. A 0.100 M solution of the weak acid HA was titrated with 0.100 M NaOH. The pH measured when Vb = ½Ve was 4.62. Using activity coefficients, calculate pKa. The size of the A- anion is 450 pm.

Answer:

4.32 is the ratio of f at the higher temperature to f at the lower temperature

Explanation:

Using the sum of Arrhenius equation you can obtain:

ln (f₂/f₁) = Eₐ / R ₓ (1/T₁ - 1/T₂)

Where f represents the rate constant of the reaction at T₁ and T₂ temperatures. Eₐ is the energy activation (155kJ / mol = 155000J/mol) and R is gas constant (8.314J/molK)

Replacing:

ln (f₂/f₁) = 155000J/mol / 8.314J/molK ₓ (1/495K - 1/515)

Where 2 represents the state with the higher temperature and 1 the lower temperature.

ln (f₂/f₁) = 155000J/mol / 8.314J/molK ₓ (1/495K - 1/515)

ln (f₂/f₁) = 1.4626

f₂/f₁ = 4.32

Answer:

Red dye

Explanation:

In the given question, the complete question has not been provided but the propanone is used as a solvent in paper chromatography. The paper chromatography was performed for the black ink in which the black ink got separated in the red, blue and yellow colour.

From these three colours that are red, blue and yellow, the dye which is most soluble in propanone was red as red colour moved the most in the given chromatogram and the dye which travelled the most is most soluble in propanone.

Thus, red dye is the correct answer.

Answer:

a. 2,3-dimethylbutane < 2-methylpentane < n-hexane

Explanation:

The boiling point of alkanes is highly affected by the degree of branching in the molecule. Branched alkanes generally have a lower boiling point than unbranched alkanes.

The reason for the higher boiling point of unbranched alkanes is because they have greater vanderwaals forces acting between their molecules due to their larger surface area. Recall that branched alkanes have a lesser surface area compared to unbranched alkanes.

n-hexane is an unbranched alkane hence it will have the highest boiling point followed by 2-methyl pentane and lastly 2,3-dimethyl butane. The boiling point continues to decrease as the extent of branching increases.

Answer:

a. NO₂⁻ + H⁺ → HNO₂

b. HNO₂ + OH⁻ → NO₂⁻ + H₂O

Explanation:

A buffer is defined as an aqueous mixture of a weak acid and its conjugate base or vice versa.

The buffer of the problem is HNO₂/NO₂⁻ where nitrous acid is the weak acid and NO₂⁻ is its conjugate base.

a. When a acid is added to a buffer as the buffer of the problem, the conjugate base will react with the acid, to produce the weak acid, thus:

NO₂⁻ + HCl → HNO₂ + Cl⁻

Ionic equation is:

NO₂⁻ + H⁺ + Cl⁻ → HNO₂ + Cl⁻

In the net ionic equation, you avoid the ions that don't react, that is:

b. In the same way, the weak acid will react with the strong acid producing water and the conjugate base, thus:

HNO₂ + NaOH → NO₂⁻ + H₂O + Na⁺

The ionic equation is:

HNO₂ + Na⁺ + OH⁻ → NO₂⁻ + H₂O + Na⁺

And the net ionic equation is:

Answer:

Sr

Explanation:

Sr has an ionization of 550 whereas Ba has an ionization of 503

Answer:

The volume of the ballon is 325L.

Explanation:

Boyle's law express that the pressure of a gas is inversely proportional to its volume. That means if the pressure increases, the volume decreases. The formula is:

P₁V₁ = P₂V₂

Where P represents pressure and V volume of 1, initial state and 2, final state of the gas.

In the problem, the volume of the tank is 13.0L and the final pressure of the ballon is 1atm -The atmospheric pressure-. As 1atm of gas is in the ballon, the pressure of the tank is 26.0atm - 1.0atm = 25.0atm.

Replacing in Boyle's law expression:

25.0atm*13.0L = 1atmV₂

325L = V₂

The volume of the ballon is 325L.

