A 10.00-mL aliquot of vinegar requires 16.95 mL of the 0.4874 M standardized NaOH solution to reach the end point of the titration. Demonstrate how to calculate the molarity of the vinegar solution (HC2H3O2). Show complete work below. Answer: 0.8261 M.

Answer:

[tex]Ksp=1.2[/tex]

Explanation:

Hello,

In this case, as the saturated solution has 7.1 grams of sodium carbonate, the solubility product is computed by firstly computing the molar solubility by using its molar mass (106 g/mol):

[tex]Molar \ solubility=\frac{7.1gNa_2CO_3}{0.1L}*\frac{1molNa_2CO_3}{106gNa_2CO_3}=0.67M[/tex]

Next, as its dissociation reaction is:

[tex]Na_2CO_3(s)\rightleftharpoons 2Na^+(aq)+CO_3^{2-}(aq)[/tex]

The equilibrium expression is:

[tex]Ksp=[Na^+]^2[CO_3^{2-}][/tex]

And the concentrations are related with the molar solubility (2:1 mole ratio between ionic species):

[tex]Ksp=(2*0.67)^2*(0.67)\\\\Ksp=1.2[/tex]

Best regards.

Answer:

0.16 M

Explanation:

Data provided as per the question is below:-

Volume of [tex]HNO_3[/tex] = 0.22 L

The Volume of NaOH = 0.18 L

Morality of NaOH = 0.2

According to the given situation, the calculation of the concentration of the [tex]HNO_3[/tex] is shown below:-

For equivalence,

Number of the equivalent of [tex]HNO_3[/tex] = Number of equivalents of NaOH

[tex]= \frac{0.18\times0.2}{0.22}[/tex]

[tex]= \frac{0.036}{0.22}[/tex]

= 0.16363 M

or

= 0.16 M

Answer:

See explanation

Explanation:

We must remember that the D / L nomenclature refers to the orientation of the hydroxyl group on carbon 5. If the "OH" is on the right we will have a D configuration. Yes, the "OH" is on the left we will have an L configuration. (See figure 1)

Now, the orientation of this "OH" has nothing to do with the ability of the molecule to deflect polarized light. If the molecule deflects light to the left we will have the symbol "(-)" (levorotation) if the molecule deflects light to the right we will have the symbol "(+)" (dextrorotation).

So in the "D" configuration, we can have both a right (+) and a left (-) deviation.

I hope it helps!

Answer:

A) dietary fiber isn't used by the body.

Explanation:

The food we eat contains certain nutritional contents that provides energy, measured in calories (CAL) to the body. A procedure called BOMB CALORIMETRY can be used to determine the energy contents of these foods. The energy-supplying macromolecules contained in food substances we eat are carbohydrate, protein, fats etc.

Bomb calorimetry uses the method of burning the food substance in a device called bomb calorimeter, and measure the caloric content of the burnt food. Bomb calorimetry measures all the present calories in a food substance, which can include dietary fibers. Due to this reason, it is considered a poor choice in determining the number of nutritional calories in a food substance.

Dietary fibers are indigestible carbohydrates that cannot be broken down and used by the body. They pass along the alimentary canal until they are egested. Hence, they are no source of nutrients to the body. Since bomb calorimetry measures all calories including dietary fibers, it is said to overestimate the caloric content of food substances.

Answer:

.0027 M

Explanation:

We must calculate the threshold concentration of PO3−4 using Ksp and the given concentration of Ca2+:

Ca3(PO4)2(s)⇌3Ca2+(aq)+2PO3−4(aq)

Ksp=8.6×10−19=[Ca2+]3[PO3−4]2=(4.9×10−5M)3[PO3−4]2

[PO3−4]=0.0027 M

Answer:

8.90 M

Explanation:

Step 1: Given data

Step 2: Calculate the initial concentration

We have a concentrated NaCl solution and we want to prepare a diluted one. We will use the dilution rule.

