiate reference table ΔHºc  and solve the problem:

ΔHºc  = - 891 kJ/mol

Qtransfer = - (4 x 10³ mol x  -891 kJ/mol ) = 3564 J

if the combustion were achieved with 100 % excess air, the result will still be the same. As long as the standard conditions are maintained, the heat of combustion remains constant. In fact in many cases the combustion is performed under excess oxygen to ensure complete combustion.

Final answer:

The heat transfer from the combustion chamber when 4 kmol of methane is burned completely with stoichiometric air is 3561.6 kJ. With 100 percent excess air, the heat transfer remains the same as excess air does not change the heat released from the combustion reaction.

Explanation:

Heat Transfer During Combustion of Methane

To determine the heat transfer from the combustion chamber when 4 kmol of methane (CH4) is burned with the stoichiometric amount of air, we must use the thermochemical equation for the combustion of methane. The equation indicates that when 1 mol of methane is combusted, 890.4 kilojoules of heat is released. Therefore, for 4 kmol, the heat release would be 4 kmol × 890.4 kJ/kmol = 3561.6 kJ.

If the combustion is achieved with 100 percent excess air, the amount of heat transferred would not change because the stoichiometry of the reaction remains consistent and the same amount of methane is being burned. However, additional air does not contribute to the heat of the reaction unless it causes incomplete combustion or other side reactions, which in this case is disregarded.

Since the conditions specify that reactants and products are maintained at 25°C and 1 atm, and the water in the products is in the liquid form, we can assume that no extra energy is required for heating up or cooling down the substances and the entire heat of combustion is transferred out from the system.

Answer:

The metal has a molar mass of 65.37 g/mol

Explanation:

Step 1: Data given

Mass of the metal = 0.708 grams

Volume of hydrogen = 275 mL = 0.275 L

Atmospheric pressure = 1.0079 bar = 0.9947 atm

Temperature = 25°C

Vapor pressure of water at 25 °C = 0.03167 bar = 0.03126 atm

Step 2: The balanced equation

M(s) + H2SO4(aq) ⟶ MSO4 (aq) + H2(g)

Step 3: Calculate pH2

Atmospheric pressure = vapor pressure of water + pressure of H2

0.9947 atm = 0.03126 atm + pressure of H2

Pressure of H2 = 0.9947 - 0.03126

Pressure of H2 = 0.96344 atm

Step 4: Calculate moles of H2

p*V=n*R*T

⇒ with p = The pressure of H2 = 0.96344 atm

⇒ with V = the volume of H2 = 0.275 L

⇒ with n = the number of moles H2 = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 25°C = 298 Kelvin

n = (p*V)/(R*T)

n = (0.96344 * 0.275)/(0.08206*298)

n = 0.01083 moles

Step 5: Calculate moles of M

For 1 mole of H2 produced, we need 1 mole M

For 0.0108 moles of H2 we need 0.01083 moles of M

Step 6: Calculate molar mass of M

Molar mass M = Mass M / moles M

Molar mass M = 0.708 grams / 0.01083 moles

Molar mass M = 65.37 g/mol

The metal has a molar mass of 65.37 g/mol

Answer:

The values for spin quantum number +1/2 and - 1/2

Explanation:

Principal quantum number denoted by (n) is used to describe the shell or orbits that electrons are found. Principal quantum number can assume a value of n= 1,2, 3, 4,5............ which indicates K, L, M, N, O shell respectively.

To know the maximum number of electrons in each shell, the formula (2n²) can be used. The letter 'n' denotes the values of principal quantum number 1,2,3,4

For example

All the electron in each shell will have a spin quantum number of +1/2 and - 1/2. One electron in each degenerate orbital will spin up (+1/2) while the other electron will spin down (-1/2).

Answer:

Number of moles nitric acid in the cylinder is 400.539g/mol.

Explanation:

From the given,

Weight of empty gas cylinder [tex]W_{1}= 90.0 lb= 40823.3 grams

Weight of full cylinder[tex]W_{2} = 116.5 lb= 52843.511 grams

The critical temperature = 287 K

The critical pressure 54.3 L

Molar mass of nitric acid = [tex]M_{NO}[/tex] = 30.01 g/mol

Number of moles nitric acid = [tex]n_{NO}[/tex] =?

