When solutions of and are mixed, the mixture becomes cloudy. After a few days, a white solid is observed on the bottom of the beaker with a clear liquid above it. (a) Write a balanced equation for the reaction that occurs. (b) Why is the solution cloudy at first

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:

Take a look at the attachment below

Explanation:

Hope that helps!

Answer:

- [tex]AgBr(s)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)[/tex]

- [tex]K=1.2x10^{-5}[/tex]

Explanation:

Hello,

In this case, by considering the dissolution of silver bromide:

[tex]AgBr(s)\rightleftharpoons Ag^+(aq)+Br^-(aq) \ \ \ Ksp=[Ag^+][Br^-]=7.7x10^{-13}[/tex]

And the formation of the complex:

[tex]Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)\ \ \ Kf=\frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}=1.6x10^7[/tex]

We obtain the balanced net ionic equation by adding the aforementioned equations:

[tex]AgBr(s)+Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)+Ag^+(aq)\\\\AgBr(s)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)[/tex]

Now, the equilibrium constant is obtained by writing the law of mass action for the non-simplified net ionic equation:

[tex]AgBr(s)+Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-+Ag^+\\\\K=[Ag^+][Br^-]*\frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}[/tex]

So we notice that the equilibrium constant contains the solubility constant and formation constant for the initial reactions:

[tex]K=Ksp*Kf=7.7x10^{-13}*1.6x10^{7}\\\\K=1.2x10^{-5}[/tex]

Best regards.

Answer:

185.49 grams of Zinc would react with 454g (1lb) of copper sulfate

Explanation:

Yo know the following balanced reaction:

CuSO₄(aq)+ Zn(s) →Cu(s) + ZnSO₄(aq)

You can see that by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reagents and products are part of the reaction:

Being:

the molar mass of the compounds participating in the reaction is:

Then, by stoichiometry of the reaction, the following amounts of mass of reagent and product participate in the reaction:

Now you can apply the following rule of three: if 160 grams of CuSO₄ react with 65.37 grams of Zn by this reaction stoichiometry, 454 grams of CuSO₄ with how much mass of Zn will it react?

[tex]mass of Zn=\frac{454 grams of CuSO_{4} *65.37 grams of Zn}{160 grams of CuSO_{4}}[/tex]

mass of Zn= 185.49 grams

185.49 grams of Zinc would react with 454g (1lb) of copper sulfate

Answer:

The correct answer is -1659.17 J/K.

Explanation:

The reaction given is:  

2H₂O (l) ⇔ 2H₂ (g) + O₂ (g)

In the given case, first there is a need to find ΔHreaction, which is equivalent to ΔHf (products) - ΔHf(reactants)

Based on the standard thermodynamic table, the ΔHf(H₂O) is -285.8 KJ/mol, the ΔHf(H₂) is 0 KJ/mol, and the ΔHf(O₂) is 0 KJ/mol.  

On putting the values, the ΔHreaction will be,  

ΔHreaction = 2 × ΔHf(H₂) + ΔHf(O₂) - 2 × ΔHf(H₂O)

= 2 × 0 + 0 - 2 × (-285.8 KJ/mol) = 571.6 KJ

The calculated value of ΔHreaction is for the two moles of H₂O, now for 1.73 moles of H₂O it will be,  

ΔHreaction = +571.6 KJ / 2 mol × 1.73 mol = 494.434 KJ

The temperature given in the question is 298 K, now ΔSsurrounding will be,  

ΔSsurrounding = -ΔHreaction/T = -494434 J/298 K = -1659.17 J/K.  

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:

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:potassium is more reactive than Mg because both lie in the same group and the element potassium has more electropositivity than magnesium

Explanation:

I hope it will help you

Answer: B. Potassium(K)

Explanation:

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:

4–octene.

Explanation:

To name the compound given in the question, we must determine the following:

1. Determine the functional group of the compound and locate its position by giving it the lowest possible count.

2. Locate the longest continuous carbon. This gives the parent name of the compound.

3. Combine the above to obtain the name of the compound.

Now, let us determine the name of the compound bearing in mind the information given above. This is illustrated below:

1. The functional group of the compound is double bond i.e alkene and it located at carbon 4.

2. The longest continuous carbon chain is 8. Since the compound is an alkene, the name becomes octene.

3. Therefore, the name of the compound is:

4–octene.

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:

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:

For every given mass of Vanadium, the relative number of oxygen atoms present or the mole ratio of Oxygen to Vanadium is:

A. 1:1

B. 3:2

C. 2:1

D. 5:2

Note: The question is stated more clearly below:

Vanadium (V) and oxygen (O) form a series of compounds with the following compositions: Mass % V 76.10 67.98 61.42 56.02 Mass % O 23.90 32.02 38.58 43.98 Compound Mass % N 1 33.28 2 39.94.

What are the relative numbers of atoms of oxygen in the compounds for a given mass of vanadium?