Answer:

32.062

Explanation:

The following data were obtained from the question:

Mass of isotope A (32S) = 31.97207 u

Abundance of isotope A (A%) = 95.0%

Mass of isotope B (33S) = 32.97146 u Abundance of isotope B (B%) = 0.76%

Mass of isotope C (34S) = 33.96786 u

Abundance of isotope C (C%) = 4.22%

Mass of isotope D (36S) = 35.96709 u Abundance of isotope D (D%) = 0.014%

Average atomic mass of S =..?

The average atomic mass of sulphur, S can be obtained as follow:

Average atomic mass = [(Mass of A x A%)/100] + [(Mass of B x B%)/100] + [(Mass of C x C%)/100] + [(Mass of D x D%)/100]

Average atomic mass of sulphur =

[(31.97207 x 95)/100] + [(32.97146 x 0.76)/100] + [(33.96786 x 4.22)/100] + [(35.96709 x 0.014)/100]

= 30.373 + 0.251 + 1.433 + 0.005

= 32.062

Therefore, the average atomic mass of sulphur is 32.062

Answer:

The concentration of  fructose-6-phosphate F6P ≅ 1.35 mM

Explanation:

Given that:

ΔG°′ is the  conversion of glucose-6-phosphate to fructose-6-phosphate (F6P)   = +1.67 kJ/mol = 1670 J/mol

concentration of glucose-6-phosphate at equilibrium = 2.65 mM

Assuming temperature = 25.0°C

=( 25 + 273)K

= 298 K

We are to find the concentration of fructose-6-phosphate

Using the relation;

ΔG' = -RT In K_c

where;

R = 8.314 J/K/mol

1670 = - (8.314 × 298 ) In K_c

1670 = -2477.572   × In K_c

1670/ 2477.572 =  In K_c

0.67 = In K_c

[tex]K_c = e^{-0.67}[/tex]

[tex]K_c =[/tex] 0.511

Now using the equilibrium constant [tex]K_c[/tex]

[tex]K_c = \dfrac{[F6P]}{[G6P]}[/tex]

[tex]0.511 = \dfrac{[F6P]}{[2.65]}[/tex]

F6P = 0.511 × 2.65

F6P = 1.35415

F6P ≅ 1.35 mM

Answer:

Two

Explanation:

2

rough endoplasmic reticulum

smooth endoplasmic reticulum

Answer: The change in enthalpy associated with the combustion of 322 g of ethanol is [tex]-9567.6kJ[/tex]

Explanation:

To calculate the number of moles we use the equation:

[tex]\text{Moles}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]

[tex]\text{Moles of ethanol}=\frac{322g}{46g/mol}=7moles[/tex]

The balanced chemical reaction is:

[tex]C_2H_5OH(l)+3O_2\rightarrow 2CO_2(g)+3H_2O(l)[/tex] [tex]\Delta H=-1366.8kJ/mol[/tex]

Given :

Energy released when 1 mole of ethanol is combusted = 1366.8 J

Thus Energy released when 7 moles of ethanol is combusted =[tex]\frac{1366.8}{1}\times 7=9567.6kJ[/tex]

Thus the change in enthalpy associated with the combustion of 322 g of ethanol is [tex]-9567.6kJ[/tex]

The  change in enthalpy associated with the combustion is -9567.6KJ

Since there is 322g of ethanol

Also, there is the chemical equation i.e.

C2H5OH(l)+3O2(g)⟶2CO2(g)+3H2O(l)ΔH∘c=−1366.8kJ/mol

So, the change should be

= -1366.8kJ *7/1

= -9567.6KJ

Since Energy released at the time when 1 mole of ethanol is combusted = 1366.8 J

So, here Energy released when 7 moles of ethanol is combusted

Learn more about energy here: https://brainly.com/question/17121992

Answer:

0.0025Litters

Explanation:

2.5ml= 2.5x10^-3l

2.5ml= 0.0025l

Answer:

AAAAAAAA

Explanation:

Answer:

Explanation:

1. A student mixes baking soda and vinegar in a glass. The results are shown at left. ... Yes I do belive that new substances are being formed because there is a chemical reaction between the baking soda and vinegar turning it into a bubbly substances instead of a powder and liquid.

