4. What are the potential sources of error that might cause disagreement between the activity series' prediction of reactions and your observations of reactions

Answer:

The percent composition of fluorine is 65.67%

Explanation:

Percent Composition is a measure of the amount of mass an element occupies in a compound. It is measured in percentage of mass.

That is, the percentage composition is the percentage by mass of each of the elements present in a compound.

The calculation of the percentage composition of an element is made by:

[tex]percent composition element A=\frac{total mass of element A}{mass of compound} *100[/tex]

In this case, the percent composition of fluorine is:

[tex]percent composition of fluorine=\frac{39.6 g}{60.3 g} *100[/tex]

percent composition of fluorine= 65.67%

The percent composition of fluorine is 65.67%

Answer:

The percent composition of fluorine is 65.67%

Explanation:

Percent Composition is a measure of the amount of mass an element occupies in a compound. It is measured in percentage of mass.

That is, the percentage composition is the percentage by mass of each of the elements present in a compound.

The calculation of the percentage composition of an element is made by:

In this case, the percent composition of fluorine is:

percent composition of fluorine= 65.67%

The percent composition of fluorine is 65.67%

Answer:

k= 1.5

[A] = 1 M

[B] = 3 M

m = 2

n = 1

Explanation:

rate = k[A]”[B]"

The rate of the reaction is 4.5 mol L⁻¹s⁻¹.

Rate of a reaction is defined as the change in concentration of any one of the reactants or products of the reaction, in unit time.

Here,

The concentration of A, [A] = 1 M

The concentration of B, [B] = 3 M

The partial order with respect to A, m = 2

The partial order with respect to B, n = 1

The rate constant of the reaction, k = 1.5

The rate of the reaction,

r = k[A]^m [B}^n

r = 1.5 x 1² x 3

r = 4.5 mol L⁻¹s⁻¹

Hence,

The rate of the reaction is 4.5 mol L⁻¹s⁻¹.

To learn more about rate of a reaction, click:

https://brainly.com/question/29049197

#SPJ2

The question is incomplete as some part is missing:

concentration in octanol Partition Ratio = concentration in water

a) What are the intermolecular forces of attraction between octanol molecules? Explain.

b) Which of the intermolecular forces of attraction identified in (a) account for most of the interactions between octanol molecules? Explain. Use the immiscibility in water and the data included in figures 1 and 2 as evidence to support your answer.

c) Would a compound with a partition coefficient less than one be more hydrophobic or more hydrophilic than one with a partition coefficient greater than 10? Explain.

d) Would nonane (figure 2) be more soluble in water or octanol? Explain.

e) Draw another structure for a compound with the same chemical formula as nonane (CH20) that has a lower boiling point. Explain.

f) Are any of the C atoms in the structure you drew for CH20 sp?hybridized? Explain.

Octanol Boiling point = 195°C Figure 2 Nonane (CH20) Boiling point = 151°C

Answer:

1. The forces between octanol molecules would be attractive. These forces include Vanderwaal forces, H-bonds due to the presence of highly polar O-H group.

2. H-bonding ahould account for most of the attractive forces. The O-H bond should behave like and dipole, oxygen of one molecule attracts the hydrogen of the neighbouring molecule forming D-H...A links throughout (D stands for donor of H-Bond and A for acceptor for H-Bond).

3. Partition coefficient less than 1 will be more hydrophilic, generally drugs with low partition coefficients are regarded as hydrophilic. As parition coefficient of 10 mean more of the solute is dispersed in octanol as compared to water.

4. Nonane is non polar, so it would not dissolve in water. It follows the rule like dissolves like. Polar substances dissolve in polar solvents. 1-octanol is able to bind with water through hydrogen bonds thus its soluble in water but nonane doesn't. Nonane will forms a different layer from water.

5) no all carbons in 2-methyloctane are single bonded. Thus sp3 hybrid. A sp2 hybridised carbon would have a double bond C=C.

Answer:

1) HCl (Hydrochloric acid reacts with NH3 and forms dense fumes)

2) Methane (It is from the group of alkanes and is a green house gas)

3) Deliquescent substance (It is a solid which when kept in open forms a solution after sometime)

4) Brass (It is a solid-in-solid solution used to make electrical fittings)

5) Aluminium

Question 2:

1)    Al₂O₃ + 2NaOH    ⇒   2NaAlO₂  (Sodium Aluminate) + H₂O

2)   Zn +  H₂SO₄   (dilute)  =>  ZnSO₄  (Zinc Sulphate) + H₂

Answer:

equations

Aluminium oxide and Sodium hydroxide react to form water and sodium aluminate

Zinc reacts with dilute sulphuric acid to form zinc sulphate and hydrogen gas

Answer:

4.26 %

Explanation:

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

Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is  4.50  ×  10 ⁻⁴.

Step 1: Given data

Initial concentration of the acid (Ca): 0.249 M

Acid dissociation constant (Ka): 4.50  ×  10 ⁻⁴

Step 2: Write the ionization reaction for nitrous acid

HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)

Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])

We will use the following expression.

[tex][A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4} } = 0.0106 M[/tex]

Step 4: Calculate the percent ionization of nitrous acid

We will use the following expression.

[tex]\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%[/tex]

Answer:

MgSo4.7H2O = Magnesium sulfate

Cs3PO4.H2O = Cesium Phosphate

Hope this helps!

Answer:

B. Reactant + Reactant -> Product + Product

Explanation:

Reactants are substances that- as the name suggests- reacts with other substances at the beginning of a reaction

Products are substances that are produced as a result of the reaction

Typically, when writing a chemical reaction, an arrow is used to show the direction the reaction is moving.  In this case, the arrows in options B and C suggest that the reaction only moves in one direction- forwards

And as mentioned above, reactants are the substances at the start of the reaction, they're what mixes together to form a new product.  

To keep things simple:

Products can't be at the beginning of a reaction since they weren't formed yet.

Similarly, reactants can't be part of the products since they already existed and didn't need to be made. In a lot cases, the reactants would be completely used up to make the products

As such, only one possible chemical reaction would follow that reasoning:

    Reactant + Reactant ->  Product + Product

Reactant + Reactant → Product + Product is the correctly written chemical reaction. Hence, option B is correct.

A chemical equation is a mathematical expression of the chemical reaction which represents the product formation from the reactants.

In an equation, the reactants are written on the left-hand side and the products are written on the right-hand side demonstrated by one-headed or two-headed arrows.

Hence, option B is correct.

Learn more about the chemical equation here:

https://brainly.com/question/12363199

#SPJ2

Answer:

A. None of these.  

Explanation:

A crystal structure is an arrangement of atoms or ions in a repeating three-dimensional array.

B. is wrong. Metal atoms, such as gold, arrange themselves into a crystal structure.

C. is wrong. Ionic solids, such as sodium chloride, arrange themselves into a crystal structure.

D. is wrong. Macromolecules (network solids), such as diamond, arrange themselves into a crystal structure.

The correct answer is None of these.  

Learn more about crystal structure here:-

https://brainly.com/question/23986315

#SPJ2

Answer:

[tex]3Sr^{+2}+6Cl^{-}+6Li^{+}+2PO_{4}^{3-}-->Sr_{3}(PO_{4})_{2}+6Li^{+}+6Cl^{-}\\\\3Sr^{+2}+2PO_{4}^{3-} --->Sr_{3}(PO_{4})_{2}[/tex]

First equation is the complete ionic equation.

Second equation is the net ionic equation.

Hey! :)

__________ ☆ ☆__________________________________

Answer:

The answer is Electronegativity increases up and to the right

Explanation:

When you move from left to right it increases ( in the periodic table )

But when you move down the table electronegativity decreases.

So “ Electronegativity increases up and to the right” describes the trends the best.

Hope this helps! :)

____________☆ ☆________________________________

By, BrainlyMember ^-^

Good luck!

Answer:

Explanation:

We shall find out the molecular formula of the substance .

Ration of number of atoms of C , O and H

= [tex]\frac{37.48}{12} :\frac{49.93}{16} :\frac{12.58}{1}[/tex]

= 3.12 : 3.12 : 12.58

= 1 : 1 : 4

volume of gas at NTP

= 250 x 273 / 350 mL .

= 195 mL .

Molecular weight of the substance = .2804 x 22400 / 195 g

= 32. approx

Let the molecular formula be

(COH₄)n  

n x 32 = 32

n = 1

Molecular formula = COH₄

The compound appears to be CH₃OH

a )

CO + 2H₂ = CH₃OH

28g     4g          32g

59      8

For 8 kg hydrogen , CO required = 56 kg

CO is in excess .  hydrogen is the limiting reagent .

mass of product formed

= 32 x 8 / 4

= 64 kg

b )

percentage yield = product actually formed / product to be formed theoretically  x 100

= 59.6 x 100 / 64

= 93.12 %

c )

2CH₃OH + 3O₂ = 2CO₂ + 4H₂O .

64 g                     2 x 22.4 L

Gram of gas in 1 gallon of fuel

= .7914 x 3785

= 2995.5 g

CO₂ produced at NTP by 2995.5 g CH₃OH

= 2 x 22.4 x 2995.5 / 64 L

= 2096.85 L

At 27° C and 766 mm Hg , this volume is equal to

2096.85 x 300 x 760 / 273 x 766

= 2286.18  L .

d )

C₈H₁₈  =  8CO₂

114g           8 x 22.4 L

gram of fuel per unit gallon

= .6986 x 3785

= 2644.2g

gram of CO₂ produced by 1 gallon of fuel  at NTP

= 8 x 22.4 x 2644.2 / 114

= 4156.5 L

So it produces more CO₂ .

Answer:

a. NH₃ : base

CH₃COOH (acetic acid) : acid

NH₄⁺ : conjugate acid

CH₃COO⁻ : conjugate base

b. HClO₄ (perchloric acid) : acid

NH₃ : base

ClO₄⁻ : conjugate base

NH₄⁺ : conjugate acid

Hope this helps.

Answer:

7p

Explanation:

principal quantum number is 7

n=7( principle shell)

angular momentum quantum number gives sub shell

l = 1 means it is p orbital

so answer is 7p orbital

Answer:

The calculator add the CO2 released from the use of electricity, released from driving and the CO2 from the waste that we disposed.

Explanation:

The carbon dioxide, CO2 is what the human body does not need, therefore, we breathe it out, hence taking in oxygen(respiration process). The plants need oxygen for the production of their own food.

The carbon calculator estimate the amount of CO2 that each individual releases into the atmosphere through the consideration of several factors such as the kind of food that we eat.

Therefore, if we are to use the carbon calculator to determine the amount of CO2 that each individual releases into the atmosphere we will have:

The amount of CO2 that each individual releases into the atmosphere =( CO2 released from the use of electricity) + (CO2 released from driving) + (the CO2 from the waste that we disposed).

Answer:

Option B

Explanation:

Scientific question are answered through experimentation, through testing the theory about the physical world.

Answer: its A

through observing and measuring the physical world

Explanation:

Answer:

WHEN 63.7 g OF K2SO4 IS DISSOLVED IN WATER, 0.73 MOLES OF POTASSIUM ION AND 0.366 MOLES OF SULFATE ION ARE FORMED.

Explanation:

Equation for the reaction:

K2SO4 + H20 ------->2 K+ + SO4^2-

When K2SO4 dissolves in water, potassium ion and sulfate ion are formed.

1 mole of K2SO4 produces 2 moles and 1 mole of SO4^2-

At STP, 1 mole of K2SO4 will be the molar mass of the substance

Molar mass of K2SO4 = ( 39 *2 + 32 + 16*4) g/mol

Molar mass = 174 g/mol

So therefore;

1 mole of K2SO4 contains 174 g and it produces 2 moles of potassium and 1 mole of sulfate ion

When 63.7 g is used; we have:

174 g = 2 moles of K+

63.7 g = ( 63.7 * 2 / 174) moles of K+

= 0.73 moles of K+

Forr sulfate ion, we have:

174 g = 1 mole ofSO4^2-

63.7 g = (63.7 * 1 / 174) moles of SO4^2-

= 0.366 moles of SO4^2-

In other words, when 63.7 g of K2SO4 is dissolved in water, 0.73 moles of potassium ion and 0.366 moles of sulfate ion are formed.

Answer:

10.328 m

Explanation:

normal atmospheric pressure = 101325 Pa

density of water at 25 °C = 1.0 g/cm^3 = 1000 kg/m^3

pressure = pgh

where p = density

g = acceleration due to gravity = 9.81 m/s^2

h = height of column

imputing values, we have

101325 = 1000 x 9.81 x h

height of column h = 101325/9810 = 10.328 m

Answer:

Option B. 0.136 g

Explanation:

The balanced equation for the reaction is given below:

2AgNO3(aq) + 2NaOH(aq) —> Ag2O(s) + 2NaNO3(aq) + H2O(l)

Next, we shall determine the masses of AgNO3 and NaOH that reacted and the mass of Ag2O produced from the balanced equation. This is illustrated below:

Molar mass of AgNO3 = 108 + 14 + (16x3) = 170g/mol

Mass of AgNO3 from the balanced equation = 2 x 170 = 340g

Molar mass of NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH from the balanced equation = 2 x 40 = 80g

Molar mass of Ag2O = (108x2) + 16 = 232g/mol

Mass of Ag2O from the balanced equation = 1 x 232 = 232g

Summary:

From the balanced equation above,

340g of AgNO3 reacted with 80g of NaOH to produce 232g of Ag2O.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

340g of AgNO3 reacted with 80g of NaOH.

Therefore, 0.2g of AgNO3 will react with = (0.2 x 80)/340 = 0.047g of NaOH.

From the calculations made above, only 0.047g out of 0.2g of NaOH given, reacted completely with 0.2g of AgNO3. Therefore, AgNO3 is the limiting reactant and NaOH is the excess reactant.

Now, we can calculate the mass of Ag2O produced from the reaction of 0.2g of AgNO3 and 0.2g of NaOH.

In this case, we shall use the limiting reactant because it will produce the maximum yield of Ag2O as all of it is used up in the reaction.

The limi reactant is AgNO3 and the mass of Ag2O produced can be obtained as follow:

From the balanced equation above,

340g of AgNO3 reacted to produce 232g of Ag2O.

Therefore, 0.2g of AgNO3 will react to produce = (0.2 x 232)/340 = 0.136g of Ag2O.

Therefore, 0.136g of Ag2O was produced from the reaction.

Answer:

C. An electron at this electrode has a higher potential energy than it has at a standard hydrogen electrode.

Explanation:

The standard hydrogen electrode (SHE) is used to measure the electrode potential of substances. The standard hydrogen electrode is arbitrarily assigned an electrode potential of zero. Recall that electrode potentials are always measured as reduction potentials in electrochemical systems.

For an electrode that has a negative electrode potential, electrons at this electrode have a higher potential energy compared to electrons at the standard hydrogen electrode. Electrons flow from this electrode to the hydrogen electrode.

On the other hand, a positive electrode potential implies that an electron at this electrode has a lower potential energy than it has at a standard hydrogen electrode. Hence electrons will flow from the standard hydrogen electrode to this electrode.

Answer:

The new solution is 1.4% m/V

Explanation:

The concentration of the new solution, obtained by adding 22.5 mL of distilled water to 2.5 mL of 14 % m/V sugar solution, is 1.4% m/V.

We have 2.5 mL (V₁) of a concentrated solution and add it to 22.5 mL of distilled water. Assuming the volumes are additives, the volume of the new solution (V₂) is:

[tex]2.5 mL + 22.5 mL = 25.0 mL[/tex]

We want to prepare a dilute solution from a concentrated one, whose concentration is 14% m/V (C₁). We can calculate the concentration of the dilute solution (C₂) using the dilution rule.

[tex]C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{14\% m/V \times 2.5 mL}{25.0 mL} = 1.4 \% m/V[/tex]

The concentration of the new solution, obtained by adding 22.5 mL of distilled water to 2.5 mL of 14 % m/V sugar solution, is 1.4% m/V.

You can learn more about dilution here: https://brainly.com/question/13844449

Answer:

In the attached image the Lewis equation is shown where it is shown how two oxygens react with two hydrogens to meet the octet of the electrons.

Explanation:

Hydrogen peroxide is one of the most named chemicals since it is not only sold as "hydrogen peroxide" in pharmacies but it is also one of the great weapons of immune defense cells to defend ourselves against anaerobic bacteria.

The disadvantage of this compound is that when dividing it forms free oxygen radicals that are considered toxic or aging for our body.

In the attached image below, you will see the Lewis equation is shown there. There, you will see how two oxygens react with two hydrogens to come about the octet of the electrons.

When two or more atoms bond with each other, they often form a molecule. When two hydrogens and an oxygen share electrons through covalent bonds, a water molecule is formed.

The octet rule is known as when most atoms want to gain stability in their outer most energy level by filling themselves that is the S and P orbitals of the highest energy level with eight electron.

HOOH is the compound  that is form. It is called Hydrogen peroxide. This because it is has reactive oxygen species and the simplest peroxide.

Lear more from

https://brainly.com/question/15715780

Answer:

D . Sulphur

Explanation:

the element with a 3p4 valence configuration, look in period 3 and group XVI, and that is ...

S, sulfur.

Answer:

1.1 × 10⁻³ g

Explanation:

Step 1: Given data

Henry's Law constant for methane (k): 1.4 × 10⁻³ M/atm

Volume of water (=volume of solution): 75 mL

Partial pressure of methane (P): 0.68 atm

Step 2: Calculate the concentration of methane in water (C)

We will use Henry's law.

[tex]C = k \times P = 1.4 \times 10^{-3}M/atm \times 0.68atm = 9.5 \times 10^{-4}M[/tex]

Step 3: Calculate the moles of methane in 75 mL of water

[tex]\frac{9.5 \times 10^{-4}mol}{L} \times 0.075 L = 7.1 \times 10^{-5}mol[/tex]

Step 4: Calculate the mass corresponding to 7.1 × 10⁻⁵ mol of methane

The molar mass of methane is 16.04 g/mol.

[tex]7.1 \times 10^{-5}mol \times \frac{16.04g}{mol} = 1.1 \times 10^{-3} g[/tex]

Answer

Naphthalene is a non electrolyte

If the unknown compound is an electrolyte it gives 2 or more ions in solution

( NaCl >> Na+ + Cl- => 2 ions

Ca(NO3)2 >> Ca2+ + 2 NO3- => 3 ions)

the f.p. lowering is directly proportional to the molal concentration of dissolved ions in the solution )

For naphthalene

delta T = 1.86 x m

for a salt that gives 2 ions

delta T = 1.86 x m x 2

hence the lowering in freezion point of unkown is greater then napthalene

The question is incomplete; the complete question is: Classify each molecule by whether its real bond angles are the same as or different than its model (ideal) bond angles. In other words, do the bond angles change when you switch between Real and Model mode at the top of the page? Same (angles do not change) Different (angles change) Answer Bank | H2O | CO2, SO2, XeF2, BF3 CIF3, NH3, CH4, SF4, XeF4, BrF5, PCI5,SF6

Answer:

Compounds whose real bond angle are the same as ideal bond angle;

SF6, BF3, CH4, PCI5

Compounds whose real bond angles differ from ideal bond angles;

H2O, CO2, SO2, XeF2, CIF3, NH3, SF4, XeF4, BrF5

Explanation:

According to the valence shell electron pair repulsion theory (VSEPR), molecules adopt various shapes based on the number of electron pairs on the valence shell of the central atom of the molecule. The electron pairs usually orient themselves as far apart in space as possible leading to various observed bond angles.

The extent of repulsion of lone pairs is greater than that of bond pairs. Hence, the presence of lone pairs on the valence shell of the central atom in the molecule distorts the bond angles of molecules away from the ideal bond angles predicted on the basis of valence shell electron pair repulsion theory.

For instance, methane is a perfect tetrahedron having an ideal bond angle of 109°28'. Both methane and ammonia are based on a tetrahedron, however, the presence of a lone pair of electrons on nitrogen distorts the bond angle of ammonia to about 107°. The distortion of lone pairs in water is even more as the bond angles of water is about 104°.

Answer: 10 moles/kg.

Explanation:

Given, Mass of solute = 24.2 g

Molar mass of solute = 24.2 g/mol

[tex]\text{Moles of solute =}\dfrac{\text{Mass of solute}}{\text{Molar mass of solute}}\\\\=\dfrac{24.2}{24.2}=1[/tex]

Mass of solvent = 100.0g = 0.1 kg  [1 g=0.001 kg]

[tex]\text{Molality}=\dfrac{\text{Moles of solute}}{\text{kilograms of Solvent}}\\\\=\dfrac{1}{0.1}\\\\=10\ moles/kg[/tex]

Hence, the molality of the resulting solution is 10 moles/kg.

Answer:

143pm is the radius of an Al atom

Explanation:

In a face centered cubic structure, FCC, there are 4 atoms per unit cell.

First, you need to obtain the mass of an unit cell using molar mass of Aluminium  and thus, obtain edge length and knowing Edge = √8R you can find the radius, R, of an Al atom.

Mass of an unit cell

As 1 mole of Al weighs 26.98g. 4 atoms of Al weigh:

4 atoms × (1mole / 6.022x10²³atoms) × (26.98g / mole) = 1.792x10⁻²²g

Edge length

As density of aluminium is 2.71g/cm³, the volume of an unit cell is:

1.792x10⁻²²g × (1cm³ / 2.71g) = 6.613x10⁻²³cm³

And the length of an edge of the cell is:

∛6.613x10⁻²³cm³ = 4.044x10⁻⁸cm = 4.044x10⁻¹⁰m

Radius:

As in FCC structure, Edge = √8 R, radius of an atom of Al is:

4.044x10⁻¹⁰m = √8 R

1.430x10⁻¹⁰m = R.

In pm:

1.430x10⁻¹⁰m ₓ (1x10¹²pm / 1m) =

The radius of the atom of Al in the FCC structure has been 143 pm.

The FCC lattice has been contributed with atoms at the edge of the cubic structure.

The FCC has consisted of 4 atoms in a lattice.

[tex]\rm 6.023\;\times\;10^2^3[/tex] atoms = 1 mole

4 atoms = [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles

The mass of 1 mole Al has been 26.98 g/mol.

The mass of [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles = [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles × 26.98 g

The mass of 1 unit cell of Al has been = 1.792 [tex]\rm \bold{\times\;10^-^2^2}[/tex] g.

Density = [tex]\rm \dfrac{mass}{volume}[/tex]

Volume = Density × Mass

The volume of Al unit cell = 2.71 g/[tex]\rm cm^3[/tex] × 1.792 [tex]\rm \times\;10^-^2^2[/tex] g

The volume of Al cell = 6.613 [tex]\rm \times\;10^-^2^3[/tex] [tex]\rm cm^3[/tex]

Volume = [tex]\rm edge\;length^3[/tex]

6.613 [tex]\rm \times\;10^-^2^3[/tex] [tex]\rm cm^3[/tex] = [tex]\rm edge\;length^3[/tex]

Edge length = [tex]\rm \sqrt[3]{6.613\;\times\;10^-^2^3}[/tex] cm

Edge length = 4.044 [tex]\rm \times\;10^-^8[/tex] cm

Edge length = 4.044 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m.

Edge length = [tex]\rm \sqrt{8\;\times\;radius}[/tex]

4.044 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m = [tex]\rm \sqrt{8\;\times\;radius}[/tex]

Radius = 1.430 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m.

1 m = [tex]\rm 10^1^2[/tex] pm

1.430 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m = 143 pm.

The radius of the atom of Al in the FCC structure has been 143 pm.

For more information about the FCC structure, refer to the link:

https://brainly.com/question/14934549

Answer:

The most common position for an double bond in an unsaturated fatty acid is delta 9 (Δ⁹)

Explanation:

Unsaturated fatty acids are carboxylic acids which contains one or more double bonds. The chain length as well as the number of double bonds is written separated by a colon. The positions of the double bonds are specified starting from the carboxyl carbon, numbered as 1, by superscript numbers following a delta (Δ). For example, an 18-carbon fatty acid containing a  single double bond between carbon number 9 and 10 is written as 18:1(Δ⁹).

In most monounsaturated fatty acids, the double bond is between C-9 and C-10 (Δ⁹), and the other double bonds of polyunsaturated fatty acids are generally Δ¹² and Δ¹⁵. This positioning is due to the nature of the biosynthesis of fatty acids. In the mammalian hepatocytes, double bonds are introduced easily into fatty acids at the Δ⁹ position, but cannot introduce additional double bonds between C-10 and the methyl-terminal end. However, plants are able to introduce these additional double bonds at the  Δ¹² and Δ¹⁵ positions.

Answer:

pH = 3.39

Explanation:

The equilibrium in water of ascorbic acid (With its conjugate base) is:

H₂C₆H₆O₆(aq) + H₂O(l) ⇄ HC₆H₆O₆⁻(aq) + H₃O⁺(aq)

Where the acidic dissociation constant is written as:

Ka = 7.9x10⁻⁵ = [HC₆H₆O₆⁻] [H₃O⁺] / [H₂C₆H₆O₆]

H₂O is not taken in the Ka expression because is a pure liquid.

As initial concentration of H₂C₆H₆O₆ is 2.5x10⁻³M, the equilibrium concentration of each species in the equilibrium is:

[H₂C₆H₆O₆] = 2.5x10⁻³M - X

[HC₆H₆O₆⁻] = X

[H₃O⁺] = X

Replacing in the Ka expression:

7.9x10⁻⁵ = [X] [X] / [2.5x10⁻³M - X]

1.975x10⁻⁷ - 7.9x10⁻⁵X = X²

0 = X² + 7.9x10⁻⁵X - 1.975x10⁻⁷

Solving for X:

X = -0.00048566→  False solution, there is no negative concentrations

X = 0.00040666 → Right solution

As [H₃O⁺] = X, [H₃O⁺] = 0.00040666

pH is defined as -log [H₃O⁺];

pH = -log 0.00040666,