C₁ × V₁ = C₂ × V₂

C₁ = C₂ × V₂  / V₁

C₁ = 5.00 M × 1.13 L  / 0.635 L

C₁ = 8.90 M

Answer:

[tex]\large \boxed{\text{8.90 mol/L}}[/tex]

Explanation:

We can use the dilution formula to calculate the concentration of the original solution.

[tex]\begin{array}{rcl}V_{1}c_{1} & = & V_{2}c_{2}\\\text{635 mL }\times c_{1} & = & \text{1130 mL} \times \text{5.00 mol/L}\\635 c_{1}&=& \text{5650 mol/L}\\c_{1}& = & \dfrac{5650}{635}\text{ mol/L}\\\\& = & \textbf{8.90 mol/L}\\\end{array}\\\text{The initial concentration was $\large \boxed{\textbf{8.90 mol/L }}$}[/tex]

Question is incomplete, the complete question is as follows:

A student wants to examine a substance by altering the bonds within its molecules. Which of the following properties of the substance should the student examine?

A. Toxicity, because it can be observed by altering the state of the substance

B. Boiling point, because it can be observed by altering the state of the substance

C. Toxicity, because it can be observed by replacing the atoms of the substance with new atoms

D. Boiling point, because it can be observed by replacing the atoms of the substance with new atoms

Answer:

B.

Explanation:

A student can examine a substance without altering the bonds within the molecules by examining its boiling point.

The boiling point is the property of a substance, at which the substance changes its state, which is from solid to liquid, liquid to gas and others. So, examining the boiling point will alter the bonds within the molecules as the state of substance will change.

Hence, the correct answer is "B".

Answer:

[tex]\Huge \boxed{\mathrm{Kr}}[/tex]

Explanation:

Krypton is an element in the periodic table with an atomic number of 36.

The symbol for Krypton is Kr.

Answer:

KR.

Explanation:

Use the periodic table for reference:

Answer:

e. Lithium

Explanation:

Correct list of options!

a. hydrogen  b. Helium  c. sulfur  d. Iodine  e. Lithium

Sea of electrons generally refers to metal atoms. This  is because of the delocalized nature of the electrons compared to non metals where the electrons are localized (fixed to a specific atoms).

Among all the elements in the options, the metal is option e. Lithium

Answer:

D

Explanation:

D is the answer

Answer:

thank you...

Answer:

All of the above processes have a ΔS < 0.

Explanation:

ΔS represents change in entropy of a system. Entropy refers to the degree of disorderliness of a system.

The question requests us to identify the process that has a negative change of entropy.

carbon dioxide(g) → carbon dioxide(s)

There is  a change in state from gas to solid. Solid particles are more ordered than gas particles so this is a negative change in entropy.

water freezes

There is  a change in state from liquid to solid. Solid particles are more ordered than liquid particles so this is a negative change in entropy.

propanol (g, at 555 K) → propanol (g, at 400 K)

Temperature is directly proportional to entropy, this means higher temperature leads t higher entropy.

This reaction highlights a drop in temperature which means a negative change in entropy.

methyl alcohol condenses

Condensation is the change in state from gas to liquid. Liquid particles are more ordered than gas particles so this is a negative change in entropy.

Answer: C

Explanation:

The article was sourced from the Oak National Laboratory

Which reasons did you include in your response? Check all of the boxes that apply.

1. The article does not present original research.

and

3. The article has references to primary sources.

Answer:

C

Explanation:

Which reasons did you include in your response? Check all of the boxes that apply.

The article does not present original research.

The article summarizes other research.

The article has references to primary sources.

Answer:

examples of things which contain potassium are:

green vegetables

root vegetables

fruits

potassium chloride

potassium sulphate

Explanation:

if you need a specific answer please send the options

Answer:

C

Explanation:

The answer is the one with 20 protons, 20 neutrons, and 6-8-8-2 electrons.

Answer:

1229.08975 kPa

Explanation:

Given that:

The depth of the water = 100 m below the surface of the water.

The density of the water = 1150 kgm

The mass of the water = depth of the water × density of the water

The mass of the water = 100 m × 1150 kgm

The mass of the water = 115000 kgf/m²

The mass of the water is the pressure of the water at a depth of 100 m below the surface of the water.

Since 1 kgf/m² = 0.00980665 kPa

Then  115000 kgm² = (115000  ×  0.00980665) kPa

=  1127.76475 kPa

At standard temperature and pressure , the atmospheric pressure = 101.325 kPa.

Therefore, the pressure below the surface of sea water  = 1127.76475 kPa +   101.325 kPa

= 1229.08975 kPa

Answer:

The correct answer is 1.60 Liters.

Explanation:

The given reaction:

CO (g) + H₂(g) ⇔ CH₃OH (g)

Based on the given reaction, two moles of H₂ reacts with one mole of CO and produce one mole of CH₃OH.

It is mentioned that 3.20 L of H₂ is reacted, therefore, there is a need to convert it into moles.

As 22.4 L at standard temperature and pressure is equivalent to 1 mole.

Therefore, 1 L at STP will be, 1/22.4 mole

Now 3.20 L at STP will be,

= 1/22.4 × 3.20

= 0.1428 mole

And as mentioned in the reaction that 2 moles of H₂ gives 1 mole of CH₃OH, therefore, 1 mole of H₂ will give 1/2 mole of CH₃OH

Now, 0.1428 mole of H₂ will give,

= 0.1428/2 = 0.071 mole of CH₃OH

= 0.071 × 22.4 = 1.60 L

The volume, in liters, of CH₃OH gas formed is 1.60 L

From the question,

We are to determine the volume of CH₃OH formed

The given chemical equation for the reaction is

CO(g)+ H₂(g) → CH₃OH

The balanced chemical equation for the reaction is

CO(g)+ 2H₂(g) → CH₃OH

This means

1 mole of CO reacts with 2 moles of H₂ to produce 1 mole of CH₃OH

Now, we will determine the number of moles of H₂ present in the 3.20 L H₂ at STP

1 mol of an ideal gas has a volume of 22.4 L at STP

Then,

x mole of the H₂ gas will have a volume of 3.20 L at STP

x = [tex]\frac{3.20 \times 1}{22.4}[/tex]

x = 0.142857 mole

∴ The number of mole of H₂ present is 0.142857 mole

Since

2 moles of H₂ reacts to produce 1 mole of CH₃OH

Then,

0.142857 mole of H₂ will react to produce [tex]\frac{0.142857}{2}[/tex] mole of CH₃OH

[tex]\frac{0.142857}{2} = 0.0714285[/tex]

∴ The number of moles of CH₃OH produced = 0.0714285 mole

Now, for the volume of CH₃OH formed

Since

1 mol of an ideal gas has a volume of 22.4 L at STP

Then,

0.0714285 mol of CH₃OH will have a volume of 22.4 × 0.0714285  at STP

22.4 × 0.0714285 = 1.5999984 L ≅ 1.60 L

Hence, the volume of CH₃OH gas formed is 1.60 L

Learn more here: https://brainly.com/question/13899989

Answer:

A. The atomic mass is 243 amu, and the atomic number is 95.

Answer:

- Addition of Ba(OH)2: favors the formation of a precipitate.

- Undergo a chemical reaction forming soluble species.

- Addition of CuSO4 : favors the formation of a precipitate.

Explanation:

Hello,

In this case, since the dissociation reaction of barium sulfate is:

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

We must analyze the effect of the common ion:

- By adding barium hydroxide, more barium ions will be added to the equilibrium system so the formation of solid barium sulfate will be favored (reaction shifts leftwards towards reactants).