Yes, there are new substances created from this mixture.

Answer: Mass of hydrogen produced is 0.0376 g.

Explanation:

The reaction equation will be as follows.

[tex]NaH(aq) + H_{2}O(l) \rightarrow H_{2}(g) + NaOH(aq)[/tex]

Now, formula for total pressure will be as follows.

  [tex]P_{total} = P_{H_{2}} + P_{H_{2}O}[/tex]

Hence,    [tex]P_{H_{2}} = P_{total} - P_{H_{2}O}[/tex]

                            = 755 mm Hg - 42.23 mm Hg

                            = 712.77 mm Hg

[tex]P_{H_{2}} = \frac{712.77 \times 1 atm}{760 mm Hg}[/tex]

             = 0.937 atm

Now, we will calculate the moles of [tex]H_{2}[/tex] as follows.

   [tex]P_{H_{2}}V = nRT[/tex]

   [tex]0.937 atm \times 0.505 L = n \times 0.0821 \times 308.15 K[/tex]

      n = [tex]\frac{0.473}{25.29}[/tex] mol

         = 0.0187 mol

Therefore, mass of [tex]H_{2}[/tex] will be calculated as follows.

        [tex]m_{H_{2}} = \frac{0.0187 mol \times 2.0158 g}{1 mol}[/tex]

                   = 0.0376 g

Thus, we can conclude that mass of hydrogen produced is 0.0376 g.

Answer:

both are the types of mixture and both are impure substances that donot have fixed composition and the composition of constituents  is not uniform

Answer:

Their components van be separated by physical processes

Explanation:

Out of the answers im given, it makes the most sense. I would double check before submitting though

Answer:

K₂SO₄ + Pb(C₂H₃O₂)₂ →PbSO₄ + 2KC₂H₃O₂

A chemical equation is a symbolic representation of a chemical reaction. The chemical equation for the reaction between potassium sulfate ([tex]K_2SO_4[/tex]) and lead(II) acetate ([tex]Pb(CH_3COO)_2[/tex]) can be written as follows:

[tex]K_2SO_4 + Pb(CH_3COO)_2 = PbSO_4 + 2CH_3COOK[/tex]

A basic chemical equation consists of two main parts: the reactant side (left side) and the product side (right side), separated by an arrow indicating the direction of the reaction. Reactants are substances that undergo a chemical change, while products are substances formed as a result of the reaction.

In this reaction, potassium sulfate reacts with lead(II) acetate to form lead(II) sulfate and potassium acetate. It is important to note that the equation is balanced with stoichiometric coefficients, ensuring that the number of atoms of each element is the same on both sides of the equation.

Therefore, the chemical equation for the reaction between potassium sulfate ([tex]K_2SO_4[/tex]) and lead(II) acetate ([tex]Pb(CH_3COO)_2[/tex]) can be written as follows:

[tex]K_2SO_4 + Pb(CH_3COO)_2 = PbSO_4 + 2CH_3COOK[/tex]

For more details regarding chemical equations, visit:

https://brainly.com/question/28792948

#SPJ6

Answer:

0.252 milimoles

Explanation:

To convert mass of a substance to moles it is necessary to use the molar mass of the substance.

The formula of morphine is C₁₇H₁₉NO₃, thus, its molar mass is:

C: 17*12.01g/mol = 204.17g/mol

H: 19*1.01g/mol = 19.19g/mol

N: 1*14g/mol = 14g/mol

O: 3*16g/mol = 48g/mol.