The mass of nitric acid in the cylinder = [tex]W_{NO}=W_{2}-W_{1}[/tex]

[tex]=52843.511-40823.3 =12,020.2g[/tex]

Number of moles of nitric acid =

[tex]\frac{Given\,mass}{Molar\,mass}=\frac{12,020.2}{30}=400.539g/mol[/tex]

Therefore, number of moles nitric acid in the cylinder is 400.539g/mol.

Answer:

Ksp = 1.3 x 10^(-32) = (3s)^(3) x (2s)^(2)

= 27s^(3) x 4s^(2) = 108s^(5)

Explanation:

Ca3(PO4)2(s) 3Ca^(2+)(aq) + 2PO4^(3-)(aq)

[Ca^(2+)] = 3 [Ca3(PO4)2] = 3s

[PO4^(3-)] = 2 [Ca3(PO4)2] = 2s

Ksp = 1.3 x 10^(-32) =[Ca^(2+)]^(3)×[PO4^(3-)]^(2)

Ksp = 1.3 x 10^(-32) = (3s)^(3) x (2s)^(2)

= 27s^(3) x 4s^(2) = 108s^(5)

Answer : The correct expression for equilibrium constant will be, [tex]K=[O_2]^5[/tex]

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given equilibrium reaction is,

[tex]P_4O_{10}(s)\rightleftharpoons P_4(s)+5O_2(g)[/tex]

The expression of [tex]K[/tex] will be,

[tex]K=[O_2]^5[/tex]

Therefore, the correct expression for equilibrium constant will be, [tex]K=[O_2]^5[/tex]

The equilibrium constant (K) for the given reaction can be expressed as [P4][O2]^5/[P4O10]. This expression represents the ratio of the concentrations of the products raised to their respective coefficients, divided by the concentration of the reactant. The value of K can be calculated based on the given concentrations.

The equilibrium constant (K) for the given reaction can be expressed as:

K = [P4][O2]^5/[P4O10]

This expression represents the ratio of the concentrations of the products (P4 and O2) raised to their respective coefficients, divided by the concentration of the reactant (P4O10).

For example, if the concentration of P4 is 0.1 M, the concentration of O2 is 0.2 M, and the concentration of P4O10 is 0.05 M, the value of K can be calculated as:

K = (0.1 * (0.2)^5) / 0.05 = 0.64

Answer:A

Explanation:

A nuclear reaction is balanced by ensuring that the Masses and charges of te reactants and products are exactly balanced on the left and right hand side of the reaction equation. If there are 60 mass units on the LHS and manganese has only 56 mass units then four mass units are left. If there is no charge on a neutron and there is a charge of 27 on the cobalt, then two charges are left. Four mass units and a charge of +2 corresponds to a helium which is actually an alpha particle.

In the nuclear reaction resulting from the bombardment of cobalt-59 with a neutron, a manganese-56 atom and a beta particle (electron) are produced.

The bombardment of cobalt-59 with a neutron produces a manganese-56 atom and another particle. This process involves a certain type of nuclear reaction known as beta decay. In beta decay, a neutron is converted into a proton and a beta particle, also known as an electron.

Considering the atomic number and mass number balances in this nuclear reaction, the other particle produced can be determined. Cobalt (Co) has an atomic number of 27 and a mass number of 59. When it captures a neutron and becomes manganese (Mn), it has an atomic number of 25 and a mass number of 56. So, there is a difference of two units in the atomic number and three units in the mass number. Therefore, this leftover particle that is formed due to this difference is a beta particle or electron (B), making the correct answer Option B.

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Answer:

Bismuth (Bi)

Explanation:

Elements of the group 15 are nitrogen (N), phosphorus (P), arsenic (As), antimony (Sb) and bismuth (Bi).

In the periodic table, metallic character increases down the group.

On the basis of metallic character, group 15 elements are arranged in the ascending order as follows:

N < P < As < Sb < Bi

Nitrogen and phosphorous are non-metal.

Arsenic and bismuth are metalloid.

Being last in the group, Bi is a metal and hence has strongest metallic character.

Answer:

Volume occupied by 10 grams of Lithium Oxide at STP is 7.52 L

Explanation:

STP (Standard Temperature and Pressure) : It is used while performing calculation on gases and provide standards of measurements while performing experiment.

According to IUPAC (International Union of Pure And applied Chemistry) , standard temperature is 273.15 K and Pressure is 100 kPa (0.9869 atm). At STP condition 1 mole of substance occupies 22.4 L volume .