- By adding sodium nitrate, the following reaction will undergo:

[tex]BaSO_4(s)+NaNO_3(aq)\rightarrow Ba(NO_3)_2(aq)+Na_2SO_4(aq)[/tex]

So the precipitate will turn into other soluble species.

- By adding copper (II) sulfate, more sulfate ions will be added to the equilibrium system so the formation of solid barium sulfate will be favored (reaction shifts leftwards towards reactants).

All of this is supported by the Le Chatelier's principle.

Best regards.

Answer:

B) CsCl

Explanation:

The ionic character is formed between two kinds of atoms having a large electronegativity differences e.g metals (like those in groups IA and IIA) and nonmetals (like those in groups VIA and VIIA). The formation of an ionic character involves a transfer of electrons from the less electronegative atom(metal) to the more electronegative atom (non-metal) such that the two kinds of atoms now have completely filled outer shell like the noble gases.

In CsCl, electrons are being transferred from Cs⁺ to Cl⁻ . As a result of this transfer , the atom of the metal becomes positively charged (cation) while that of the non-metal becomes negatively charged (anion).

The highest percentage of ionic character will occur as a result of smaller negatively charged (anion) and larger positively charged (cation). From the options given, CsCl have the highest percentage of ionic character.

Answer:

For the mentioned reaction, the balanced chemical equation is:  

KBr (aq) + AgNO3 (s) ⇒ KNO3 (aq) + AgBr (s)

The number written in front of the ion, atoms, and molecules in a chemical reaction so that each of the elements on both the sides of reactants and products of the equation gets balanced is known as the stoichiometric coefficient.  

From the mentioned balanced equation, the stoichiometric coefficient before KBr is 1, AgNO3 is 1, KNO3 is 1, as well as before AgBr is also 1. Thus, it is clear that 1 mole of potassium bromide reacts with 1 mole of silver nitrate to produce 1 mole of potassium nitrate and 1 mole of silver bromide.  

Answer:

I think its 1.2 cause I divided 15.5 with 12 and got 1.2 as an answer

The given question is incomplete, the complete question is:

How does the spontaneity of the process below depend on temperature? PCI5(9)+H2O(g)POCI3(g) +2HCI(g) -126 kJ mol1, AS = 146 J K-'mol1 ΔΗ Select the correct answer below: nonspontaneous at all temperatures spontaneous at all temperatures spontaneous at high temperatures and nonspontaneous at low temperatures spontaneous at low temperatures and nonspontaneous at high temperatures

Answer:

The correct answer is spontaneous at all the temperatures.

Explanation:

Gibbs Free energy is an essential relation that determines the spontaneity of any reaction, that is, ΔG = ΔH - TΔS

When ΔG is less than zero, that is, negative, the reaction is considered to be in spontaneous state. Based on the given information, ΔH = -126 kJ/mol

= -126000 J/mol, it is negative

ΔS = 146 J/K/mol, it is positive

Now, ΔG = ΔH-TΔS

= (-ve) - T (+ve), Thus, when ΔH, is -ve, ΔS is +ve, -TΔS is -ve, the ΔG will be -ve. Therefore, reaction will be spontaneous at all the temperatures.  

Answer:

A precipitate will form.

[Ag⁺] = 2.8x10⁻⁵M

Explanation:

When Ag⁺ and CrO₄²⁻ are in solution, Ag₂CrO₄(s) is produced thus:

Ag₂CrO₄(s) ⇄ 2 Ag⁺(aq) + CrO₄²⁻(aq)

Ksp is defined as:

Ksp = 1.1x10⁻¹² = [Ag⁺]² [CrO₄²⁻]

Where the concentrations [] are in equilibrium

Reaction quotient, Q, is defined as:

Q = [Ag⁺]² [CrO₄²⁻]

Where the concentrations [] are the actual concentrations

If Q < Ksp, no precipitate will form, if Q >= Ksp, a precipitate will form,

The actual concentrations are -Where 500mL is the total volume of the solution-:

[Ag⁺] = [AgNO₃] = 0.12M ₓ (250mL / 500mL) = 0.06M

[CrO₄²⁻] = [Na₂CrO₄] = 0.33M × (250mL / 500mL) = 0.165M

And Q = [0.06M]² [0.165M] = 5.94x10⁻⁴

As Q > Ksp; a precipitate will form

In equilibrium, some Ag⁺ and some CrO₄⁻ reacts decreasing its concentration until the system reaches equilibrium. Equilibrium concentrations will be:

[Ag⁺] = 0.06M - 2X

[CrO₄²⁻] = 0.165M - X

Where X is defined as the reaction coordinate

Replacing in Ksp expression:

1.1x10⁻¹² = [0.06M - 2X]² [0.165M - X]

Solving for X:

X = 0.165M → False solution. Produce negative concentrations.

X = 0.0299986M

Replacing, equilibrium concentrations are:

[Ag⁺] = 0.06M - 2(0.0299986M)

[CrO₄²⁻] = 0.165M - 0.0299986M

[CrO₄²⁻] = 0.135M

Answer:

pH = 8.72

Explanation:

This is like a titration of a weak acid and a strong base, in this case, we are at the equivalence point plus we have the same mmoles of acid and base. We have completely neutralized the acid.

CH₃COOH      +     OH⁻        ⇄    CH₃COO⁻   +   H₂O

0.1M . 15 mL      0.1M . 15 mL

We only have (0.1M . 15 mL) mmoles of acetate ion. → 1.5 mmoles

As this compound acts like a base, we propose this equilibrium:

CH₃COO⁻   +  H₂O  ⇄  CH₃COOH      +     OH⁻   Kb

We need to work with Kb and we know, that Kw = Ka. Kb so, Kb = Kw/Ka

Kb = 1×10⁻¹⁴ /1×10 ⁻⁵ = 5.55×10⁻¹⁰

Concentration of CH₃COO⁻ → 1.5 mmol / 30mL (volumes of the solutions are additive) = 0.05M

So: [CH₃COOH] . [OH⁻] / [CH₃COO⁻] = Kb

x²/ 0.05-x = 5.55×10⁻¹⁰

We can avoid the quadractic equation because Kb is so small

[OH⁻] = √(5.55×10⁻¹⁰ . 0.05) = 5.27×10⁻⁶

pOH = - log [OH⁻]  → 5.28

pH = 14 - pOH = 8.72

The pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid should be 8.72.

Since the following equation should be used.

CH₃COOH      +     OH⁻        ⇄    CH₃COO⁻   +   H₂O

0.1M . 15 mL      0.1M . 15 mL

Now

(0.1M . 15 mL) mmoles of acetate ion. → 1.5 mmoles

So,

CH₃COO⁻   +  H₂O  ⇄  CH₃COOH      +     OH⁻   Kb

Now

Kw = Ka. Kb

Kb = Kw/Ka

And,

Kb = 1×10⁻¹⁴ /1×10 ⁻⁵

= 5.55×10⁻¹⁰

Now

[CH₃COOH] . [OH⁻] / [CH₃COO⁻] = Kb

x²/ 0.05-x = 5.55×10⁻¹⁰

Now

[OH⁻] = √(5.55×10⁻¹⁰ . 0.05) = 5.27×10⁻⁶

pOH = - log [OH⁻]  → 5.28

pH = 14 - pOH

= 8.72

Hence, The pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid should be 8.72.

Learn more about an acid here: https://brainly.com/question/4519963

Answer:

[tex]\% m/m= 14.3\%[/tex]

Explanation:

Hello,

In this case, the by mass percent is computed as shown below:

[tex]\% m/m=\frac{m_{solute}}{m_{solute}+m_{solvent}} *100\%[/tex]

Whereas the solute is the sugar and the solvent the water, therefore, the concentration results:

[tex]\% m/m=\frac{63g}{63g+378g} *100\%\\\\\% m/m= 14.3\%[/tex]

Best regards.

Answer:

Explanation:

Answer in attached file .

Answer:

30.12 mL.

Explanation:

We'll begin by calculating the molarity of the phosphoric acid. This can be obtained as follow:

Phosphoric acid H3PO4 will dissociate in water as follow:

H3PO4(aq) <==> 3H^2+(aq) + PO4^3-(aq)

From the balanced equation above,

1 mole of H3PO4 produces 3 moles of H+.

Therefore, XM H3PO4 will produce 8.70 M H+ i.e

XM H3PO4 = 8.70/3

XM H3PO4 = 2.9 M.

Therefore, the molarity of the acid solution, H3PO4 is 2.9 M.

Next, we shall write the balanced equation for the reaction. This is illustrated below:

2H3PO4 + 3Ba(OH)2 —> Ba3(PO4)2 + 6H2O

From the balanced equation above, we obtained the following:

Mole ratio of the acid, H3PO4 (nA) = 2

Mole ratio of the base, Ba(OH)2 (nB) = 3

Data obtained from the question include the following:

Molarity of base, Ba(OH)2 (Mb) = 6.50 M

Volume of base, Ba(OH)2 (Vb) =.?

Molarity of acid, H3PO4 (Ma) = 2.9 M

Volume of acid, H3PO4 (Va) = 45 mL

The volume of the base, Ba(OH)2 Needed for the reaction can be obtained as follow:

MaVa /MbVb = nA/nB

2.9 x 45 / 6.5 x Vb = 2/3

Cross multiply

2 x 6.5 x Vb = 2.9 x 45 x 3

Divide both side by 2 x 6.5

Vb = (2.9 x 45 x 3) /(2 x 6.5)

Vb = 30.12 mL

Therefore, the volume of the base, Ba(OH)2 needed for the reaction is 30.12 mL

Answer:

rate = [NO]²[H₂]

Explanation:

2NO + H2 ⟶N2 + H2O2 (slow)

H2O2 + H2 ⟶2H2O (fast)

From the question, we are given two equations.

In chemical kinetics; that is the study of rate reactions and changes in concentration. The rate law is obtained from the slowest reaction.

This means that our focus would be on the slow reaction. Generally the rate law is obtained from the concentrations of reactants in a reaction.

This means our rate law is;

rate = [NO]²[H₂]

Answer:

ethyl 4-bromobenzoate

Explanation:

In this question, we can start with the Index of Hydrogen Deficiency (I.H.D):

[tex]I.H.D=\frac{2C+2+N-H-X}{2}=\frac{(2*9)+2+0-9-1}{2}~=~5[/tex]

This indicates, that we can have a benzene ring (I.H.D = 4) and a carbonyl group (I.H.D = 1), for a total of 5.

Additionally, in the 1H-NMR info, we have a triplet 1.39 (3H) followed by a doublet 4.38 (2H), this indicates the presence of an ethyl group ([tex]CH_3-CH_2-[/tex]). Also, in the formula, we have 2 oxygens if we have carbonyl group with 2 oxygens we have a high probability to have an ester group.

[tex]O=C-O-CH_2-CH_3[/tex]

Now, if we add this to the benzene ring and the "Br" atom that we have in the formula, we will have ethyl 4-bromobenzoate.

See figures 1 and 2 to further explanations.

I hope it helps!

Answer:

water molecules

Explanation:

Redox reactions are carried out under acidic or basic conditions as the case may be.

If the reaction is carried out in an acid medium, then we must balance the hydrogen ions on the lefthand side of the reaction equation with water molecules on the righthand side of the reaction equation.

For instance, the equation for reduction of MnO4^- under acidic condition is shown below;

MnO4^-(aq) + 5e + 8H^+(aq) --------> Mn^2+(aq) + 4H2O(l)