204.17 + 19.19 + 14 + 16 = 285.36g/mol

Thus, moles of 71.891 mg = 0.071891g:

0.071891g × (1mol / 285.36g) = 2.5193x10⁻⁴ moles

As 1 mole = 1000 milimoles:

2.5193x10⁻⁴ moles = 0.252 milimoles

Answer:

t = 0.049 mins or 2.94 secs

Explanation:

For a simple second order reaction, the integrated law which describes the concentration of reactants at a given time t, is as follows: 1/[A] = 1/[A]o + Kt;

Where [A] is concentration of reactant at time, t, [A]o is initial concentration of A; K is rate constant; t is time at a given instant.

Using the integrated rate law:

I/[NO2]t - 1/[NO2]o = Kt

Where K = 3.40 L/mol/min

[NO2]t = 1.5 mol/L

[N02]o = 2.0 mol/L

t = ?

Making t subject of formula;

t = (1/[NO2]t - 1/[NO2]o) / K

t = (1/1.5 - 1/2.0)/3.40

t = 0.049 mins or 2.94 secs

Answer:

1s²,2s²,2p⁶,3s²,3p⁶,4s²,3d¹⁰,4p⁶,5s⁰,4d¹⁰.

Explanation:

Palladium is a chemical element with the symbol Pd and atomic number 46.

The electronic configuration is;

[Kr] 4d¹⁰

The full electronic configuration observed for palladium is given as;

1s²,2s²,2p⁶,3s²,3p⁶,4s²,3d¹⁰,4p⁶,5s⁰,4d¹⁰.

The reason for for the anomlaous electron configuration is beacuse;

1. Full d orbitals are more stable than partially filled ones.

2. At higher energy levels, the levels are said to be degenerated which means that they have very close energies and then electrons can jump from one orbital to another easily.

Answer:

0. 000115

Explanation:

A percentage is defined as a ratio with a basis of 100 as total substance. Convert a percentage to decimal implies to divide the percentage in 100 because decimal form has as basis 1.

For the isotopic forms:

1H: 99.98% → As percent.

99.98% / 100 = 0.9998 → As decimal form.

2H: 0.0115% → As percent.

0.0115% / 100 = 0. 000115→ As decimal form.

The percent should be 0. 000115

For the isotopic forms:

1H: 99.98% → As percent.

Now

[tex]99.98\% \div 100[/tex]= 0.9998 → As decimal form.

Now

2H: 0.0115% → As percent.

And,

[tex]0.0115\% \div 100[/tex]= 0. 000115→ As decimal form.

Learn more: https://brainly.com/question/6789603?referrer=searchResults

Answer:

4.74

Explanation:

It is possible to find pH of a buffer (The mixture of a weak acid: Acetic acid, with its conjugate base: Sodium acetate) using Henderson-Hasselbalch equation:

pH = pKa + log₁₀ [A⁻] / [HA]

Where pKa is -log Ka of the weak acid,  [A⁻] concentration of the conjugate base and [HA] concentration of the weak acid

pKa of acetic acid is -log 1.8x10⁻⁵ = 4.74

The concentration of both, acetic acid and sodium acetate is 0.1M. Replacing in H-H equation:

pH = pKa + log₁₀ [A⁻] / [HA]

pH = 4.74 + log₁₀ [0.1] / [0.1]

pH = 4.74 + log₁₀ 1

pH = 4.74

Theoretical pH is 4.74

Answer:

35.578g or 36g if you round

Explanation:

Q=mc ∆∅ where ∅ is temperature difference

1160= m x 1.716 x (42-23)

m = 1160/ 1.716 x19

m=35.578g

m = 36g to nearest whole number

Answer: C. 36 g

Explanation: I got this right on Edmentum.

Answer:

Concentration of the H₂SO₄ solution is 0.0737 M

Explanation:

Equation of the neutralization reaction between the acid, H₂SO₄, and the base, NaOH, is given below:

H₂SO₄ + 2NaOH -----> Na₂SO₄ + 2H₂O

From the above equation, one mole of acid requires 2 moles of base for complete neutralization which occurs at phenolphthalein endpoint.

mole ratio of acid to base, nA/nB = 1:2

Concentration of the base, Cb = 0.1311 M

Volume of base, Vb, = 28.11 mL

Concentration of acid, Ca = ?

