As heat records across europe became toast, england topped what torturous temp for the first time?.

Nh4cl is sometimes preferred instead of hcl or h2so4 for "acid" work-up after grignard reactions, particularly when the expected and desired product is a tertiary alcohol because NH4+ is a much milder acid than HCl or H2SO4, which achieve the protonation of

the oxyanion to yield the alcohol while minimizing the risk of dehydration.

Ammonium chloride (NH4Cl) is the used as reagent that quenches the magnesium alkoxide product of the Grignard addition.

It is a proton source without being acidic as in acidic medium the protonation of the tertiary alcohol product and elimination to the alkene.

In the presence of HCl or any other strong acid protonation proceed and form alkene but not with ammonium chloride.

Thus from above we concluded that Nh4cl is preferred instead of hcl or h2so4 for "acid" work-up after grignard reactions.

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The pH at 25°c of a 0. 0077 m solution of a weak base with a kb of 6. 2 × 10−9 is 8.8.

The base dissociation constant is termed as Kb. Throughout a base split into ts constituent ions in water is determined by its base dissociation constant.

Kb = [B+] [OH-]/[BOH]

Now, let the concentration of [B+] = [OH-] = x

Given,

Kb = 6. 2 × 10^-9

6. 2 × 10^-9 = x^2/(0.0077-x)

x = 6.909 × 10^(-6)

[B+] = [OH-] = 6.909 × 10^(-6)

As we know that,

pOH = -log [OH-]

pOH = -log( 6.909 × 10^(-6))

pOH = 5.2

As we also know that,

pOH + pH = 14

pH = 14 - pOH

pH = 14 - 5.2

pH = 8.8

Thus we calculated that the pH of the solution is 8.8.

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Answer:

112 L

Explanation:

Since the pressure is being held constant, you can use the following variation of the Ideal Gas Law to find the new volume:

[tex]\frac{V_1}{T_1N_1}=\frac{V_2}{T_2N_2}[/tex]

In this equation, "V₁", "T₁", and "N₁" represent the initial volume, temperature, and moles. "V₂", "T₂", and "N₂" represent the final volume, temperature, and moles.

V₁ = 87.2 L                             V₂ = ? L

T₁ = 425 K                              T₂ = 273 K

N₁ = 2.5 moles                       N₂ = 2.5 + 2.5 = 5.0 moles

[tex]\frac{V_1}{T_1N_1}=\frac{V_2}{T_2N_2}[/tex]                                                   <----- Formula

[tex]\frac{87.2 L}{(425K)(2.5 moles)}=\frac{V_2}{(273 K)(5.0 moles)}[/tex]                    <----- Insert values

[tex]\frac{87.2 L}{1062.5}=\frac{V_2}{1365}[/tex]                                                    <----- Simplify denominators

[tex]0.08207=\frac{V_2}{1365}[/tex]                                                 <----- Simplify left side

[tex]112L={V_2}[/tex]                                                        <----- Multiply both sides by 1365

The molarity of the diluted solution is 4.16 M

Given:

volume of water = 45ml

volume of k2so4 = 250ml

Molarity of k2so4 = 0.75 M

To Find:

molarity of the diluted solution

Solution: Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per liters of a solution. Molarity is also known as the molar concentration of a solution

M1V1 = M2V2

45*M1 = 250*0.75

M1 = 250*0.75/45

M1 = 4.16 M

So, Molarity of given solution is 4.16 M

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Answer:

15.94 g

Explanation:

80 mins is 4 half lives ( 20 min each)

255 * 1/2 * 1/2 * 1/2 * 1/2   = 255 * 1/2^4 = 15.94 g

Answer:

44.1 L

Explanation:

Since volume is being held constant, we can use the following variation of the Ideal Gas Law to find the new pressure.

[tex]\frac{P_1}{T_1N_1}=\frac{P_2}{T_2N_2}[/tex]

In the equation, "P₁", "T₁", and "N₁" represent the initial pressure, temperature, and moles. "P₂", "T₂", and "N₂" represent the final pressure, temperature, and moles. Your answer should have 3 sig figs to match the sig figs of the given values.

P₁ = 234 atm                                 P₂ = 68.0 atm

T₁ = 295 K                                     T₂ = 280 K

N₁ = 144 moles                              N₂ = ? moles

[tex]\frac{P_1}{T_1N_1}=\frac{P_2}{T_2N_2}[/tex]                                              <----- Equation

[tex]\frac{234 atm}{(295 K)(144 moles)}=\frac{68.0 atm}{(280 K)N_2}[/tex]                        <----- Insert values

[tex]\frac{234 atm}{42480}=\frac{68.0 atm}{(280 K)N_2}[/tex]                                      <----- Multiply 295 and 144

[tex]0.00551=\frac{68.0 atm}{(280 K)N_2}[/tex]                                    <----- Simplify left side

[tex]1.54=\frac{68.0 atm}{N_2}[/tex]                                            <----- Multiply both sides by 280

[tex](1.54)N_2={68.0 atm}[/tex]                                   <----- Multiply both sides by N₂

[tex]N_2 = 44.1L[/tex]                                                <----- Divide both sides by 1.54

There will be approximately 49.7 moles of air in the tank at the end of the dive.

The Ideal Gas Law, which asserts the following, can be used to solve this problem:

PV = nRT

Where:

P = pressure

V = volume

n = moles of gas

R = ideal gas constant

T = temperature

Let us first use the given data to obtain the initial amount of moles of air in the tank:

[tex]P_1[/tex]= 234 atm

V = 15.0 L

[tex]T_1[/tex] = 295 K, and

R = 0.0821 L/(Kmol) atm.

Using the ideal gas law equation, we can solve for [tex]n_1[/tex]:

[tex]n_1 = (P_1 * V) / (R * T_1)[/tex]

= (234 atm * 15.0 L) / (0.0821 L·atm/(K·mol) * 295 K)

≈ 189.6 moles

Using the new pressure and temperature, we can determine the exact amount of air in the tank:

P2=68.0 atm and T2=280 K

We can determine [tex]n_2[/tex] by applying the same formula:

[tex]n_2 = (P_2 * V) / (R * T_2)[/tex]

= (68.0 atm * 15.0 L) / (0.0821 L·atm/(K·mol) * 280 K)

≈ 49.7 moles

As a result, there will be approximately 49.7 moles of air in the tank at the end of the dive.

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Cyclic bromonium ion is the reactive intermediate that is formed in the reaction of an alkene with br2 and water. the correct answer is option(d).

A reactive intermediate is a highly reactive, high-energy, brief-lived molecule that, when produced in a chemical reaction, will swiftly transform into a stable molecule. They may occasionally be divided and stored. For instance, low temperatures and Matrix Isolation. Carbocations, carbanions, free radicals, carbenes, nitrenes, and benzyne are six different categories of reaction intermediates. These intermediaries are frequently produced when a chemical substance is chemically broken down.

Reactive intermediates can be used to explain the process of a chemical reaction. Reactive intermediates are high-energy, stable products that are present only in one of the intermediate phases of most chemical reactions, which typically involve more than one elementary step.

The complete question is:

What type of reactive intermediate is formed in the reaction of an alkene with Br2 and water to give a bromohydrin?

a.carbocation

b.carbanion

c.radical

d.cyclic bromonium ion

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Answer:

.24 mol / liters

Explanation:

38. 2 ml = .0382 liters

.0382 liters times ( 0. 163 m KOH/1 liter) times ( 1 mol H2SO4/  1 mol koh) =

.006/.025 liters = .24 mol / liters

chegg

4.8 × 10^9 M  is the concentration of n2 gas at equilibrium when the concentration of no2 is twice the concentration of o2 gas.

In some gas-producing fields, such as those in the US Midwest, North Sea, Eastern Europe, and South East Asia, nitrogen (N2) may naturally present in significant concentrations. Around 15% of the world's non-associated gas deposits have nitrogen content levels that are too high to be considered pipeline-quality gas (usually 3–4 mol%). Because it is an inert gas, nitrogen cannot support burning. Consequently, unstable combustion could result from burning a gas that contains too much nitrogen. The recovery of NGL content will be lowered when high nitrogen gas is treated in a gas plant because nitrogen acts as stripping gas. The compression technology and size of the transmission pipeline will rise with increased nitrogen concentration. High nitrogen levels are undesirable in an LNG plant because they lower the LNG temperature, which increases the energy required to liquefy natural gas and also increases the amount of boil-off gas from the LNG storage tanks.

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Considering the definition of atomic mass, isotopes and atomic mass of an element, the % composition of each isotope is:

First of all, the atomic mass (A) is obtained by adding the number of protons and neutrons in a given nucleus of a chemical element.

The same chemical element can be made up of different atoms, that is, their atomic numbers are the same, but the number of neutrons is different. These atoms are called isotopes of the element.

On the other hand, the atomic mass of an element is the weighted average mass of its natural isotopes. In other words, the atomic masses of chemical elements are usually calculated as the weighted average of the masses of the different isotopes of each element, taking into account the relative abundance of each of them.

In this case, the first isotope Gallium - 69 has an atomic mass of 69 and a percent natural abundance of X%. The second isotope Gallium - 71 has an atomic mass of 71 and a percent natural abundance of (100-X)%.

On the other hand, the Relative Atomic Mass of Gallium is 69.8 .

Then, the value of X can be calculated as:

69× X+ 71× (1-X)= 69.8

Solving

69× X+ 71× 1- 71× X= 69.8

69× X+ 71- 71× X= 69.8

69× X- 71× X= 69.8 - 71

- 2× X= -1.2

X= (-1.2)÷ (-2)

X=0.6 which expressed as a percentage is X%= 60%.

So, the % compositon of Gallium - 71 is calculated as (100-X)%=(100 -60)%= 40%

Finally, the % composition of each isotope is:

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Mass of copper would be=92.46 g

I = Current = 19.5 A

t = 4 hours =4×60×60=14400 s

F = Faraday constant = 96485.33 C/mol

Molar mass of copper = 63.546 g/mol

A charge is given by

Q=19.5×14400=280880 C

Moles of electrons are given by Q/F=280880/96485.33=2.91 mol

Moles of copper is=1/2×2.91=1.455 mol

Mass of copper would be=1.455×63.546=92.46 g

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The mass of Copper electroplated is 68.76 g

The process of plating a metal onto another is known as electroplating.

It is often used to prevent corrosion of metal or for the decorative purposes

In this process, electric current is passed through an aqueous solution containing dissolved cations.

The dissolved cations are reduced developing a thin metal coating on the electrode.

At cathode,

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

Current, I = 14.5 A

Time, t = 4 hrs = 4×60×60 = 14400 sec

Charge, q = It = 14.5×14400= 208800 C

Copper metal deposited by 2×96487 C = 63.55 g

Copper metal deposited by 208800 C = [tex]\frac{63.55 \times208800}{2\times96487}[/tex]

                                                         = 68.76g

Hence, The mass of Copper electroplated is 68.76 g

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The δs∘rxn  for the reaction [tex]2NO(g) + O_{2}[/tex] → [tex]2NO_{2} (g)[/tex] will be -146 J/K.

Entropy would be a measurable physical characteristic and a scientific notion that is frequently connected to a condition of disorder, unpredictability, or uncertainty.

Entropy would be a measurement of the system's unpredictability or disorder. The entropy increases as randomness do. It has broad properties as well as a state function. It has the unit [tex]JK^{-1} mol^{-1}[/tex].

Entropy of the reaction can be calculated by the reaction.

Δ[tex]S^{0} rxn[/tex] = 2 mol × [tex]S^{0} (NO_{2} (g) - 2 mol[/tex] × [tex]S^{0} NO (g)[/tex] - 1 mol × [tex]S^{0} (O_{2} )[/tex]

Δ[tex]S^{0} rxn[/tex]  = 2 mol × 240 J/mol.K - 2 mol × 210 J/mol.K-1 mol  ×205.2 J/mol.K

Δ[tex]S^{0} rxn[/tex]  = -146.8 J/K

Therefore, the δs∘rxn  for the reaction [tex]2NO(g) + O_{2}[/tex] → [tex]2NO_{2} (g)[/tex] will be -146 J/K.

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The partial pressure of gas C is 0.54atm. The correct option is D.

The pressure that one of the gases in a mixture would exert if it were in the same volume on its own.

The total pressure of A, B and C is 6.11 atm.

The partial pressure of A is 1.68 atm

The partial pressure of B is 3.89 atm.

Total pressure = sum of partial pressures of all the gasses in that mixture

6.11atm = 1.68atm + 3.89atm + Pc

Pc = 6.11atm - (1.68 + 3.89) = 0.54atm

Thus, the partial pressure of C is 0.54atm. The correct option is D.

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The ka of the acid will be-

HA(aq)+OH−(aq)→A−(aq)+H2O(l)

You have the half-equivalence point when

[HA]=[A−]

it suggests that

log([HA][A−])=log(1)=0

As a result, it can be said that the pH of the solution and the pKa of the weak acid are equivalent at the half-equivalence point.

At the halfway point of equivalence: pH=pKa

The acid dissociation constant of the weak acid, Ka, determines the pKa. pKa=log(Ka), which indicates that Ka=10pKa.

Ka=10pH will be present when the two points are half equal.

Enter your value to determine Ka=105.67=2.1106.

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A membrane is impermeable to charged molecules and does not allow the passage of ions because the charged molecules possess polarity and the membrane possesses hydrophobic interiors.

Charged hydrophilic molecules of all sizes, from small molecules to giant enzymes, can now be made more soluble via a technique called hydrophobic ion pairing. Hydrophobic molecules having hydrophilic moieties are ionically coupled with charged hydrophilic molecules, and the resulting uncharged complex is water-insoluble and will precipitate in aqueous conditions.

With polar solvents like water or alcohol, hydrophilic compounds can easily establish hydrogen bonds. Ionic (charged) groups with oxygen or nitrogen atoms make up the chemical structure of hydrophilic compounds. Typically, a substance's hydrophilicity is determined by its polarity. A molecule will be hydrophobic overall even if it has polar covalent bonds and these bonds are distributed symmetrically.

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The given bond is polar covalent bond.

Polar covalent bonds are covalent bonds in which the electrons are shared unequally. Nonpolar covalent bonds are covalent bonds with an equal distribution of electrons. Chemists utilise electronegativity, a relative measurement of how strongly an atom attracts electrons as it forms a covalent connection, to assess the relative polarity of a covalent bond.

Polarity characterises io3-. If a molecule's dipole moment is greater than 0, it is considered to be polar. The three I-O bonds in this combination are polar due to the difference in electronegativity between the I and O atoms. The three I-O bond moments point toward I atom because I is more electronegative than O atom.

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The original molarity of an aqueous solution of ammonia whose pH is 11. 02 at 25°C is 0.556 M

Calculation ,

Given : pH = 11.5

[tex][H^{+} ][/tex] = [tex]10^{-11.5}[/tex]

[tex][OH^{-} ][/tex] = [tex]K_{w} /[H^{+} ][/tex] = [tex]10^{-14}[/tex]/  [tex]10^{-11.5}[/tex] = [tex]10^{-2.5}[/tex] M

Kb = [tex][NH_{4}^{+} ][OH^{-} ]/[NH_{4} OH][/tex] = [tex](10^{-2.5})^{2} /C[/tex] = 1.8×[tex]10^{-5}[/tex] M

C = 1/1.8 = 0.566 M

So, molarity of an aqueous solution of ammonia is 0.566 M

pH ia a measure of hydrogen ion concentration , a measure of acidity or alkalinity of the solution .pH scale usually range from 0 to 14 Aqueous solutions at 25°C witha pH less rhan 7 are acidic  , whereas , those with pH greater than 7 are basic or alkaline. pOH ia a measure of hydroxide ion concentration , a measure of  alkalinity of the solution .

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The correct answer is  C8H8O2.

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The density of rock weighing 1.705 kg is 3.213 gm/ml.

Explanation:

Given;

Volume of rock = 334.5 ml

Weight of rock = 1.075 kg = 1075 gm

By the formula,

Density = Weight in gm/Volume in ml

              = 1075/334.5 gm/ml = 3.2137 gm/ml

Therefore, density of rock is 3.213 gm/ml.

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The pH at 25°c of a 0. 11 m solution of a weak acid that has ka = 9. 2 × 10−6 is 3.

The base dissociation constant is termed as Kb. Throughout a base split into ts constituent ions in water is determined by its base dissociation constant.

Kb = [H+] [A-]/[HA]

Now, let the concentration of [H+] = [A-] = x

Given,

Ka = 9.2 × 10−6.

Firstly we will calculate the value of the concentration of [H+]

pKa = x^2/(0.11-x)

9.2 × 10−6 = x^2/(0.11-x)

x^2 = 1.012 × 10−6

x = 1.002 × 10^−3.

The concentration of [H+] = [A-] = 1.002 × 10^−3.

Now, we will find pH as

pH = -log[H+]

pH = -log(1.002 × 10^−3)

pH = 3

Thus we calculated that the pH of the solution is 3.

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The weight of the 3 lb roast purchased by the housewife is 1,360.78 grams.

The weight of an object is the force acting on the object due to gravity.

The weight of an object varies from place to place across the globe due to difference in acceleration due to gravity.

1 lb ------------> 453.592 grams

3 lb ------------> ?

= (3 x 453.592)

= 1,360.78 grams

Thus, the weight of the 3 lb roast purchased by the housewife is 1,360.78 grams.

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The temperature at which a liquid's vapor pressure equals the pressure of the gas above it is known as the boiling point of the liquid. The temperature at which a liquid's vapor pressure equals one atmosphere is considered the liquid's typical boiling point (760).

When the vapor pressure of a liquid equals the pressure of the gas above it, the liquid boils. The temperature at which a liquid will boil decreases with decreasing gas pressure above the liquid.

The Macro Perspective.A liquid's vapor pressure rises as it warms until it reaches the same level as the gas above it. Within the bulk liquid, vaporize liquid (i.e., gas) bubbles form, rise to the surface, burst, and release the gas. (A bubble's internal vapor pressure is sufficient to prevent the bubble from collapsing at boiling temperature.)The liquid's molecules must overcome their forces of attraction in order to create vapor.

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Fe is the limiting reactant.

The balanced chemical equation between iron and oxygen to produce iron (III) oxide is

4Fe(s) + 302(g) - ---> 2Fe2O3(s)

Mass of Fe = 227.8 g

Moles of Fe = 227.8gFe*Imol Fe/55.85g Fe = 4.079mol Fe

Mass of oxygen = 128 g

Moles of O2 = 128g02 * 1molo/32g02 = 4molO2

Calculating the limiting reactant: The reactant that produces the least amount of product will be the limiting reactant.

Mass of iron (III) oxide produced from Iron = 4.079mol Fe*2molFe2O3/4molFe*159.69g Fe2O3/1 mol Fe2O3 = 325.7gFe2O3

Mass of iron (III) oxide produced from oxygen = 4molO2 *2mol Fe2O3/3molO2*159.69gFe2O3/1mol Fe2O3 = 425.84gFe2O3

Iron (Fe) produces the least amount of the product iron (III) oxide. So, Fe is the limiting reactant.

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The number of π molecular orbitals in a molecule is always equal to the number of p orbitals used to construct the π bonds.

An electron's position and wave-like behavior within a molecule are described by a mathematical function called a molecular orbital. Chemical, as well as physical properties like the probability of locating an electron in a particular area, can be determined using this function.

A molecular orbital would be created when two atomic orbitals cross one other along the internuclear axis. A molecular orbital is created when two atomic orbitals cross each other sideways.

Therefore, the number of π molecular orbitals in a molecule is always equal to the number of p orbitals used to construct the π bonds.

Hence, the correct answer will be option (a).

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Answer:

instantaneous rate would be the term.

A polar molecule is a water. As a result, polar solutes such as ionic compounds and polar molecular compounds will dissolve in them.

Types of dissolved substances that may be present in water​:

Ionic chemicals and polar molecular compounds, for example, will dissolve in water since it is a polar molecules.

When it comes to organic compounds, water will dissolve any that have a carbon to oxygen atom ratio of less than or equal to 5.

[tex]$\frac{C}{O} \leq 5$[/tex]

Because of this, vitamin A is not water-soluble. Vitamin C is soluble, though.

A polar molecule is a water. As a result, polar solutes such as ionic compounds and polar molecular compounds will dissolve in them.

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Answer:

Hypernantremia due to excess water loss

A rare disorder that causes the body to make too much urine

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Hypernatremia due to excess water loss is a diabetes insipidus. Hence option A is correct.

Hypernatremia is defined as an increase in serum sodium levels to a level greater than 145 mmol/L. Water leaks from the tissues into the bloodstream as a result of hypernatremia, which can induce cell shrinkage. Additionally, it affects the hormone levels that regulate salt levels, leading to an increase in thirst and more concentrated urine production.

Diabetes insipidus is defined as a rare illness that makes you almost insatiably thirsty and produces copious amounts of urine. A brain tumor that affects the hypothalamus or pituitary gland is one of the three most frequent causes of cranial diabetes insipidus. a severe head injury that damages the pituitary or hypothalamus. complications resulting from pituitary or brain surgery.

Thus, Hypernatremia due to excess water loss is a diabetes insipidus. Hence option A is correct.

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The pH of the solution is the negative logarithm of a proton or the hydrogen ion concentration. The pH of 0.39 M acetic acid solution (CH₃COONa) is 2.58.

The pH has been said to be the hydrogen ion concentration that can also be given by the pOH.

Given,

The acid dissociation constant Ka = 1.8 × 10⁻⁵

Concentration of acetic acid (C) = 0.39 M

The hydrogen ion concentration from Ka and molar concentration are calculated as:

H⁺ = √ Ka × C

= √1.8 × 10⁻⁵ × 0.39

= √0.00000702

= 0.0026

Now, pH from hydrogen ion is calculated as,

pH = - log [H⁺]

= - log [0.0026]

= 2.58

Therefore, the pH of acetic acid is 2.58.

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The final answer is 29.8 mL

Molar mass HCl = 36.5 g/mol

Mol HCl in 1.85 g = 1.85 g / 36.5 g/mol = 0.05068 mol

The molarity of the HCl solution is 0.0507 mol /L

Mol HCl in 50 mL = 50 mL / 1000 mL/L * 0.0507 mol /L = 0.002535 mol

This will require 0.002535 mol NaOH

1000 mL contains 0.0850 mol

The volume that contains 0.002535 mol

Volume = 0.002535 mol / 0.0850 mol /L *1000 mL/L = 29.8 mL

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The volume of 0.0850 M NaOH is required to titrate 25.0 ml of 0.0720 M is 21.176 ml.

Normality is defined as the number of equivalents per litre of the solution

It is given by

Normality = number of equivalents / 1 L of solution

Molarity is defined as the number of moles of solute present in one litre of solution

Molarity = moles of solute / 1 litre of solution

Relationship between molarity and normality

Normality = Molarity × Acidity or Basicity of a salt

Here, the Acidity Of NaOH is 1 and the basicity of HBr is also 1.

Thus, Normality = Molarity

We know, N₁V₁ = N₂V₂

We can also write it as M₁V₁ = M₂V₂

V1 = [tex]\frac{M_2V_2}{M_!}[/tex]

Substitute M₁ = 0.0850 M, M₂ = 0.0720 M, V₂= 25 ml

V1 = 21.176 ml

Hence, The volume of 0. 0850 m NaOH are required to titrate 25. 0 ml of 0. 0720 m is 21.176 ml.

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Molality of ethylene glycol, C₂H₄(OH)₂, in a solution prepared from 2. 331×10³ g of ethylene glycol and 2.00×10³ g of water, H₂O is 47.6m

Ethylene Glycol is known as C₂H₄(OH)₂. It is added in water to prepare an Antifreeze solution.

Given,  

Mass of Ethylene Glycol = 2.331 × 10³ g = 2.331kg

Mass of Water = 2.00 × 10³ g

Since, Ethylene Glycol is in excess. Hence, it acts as a solvent and water acts as a solute.

We know, Molar Mass of Water = 18g

Hence, Moles of  Water = Given mass of water / Molar Mass of Water

⇒             Moles of  Water = 2000 / 18

⇒            Moles of  Water = 111.1

Molality is defined as the moles of solute present in a given solvent in kg.

∴ Molality = Moles of  Solute / Mass of Solvent (in kg)

Molality = Moles of  Water / Mass of Ethylene Glycol

⇒ Molality = 111.1 / 2.331

⇒ Molality = 47.6m

Molality of ethylene glycol, C₂H₄(OH)₂, in a solution prepared from 2. 331×10³ g of ethylene glycol and 2.00×10³ g of water, H₂O is 47.6m

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