Molar mass of Lithium Oxide = 29.8 g/mol

[tex]Li_{2}O[/tex] = 2(6.9) + 15.99 = 29.8

Mass of 1 mole [tex]Li_{2}O[/tex] = 29.8 g

1 mole [tex]Li_{2}O[/tex] occupies, Volume = 22.4 L

29.8 g [tex]Li_{2}O[/tex] occupies, V = 22.4 L

1 g [tex]Li_{2}O[/tex] occupies ,V = [tex]\frac{22.4}{29.8}[/tex]

1 g [tex]Li_{2}O[/tex] occupies ,V = 0.7516 L

10 g tex]Li_{2}O[/tex] occupies ,V = [tex]0.7516 \times10[/tex] L

V = 7.52 L

So, volume occupied by Lithium Oxide At STP is 7.52 L

Answer:

2,53x10⁻³ moles of NaOH

Explanation:

The reactions of a diprotic acid with NaOH are:

H₂X + NaOH → HX⁻ + H₂O + Na⁺

HX⁻ + NaOH → X²⁻ + H₂O + Na⁺

Where the complete first reaction gives the first equivalence point and the complete second reaction gives the second equivalence point.

The total volume spent of NaOH to reach the second equivalence point is:

7,00mL + 14,07mL = 21,07 mL = 0,02107L

As molar concentration of NaOH is 0,120M, the moles used to reach the second equivalence point are:

0,02107L×(0,120mol/L) = 2,53x10⁻³ moles of NaOH

I hope it helps!

Answer:

0.0714 M for the given variables

Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: [tex]m_{(AcO)_2Cu} = 0.972 g[/tex]

Volume of the sodium chromate solution: [tex]V_{Na_2CrO_4} = 150.0 mL[/tex]

Molarity of the sodium chromate solution: [tex]c_{Na_2CrO_4} = 0.0400 M[/tex]

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

[tex](CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)[/tex]

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

[tex]n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol[/tex]

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

[tex]n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol[/tex]

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

[tex]Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)[/tex]

Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:

[tex]n_{(AcO)_2Cu} = 0.0053515 mol[/tex]

According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:

[tex]n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol[/tex]

The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:

[tex]c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M[/tex]

The final molarity of the acetate anion in the solution, after copper(II) acetate is dissolved in a sodium chromate solution, is 0.1 M.

The acetate anion, CH3 CO₂¯, found in copper(II) acetate, is the conjugate base of acetic acid. When copper(II) acetate is dissolved in a solution of sodium chromate, it yields a solution of inert cations and weak base anions, resulting in a basic solution. Given the initial acetate concentration, if the newly formed acetate ion gives a final acetate concentration of (1.0 × 10−²) + (0.01 × 10−²) = 1.01 × 10-² mol NaCH3 CO₂, this would mean there are both 9.9 × 10-3 mol/L of copper(II) acetate and 1.01 × 10-2 mol/L of sodium chromate.

To compute the molarity of acetate anion, you can divide the moles of the acetate by the volume of the solution, yielding a final molarity of the acetate anion in the solution. Hence the molarity of acetate anion in the solution would be 1.01 x 10^-2 mol /0.101 L = 0.1 M.

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Final answer:

The true statements about the electron transport chain are (a) it uses O₂ and NADH or FADH2 as reactants, (c) it involves a series of reactions moving electrons through carriers, and (e) Coenzyme Q and cytochrome c are components of the chain. Statements (b) and (d) are false.

Explanation:

Among the given statements about the electron transport chain (ETC), the following are true:

The electron transport chain is a series of electron carriers that transfer electrons from NADH and FADH2 to oxygen, with the formation of water and ATP as end products. It involves multiprotein complexes and shuttle electron carriers in the inner mitochondrial membrane and is a crucial step in aerobic respiration. The electron transport chain is not an anaerobic process (d is false), and it produces significantly more than two ATP as stated in (b), which is also false.

The equilibrium constant for the reaction 2 K(s) + 2 H2O(l) ↔ 2 KOH(aq) + H2(g) is Kc = [KOH]^2[H2], where Kc represents the ratio of the concentration of the products to the reactants.

In the given chemical reaction, 2 K(s) + 2 H2O(l) ↔ 2 KOH(aq) + H2(g), the equilibrium constant, denoted by Kc, represents the ratio of the concentrations of the products to the reactants. Compounds in the solid or liquid state, like K(s) and H2O(l) in this case, are usually not included in the Kc expression. Hence, the equilibrium constant for this reaction would be expressed as:

Kc = [KOH]^2[H2]

This equation indicates that the equilibrium constant is the square of the concentration of KOH, times the concentration of H2. It is important to note that the coefficients become exponents in the equilibrium constant expression, hence [KOH] is squared.

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Final answer:

In chemistry, students are often asked to express equilibrium constants for reactions. This involves determining the respective equilibrium constant expression for each reaction given. Each equilibrium constant formula is unique to the specific reaction and helps understand the extent of the reaction at equilibrium.

Explanation:

Equilibrium Constant (K) Expressions:

For reaction a: K = [NO₂]² / [NO]²[O₂]

For reaction b: K = [NO₂]²[H₂O] / [N₂H₂][O₂]³

For reaction c: K = [H₂][O₂]² / [O₂][H₂O]

Answer:

Dispersion forces

Relative molecular mass

Explanation:

Alkanes experience only dispersion forces. Dispersion forces increase with increasevin the relative molecular mass of the compounds. Hence a higher relative molecular mass implies greater dispersion forces and a greater boiling point.

Methane (CH4) and Propane (CH3CH2CH3) have low boiling points because they only experience weak dispersion forces, which increase with increasing molecular size. This is due to their small size which leads to weaker forces and, thereby, lower boiling points.

Methane (CH4) and Propane (CH3CH2CH3) have the lowest boiling points due to their experience of London dispersion forces, which increase with increasing molecular size. Dispersion forces are weak intermolecular forces that exist between all molecules, regardless of polarity, which are stronger in larger molecules due to greater electron cloud polarizability. Therefore, the reason why compounds like CH4 and CH3CH2CH3 have low boiling points is that they are small in size, leading to weaker dispersion forces and, thus, lower boiling points.

For example, the molar masses of CH4, SiH4, GeH4, and SnH4 are approximately 16 g/mol, 32 g/mol, 77 g/ mol, and 123 g/mol, respectively. Therefore, it can be inferred that CH4 with the lowest molar mass out of the group will have the lowest boiling point, and SnH4 with the highest molar mass would exhibit the highest boiling point.

Ultimately, the relationship between molecular size and type of forces impacts the boiling point of molecular compounds. Nonpolar compounds like CH4 and CH3CH2CH3 only experience dispersion forces, the effects of which increase with increasing molecular size.

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Answer:

standard entropy of vaporization of ethanol = 142.105 J/K-mol

Explanation:

given data

enthalpy of vaporization of ethanol = 40.5 kJ/mol = 40.5 × [tex]10^{3}[/tex] J/mol

entropy of vaporization of ethanol boiling point = 285 K

to find out

standard entropy of vaporization of ethanol

solution

we get here standard entropy of vaporization of ethanol that is expess as

standard entropy of vaporization of ethanol ΔS = [tex]\frac{\Delta H}{T}[/tex] .............1

here ΔH is enthalpy of vaporization of ethanol and  T is temperature

put value in equation 1

standard entropy of vaporization of ethanol ΔS =  [tex]\frac{40.5*10^3}{285}[/tex]

standard entropy of vaporization of ethanol = 142.105 J/K-mol

Answer:

The standard entropy of vaporization is 142 J/mol*K

Explanation:

Step 1: Data given

The boiling point = 285 Kelvin

The standard molar enthalpy of vaporization of ethanol at its boiling point is 40.5 kJ/mol

Step 2: Calculate the standard entropy of vaporization

ΔS = ΔH /T

⇒ with ΔS = the change in entropy

⇒ with ΔH = enthalpy of vaporization of ethanol = 40.5 kJ/mol = 40.5 *10³ J/mol

⇒ with T = the temperature =285 Kelvin

ΔS = (40.5 * 10³ J/mol)/  285 Kelvin

ΔS = 142.1 ≈ 142 J/mol*K

The standard entropy of vaporization is 142 J/mol*K

(Note: The boiling point of ethanol is not 285K but 352 K)

Answer:

M = 0.729 M

Explanation:

In this case, we have a solution of Zn(NO₃)₂.

The chemist wants to prepare a dillute solution of this reactant.

The stock solution of the nitrate has a concentration of 2.5 M, and he wants to prepare 240 mL of a more dillute concentration of the same solution. He adds 70 mL of the stock and complete with water until it reach 240 mL.

We want to know the concentration of this dilluted solution.

As we are working with the same solution, we can assume that the moles of the stock solution will be conserved in the dilluted solution so:

n1 = n2  (1)

and we also know that:

n = M*V (2)

If we replace this expression in (1) we have:

M1*V1 = M2*V2

Where 1, would be the stock solution and 2, the solution we want to prepare.

So, we already know concentration and volume used of the stock solution and the desired volume of the dilluted one, therefore, all we have to do is replace the given data in (2) and solve for the concentration which is M2:

2.5 * 70 = 240M2

M2 = 2.5 * 70 / 240

This is the concentration of the solution prepared.

Answer:

72.03 %

Explanation:

Total mass of dolomite = 9.66 g

Let the mass of Magnesium carbonate = x g

The mass of calcium carbonate = 9.66 - x g

Calculation of the moles of Magnesium carbonate as:-

Molar mass of Magnesium carbonate = 122.44 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

[tex]Moles= \frac{x\ g}{84.3139\ g/mol}=\frac{x}{84.3139}\ mol[/tex]

Calculation of the moles of calcium carbonate as:-

Molar mass of calcium carbonate = 100.0869 g/mol

Thus,

[tex]Moles= \frac{9.66 - x\ g}{100.0869\ g/mol}=\frac{9.66 - x}{100.0869}\ mol[/tex]

According to the reaction shown below:-

[tex]MgCO_3\rightarrow MgO+CO_2[/tex]

[tex]CaCO_3\rightarrow CaO+CO_2[/tex]

In both the cases, the oxides formed from the carbonates in the 1:1 ratio.

So, Moles of MgO = [tex]\frac{x}{84.3139}\ mol[/tex]

Molar mass of MgO = 40.3044 g/mol

Thus, Mass = Moles*Molar mass = [tex]\frac{x}{84.3139}\times 40.3044 \ g[/tex]

Moles of CaO = [tex]\frac{9.66 - x}{100.0869}\ mol[/tex]

Molar mass of CaO = 56.0774 g/mol

Thus, Mass = Moles*Molar mass = [tex]\frac{9.66 - x}{100.0869}\times 56.0774 \ g[/tex]

Given that total mass of the oxide = 4.84 g

Thus,

[tex]\frac{x}{84.3139}\times 40.3044 +\frac{9.66 - x}{100.0869}\times 56.0774=4.84[/tex]

[tex]\frac{40.3044x}{84.3139}+56.0774\times \frac{-x+9.66}{100.0869}=4.84[/tex]

[tex]-694.1618435x+45673.48749\dots =40843.38968\dots[/tex]

[tex]x=\frac{4830.09780\dots }{694.1618435}[/tex]

[tex]x=6.9582[/tex]

Thus, the mass of Magnesium carbonate = 6.9582 g

[tex]\%\ mass=\frac{Mass_{MgCO_3}}{Total\ mass}\times 100[/tex]

[tex]\%\ mass=\frac{6.9582}{9.66}\times 100=72.03\ \%[/tex]

Answer:

(a) [tex]KOH[/tex] acts as base.

[tex]H_2SO_4[/tex] acts as acid.

(b) [tex]NH_3[/tex] acts as base.

[tex]HCl[/tex] acts as acid.

Explanation:

Arrhenius theory:-

The Arrhenius theory was introduced introduced by Swedish scientist named Svante Arrhenius in 1887.

According to the theory, acids are the substances which dissociate in the aqueous medium to produce electrically charged atoms ( may be molecule). Out of these species furnished, one must be a proton or the hydrogen ion, [tex]H^+[/tex].

Base are the substances which dissociate in the aqueous medium to produce electrically charged atoms ( may be molecule). Out of these species furnished, one must be a hydroxide ion, [tex]OH^-[/tex].

Thus, according to the reaction:-

(a)

[tex]2KOH_{(aq)}+H_2SO_4_{(aq)}\rightarrow K_2SO_4_{(aq)}+2H_2O_{(l)}[/tex]

[tex]KOH[/tex] dissociates as:-

[tex]KOH\rightarrow K^++OH^-[/tex] and hence, acts as base.

[tex]H_2SO_4[/tex] dissociates as:-

[tex]H_2SO_4\rightarrow 2H^++SO_4^{2-}[/tex] and hence, acts as acid.

(b)

[tex]NH_3_{(g)}+HCl_{(g)}\rightarrow NH_4Cl_{(s)}[/tex]

[tex]NH_3[/tex] dissociates as:-

[tex]NH_3+H_2O\rightarrow NH_4^++OH^-[/tex] and hence, acts as base.

[tex]HCl[/tex] dissociates as:-

[tex]HCl\rightarrow H^++Cl^{-}[/tex] and hence, acts as acid.