Explanation:

Number of moles in 100 g mass = % mass / molar mass

Molar mass of Vanadium, V = 51 g/mol

Molar mass of oxygen atom, O = 16 g/mol

1. Percentage mass of V and O is 76.10% and 23.90% respectively.

Number of moles of each atom;

V = 76.10/51.0 = 1.5 moles

O = 23.9/16 = 1.5 moles

Mole ratio of oxygen to vanadium = 1.5/1.5 = 1 : 1

2. Percentage mass of V and O is 67.98% and 32.02% respectively

Number of moles of each atom:

V = 67.98/51 = 1.33

O = 32.02/16 = 2

Mole ratio of oxygen to vanadium = 2/1.33 = 1.5 : 1 = 3 : 2

3. Percentage mass of V and O is 61.42% and 38.58% respectively

Number of moles of each atom:

V = 61.42/51 = 1.2

O = 38.58/16 = 2.4

Mole ratio of oxygen to vanadium = 2.4/1.2 = 2 : 1

4. Percentage mass of V and O is 56.02% and 43.98% respectively

Number of moles of each atom:

V = 56.02/51 = 1.10

O = 43.98/16 = 2.75

Mole ratio of oxygen to vanadium = 2.75/1.10 = 2.5 : 1 = 5 : 2

Mass of the Vanadium, number of O2 atoms present, or the mole ratio of   1:1 , 3:2 , 2:1 , 5:2 . As Vanadium (V) and oxygen (O) form a series of compounds is given with masses of 76.10 67.98, 23.90 32.02, 33.28 2 39.94, etc.

Learn more about the Vanadium (V) and oxygen (O).  

brainly.com/question/2145642.

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:

The correct answer is option A, that is, the organic layer.

Explanation:

One knows that non-polar solvent dissolves non-polar substances and the polar solvent dissolves polar substances, that is, like dissolved like. The toluic acid or benzoic acid refers to the carboxylic acids, which are non-polar or are very less polar in characteristics.  

Therefore, they possess the tendency to remain in the organic solvent, which is non-polar like diethyl ether. The toluic acid's or benzoic acid's non-polar characteristic is because of the existence of huge hydrophobic rings of benzene. Hence, the correct answer is option A, that is, the organic layer.  

Answer:

Alkene form hexan-1-ol with oxidation in presence of NaOH with highest yield  

Explanation:

Answer:

2.41065 grams

Explanation:

Here we have to apply molarity, particularly in reference to the equation molarity = moles of solute / volume. I would like to rewrite this formula, but with respect to the units - grams = moles / Liters,

We can use molarity to determine the number of moles. After doing so, we can determine the mass of the solute with respect to the formula moles = mass / molar mass. The molar mass of NaCl is 58.44 grams.

_______________________________________________________

275 mL = 0.275 L,

Number of Moles of NaCl = 0.150 * 0.275 = 0.04125 moles,

Mass = 0.04125 * 58.44 = 2.41065 grams,

Solution - Mass of NaCl = 2.41065 grams

Hope that helps!

Answer:

2.41g

Explanation:

Data obtained from the question include the following:

Molarity of NaCl = 0.150M

Volume = 275mL

Mass of NaCl =..?

Next, we shall determine the number of mole of NaCl in the solution. This is illustrated below:

Molarity of a solution is simply defined as the mole of solute per unit litre of the solution. Mathematically, the molarity is expressed as:

Molarity = mole /Volume

Molarity = 0.150M

Volume = 275mL = 275/1000 = 0.275L

Mole =..?

Molarity = mole /Volume

0.150 = mole /0.275

Cross multiply

Mole = 0.15 x 0.275

Mole = 0.04125 mole

Therefore, the number of mole of solute, NaCl in the solution is 0.04125 mole.

Finally, we shall convert 0.04125 mole of NaCl to farms. This is illustrated below:

Molar mass of NaCl =23 + 35.5 = 58.5g/mol

Mole of NaCl = 0.04125 mole

Mass of NaCl =..?

Mole = mass/molar mass

0.04125 = Mass /58.5

Cross multiply

Mass = 0.04125 x 58.5

Mass = 2.41g

Therefore, the mass of the solute, NaCl needed to prepare the solution is 2.41g

Strong acids are completely ionised and weak acids are partly ionised

Answer:

Como forman los iones en soluciión

Explanation:

Los ácidos fuertes y las bases fuertes se refieren a especies que se disocian completamente para formar los iones en solución. Por el contrario, los ácidos y bases débiles se ionizan solo parcialmente y la reacción de ionización es reversible.

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:

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:

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:

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:

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:

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:

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:

Reactant concentration, the physical states of the reactants, surface area, temperature and the presence of catalyst.

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:

a) HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻

b) HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂

c) C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻

Explanation:

a) In a HC₂H₃O₂/C₂H₃O₂⁻ buffer system, the following reactions take place:

HC₂H₃O₂ + H₂O ⇄ C₂H₃O₂⁻ + H₃O⁺

C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻

Thus, the species present are: HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻.

b) When LiOH is added to the buffer system, it is partially neutralized according to the following equation.

HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂

c) When HBr is added to the buffer system, it is partially neutralized according to the following equation.

C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻

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.

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