Volume of acid, Va + 25.0 mL

Using the formula, CaVa/CbVb = nA/nB

making Ca subject of the formula, Ca = Cb*Vb*nA/Va*nB

substituting the values into the equation

Ca = (0.1311 * 28.11 * 1) / 25.0 * 2 = 0.0737 M

Therefore, concentration of the H₂SO₄ solution is 0.0737 M

Answer:

31.2g

Explanation:

The following data were obtained from the question:

Volume of bromine = 10mL

Density of bromine = 3.12 g/mL

Mass of bromine =...?

The Density of the substance is related to it's mass and volume by the following equation:

Density = Mass /volume

With the above equation, we can calculate the mass of bromine as follow:

Density = Mass /volume

Volume of bromine = 10mL

Density of bromine = 3.12 g/mL

Mass of bromine =...?

Density = Mass /volume

3.12 = Mass /10

Cross multiply

Mass of bromine = 3.12 x 10

Mass of bromine = 31.2g

Therefore, the mass of bromine is 31.2g

Answer and Explanation:

Sodium hydroxide (NaOH) is a strong base and nitric acid (HNO₃) is a strong acid. That means that they dissociates in water by giving the ions:

NaOH ⇒ Na⁺(ac) + OH⁻(ac)

HNO₃ ⇒ H⁺(ac) + NO₃⁻(ac)

The reaction between an acid and a base is called neutralization. In this case, HNO₃ loses its proton and it is converted in NO₃⁻ (nitrate anion). NaOH loses its hydroxyl anion (OH⁻) by giving Na⁺ cations.

Na⁺ cations with NO₃⁻ anions form the salt NaNO₃ (sodium nitrate); whereas H⁺ and OH⁻ form water molecules. The complete equation is the following:

HNO₃(ac) + NaOH(ac) ⇒ NaNO₃(ac) + H₂O(l)

The ionic equation is:

H⁺(ac) + NO₃⁻(ac) + Na⁺(ac) + OH⁻(ac) ⇄ Na⁺(ac) + NO₃⁻(ac) + H₂O(ac)

If we cancel the repeated ions at both sides of the equation, it gives the following ionic reaction:

H⁺(ac) + OH⁻(ac) ⇄ H₂O(ac)

Answer:

Take a look at the attachment below

Explanation:

Hope that helps!

Answer:

Explanation:

use this equation and solve for wavelength (λ)

λ = h/mv.

where h is plancks constant 6.63 × 10−34 J·s

m = mass of lectron

v = velcoeity of electron

Answer:

0.9180 M

Explanation:

Step 1: Write the balanced equation

H₂SO₄(aq) + 2 KOH(aq) ⇒ K₂SO₄(aq) + 2 H₂O(l)

Step 2: Calculate the reacting moles of KOH

27.00 mL of 1.700 M KOH react. The reacting moles of KOH are:

[tex]0.02700L \times \frac{1.700mol}{L} = 0.04590mol[/tex]

Step 3: Calculate the reacting moles of H₂SO₄

The molar ratio of H₂SO₄ to KOH is 1:2. The reacting moles of H₂SO₄ are 1/2 × 0.04590 mol = 0.02295 mol.

Step 4: Calculate the molarity of H₂SO₄

0.02295 moles of H₂SO₄ are in 25.00 mL of solution. The molarity of the acid solution is:

[tex]M = \frac{0.02295 mol}{0.02500} = 0.9180 M[/tex]

Answer:

28g.

Explanation:

Hello,

In this case, considering the statement, we can infer that the monoatomic atomic mass of B is 14 g in one one mole. In such a way, since it is diatomic, we can notice that one mole of B2, is having 28 g of B2, as monoatomic atomic mass is considered twice.

Regards.

Answer:

it was based on the studies by emission line spectra of various elements

Explanation:

Answer:

The answer is: emission line spectra of various elements

Explanation: