Consider the reaction for the decomposition of hydrogen disulfide: 2H2S(g)⇌2H2(g)+S2(g), Kc = 1.67×10−7 at 800∘C A 0.500 L reaction vessel initially contains 0.163 mol of H2S and 5.00×10−2 mol of H2 at 800∘C. Find the equilibrium concentration of [S2].

Answers

Answer 1

Final answer:

The equilibrium concentration of [[tex]S_2[/tex]] is [tex]1.12 \times 10^{-3} M[/tex].

Explanation:

To find the equilibrium concentration of [[tex]S_2[/tex]], we need to use the equilibrium constant expression [tex]\( K_c = \frac{{[H_2]^2[S_2]}}{{[H_2S]^2}} \)[/tex]. Given the initial concentrations of [tex]H_{2}S[/tex] and [tex]H_2[/tex], and assuming x moles of [tex]H_{2}S[/tex] decompose, we can set up an ICE table and solve for the equilibrium concentrations. Substituting the equilibrium concentrations into the equilibrium constant expression and solving for [[tex]S_{2[/tex]], we find the equilibrium concentration to be [tex]1.12 \times 10^{-3} M[/tex]. This indicates the concentration of [tex]S_2[/tex] at equilibrium after the reaction has reached its dynamic equilibrium state at 800°C.


Related Questions

Ca3(PO4)2 + 3 H2SO4 ⟶ 2 H3PO4 + 3 CaSO4

Joaquin needs to react [m] grams of calcium phosphate. He will need to measure out ____ grams H2SO4 for this reaction.

Answers

Answer:

The answer to your question is below

Explanation:

Balanced chemical reaction

               Ca₃(PO₄)₂  +  3H₂SO₄   ⇒   2H₃SO₄  +  3CaSO₄

To answer this question just calculate the molar mass of both reactants.

Molar mass of Ca₃(PO₄)₂ = (3 x 40) + (2 x 31) + (8 x 16)

                                         = 120 + 62 + 128

                                         = 310 g

Molar mass of 3H₂SO₄ = 3[(2 x 1) + (1 x 32) + (4 x 16)]

                                      = 3[2 + 32 + 64]

                                      = 3[98]

                                      = 294 g

Conclusion

310 g of Ca₃(PO₄)₂ will react with 294 g of 3H₂SO₄

Paradichlorobenzene, C6H4Cl2, is a component of mothballs. A solution of 2.00 g in 22.5 g of cyclohexane boils at 82.39 ∘C. The boiling point of pure cyclohexane is 80.70 ∘C. Calculate Kb for cyclohexane.

Answers

Answer:

The Kb for cyclohexane is 2.79 °C/m

Explanation:

Step 1: Data given

Mass of Paradichlorobenzene = 2.00 grams

Mass of cyclohexane = 22.5 grams

Boiling point of the solution = 82.39 °C

Boiling point of pure cyclohexane = 80.70 °C

Molar mass of Paradichlorobenzene = 147 g/mol

Step 2: Calculate moles Paradichlorobenzene

Moles Paradichlorobenzene = mass / molar mass

Moles Paradichlorobenzene = 2.00 grams / 147 g/mol

Moles Paradichlorobenzene = 0.0136 moles

Step 3: Calculate molality

Molality = moles Paradichlorobenzene / mass cyclohexane

Molality = 0.0136 moles / 0.0225 kg

Molality = 0.605 molal

Step 4: Calculate Kb

Kb = change in boiling point / molality of solution  

 ⇒ Change in boiling point = 82.39 - 80.70 = 1.69 °C

⇒ molality = 0.605 molal

Kb = 1.69 °C / 0.605 molal = 2.79 °C/m

The Kb for cyclohexane is 2.79 °C/m

Answer:

The Kb for cyclohexane is 2.80°C/m

Explanation:

delta T = 82.39 - 80.70 = 1.69 °C

Moles C6H4Cl2 = 2.00 g/ 147.0 g/mol= 0.0136

molality = 0.0136 mol / 0.0225 Kg = 0.604

1.69 = 0.604 x kf

kf = 2.80

The Kb for cyclohexane is 2.80°C/m

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When measuring Potassium with an ion-selective electrode by means of a liquid ion-exchange membrane, what antibiotic will be incorporated into the membrane?

Answers

Answer:

Valinomycin

Explanation:

This antibiotic is acquired from the Streptomyces species cells. Valinomycin is selective to potassium and inhibits sodium ions from entering the cell. This antibiotic allows the potassium ions to move down the electrochemical potential gradient of  the lipid membranes.

Final answer:

Valinomycin is the antibiotic incorporated into the liquid ion-exchange membrane when measuring Potassium with an ion-selective electrode. It binds specifically to potassium ions, thus allowing accurate detection and measurement.

Explanation:

When measuring Potassium with an ion-selective electrode by means of a liquid ion-exchange membrane, the antibiotic incorporated into the membrane is valinomycin.

Valinomycin is a relatively specific antibiotic used in these measurements because it binds selectively with potassium ions, allowing for accurate detection and measurement. The mechanism operates such that when the potassium is bound by the valinomycin, it causes a change in the membrane's potential. This change can be measured and is proportional to the concentration of potassium in the solution.

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When a solution of AgNO3 is mixed with a solution of NaBr (multiple options could be correct):

1) NaNO3 precipitate will spontaneously form.


2) After a few minutes pass, the concentration of Ag+ and Br- will be lower than when the two solutions were first mixed.


3) After a few minutes pass, the concentration of Ag+ and Br- will be the same as when the two solutions were first mixed


4) The concentration of Ag+ and Br- are momentarily greater than in a saturated solution of AgBr.


5) AgBr precipitate will spontaneously form.

Answers

Answer:

After a few minutes pass, the concentration of Ag+ and Br- will be lower than when the two solutions were first mixed.

AgBr precipitate will spontaneously form.

Explanation:

After the net ionic equation AgBr forms

Final answer:

The reaction between AgNO3 and NaBr leads to the formation of AgBr, a precipitate, and NaNO3, which remains dissolved in the solution. The concentration of Ag+ and Br- ions in the resulting solution will therefore be decreased.

Explanation:

When a solution of AgNO3 is mixed with a solution of NaBr, a double displacement reaction or metathesis occurs, producing AgBr and NaNO3. According to solubility rules, AgBr is a precipitate, hence the options 2 and 5 are correct: After a few minutes, the concentration of Ag+ and Br- ions will be lower than when the two solutions were first mixed, and AgBr precipitate will spontaneously form. On the other hand, NaNO3 is soluble in water, so it remains dissolved and no precipitate of NaNO3 forms, which makes option 1 incorrect. Options 3 and 4 are also incorrect because the concentration of Ag+ and Br- ions will decrease as they form a precipitate and won't exceed the saturation point of AgBr.

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What form does nitrogen take in the atmosphere?

N

NH4+

N2

NO3–

Answers

Answer:

N2

Explanation:

The nitrogen element exists as N₂ molecule in the atmosphere which is a major constituent of air.

What is an element?

It is defined as a substance which cannot be broken down further into any other substance. Each element is made up of its own type of atom. Due to this reason all elements are different from one another.

Elements can be classified as metals and non-metals. Metals are shiny and conduct electricity and are all solids at room temperature except mercury. Non-metals do not conduct electricity and are mostly gases at room temperature except carbon and sulfur.

The number of protons in the nucleus is the defining property of an element and is related to the atomic number.All atoms with same atomic number are atoms of same element.Elements combine to give compounds.

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Which of the following electron configurations for neutral atoms is correct? Li nitrogen: 1s 2, 2s 2, 2p 3 the third sub-level silicon: 1s 2, 2s 2, 3s 2, 2p 6 3p 2 helium: 1s 1, 1p 1

Answers

ans is helium.........

Answer:

The answer is: helium: 1s 1, 1p 1

Explanation:

What pressure does 3.54 moles of chlorine gas at 376 k exert on the walls of it 51.2 l container? I'm Lazy sooooo

Answers

Answer:

The answer to your question is P = 2.13 atm

Explanation:

Data

Pressure = ?

number of moles = 3.54

Temperature = 376 °K

Volume = 51.2 L

R = 0.08205 atm L/mol°K

Formula

PV = nRT

- Solve for P

  P = nRT / V

- Substitution

  P = (3.54)(0.08205)(376) / 51.2

- Simplification

  P = 109.21 / 51.2

Result

 P = 2.13 atm  

Answer:

2.13

Explanation:

I just did the problem on acellus and got it right

The histogram shown below represents the weights (in kg) of 47 female and 97 male cats. Approximately % of these cats weigh less than 2.5kg. Approximately % of these cats weigh between 2.5 and 2.75kg. Approximately % of these cats weigh between 2.75 and 3.5kg.

Answers

Answer:

Approximately % of these cats weigh less than 2.5kg

the percentage = 61/144 × 100 = 6100/144 = 42.3611111111  ≈ 42.36%

Approximately % of these cats weigh between 2.5 and 2.75 kg

percentage = 20/144 ×100 = 2000/144 = 13.8888888889  ≈ 13.90%

Approximately % of these cats weigh between 2.75 and 3.5kg.

percentage = 54/144 × 100 = 5400/144 = 37.5 %

Explanation:

The picture below is the histogram used .

The horizontal is the weight of the cat . The vertical is the number of cat. The cats have 47 female and 97 male . The total cats is 47 + 97 = 144.

Approximately % of these cats weigh less than 2.5kg

Cat that weighs less than 2.5 kg is the sum of the first bar and the second bar. The sum is 29 + 32 = 61

61 cat weighs less than 2.5 kg

the percentage = 61/144 × 100 = 6100/144 = 42.3611111111  ≈ 42.36

Approximately % of these cats weigh between 2.5 and 2.75 kg

cat that weigh between 2.5 and 2.75 is 20

percentage = 20/144 ×100 = 2000/144 = 13.8888888889  ≈ 13.90

Approximately % of these cats weigh between 2.75 and 3.5kg.

The number of cat that fall under this category is 27 + 12 + 15 = 54

percentage = 54/144 × 100 = 5400/144 = 37.5

The approximately % of cats weigh less than 2.5 kg has been 42.36 %.

The approximately % of cats weighing between 2.5 and 2.75 kg has been 13.88 %.

The approximately % of cats weighing between 2.75 and 3.5 kg has been 37.5 %.

The histogram has been the representation of the data in a user defined condensed format. The total number of cats has been the sum of male and female cats.

The given male cats can be, [tex]C_M=97[/tex]

The given female cats, [tex]C_F=47[/tex]

The total cats (C) can be given as:

[tex]C=C_M\;+\;C_F\\C=97\;+\;47\\C=144[/tex]

The total number of cats has been 144.

The percentage of cats weigh less than 2.5 kg has been given as:

From the histogram, the number of cats weighing less than 2.5 kg, [tex]C_>_2_._5=61[/tex]

The % of cats weighing less than 2.5 kg ([tex]C_>_2_._5\;\%[/tex]) has been:

[tex]C_>_2_._5\;\%=\dfrac{61}{144}\;\times\;100\\ C_>_2_._5\;\%=42.36\%[/tex]

The approximately % of cats weigh less than 2.5 kg has been 42.36 %.

The percentage of cats weigh between 2.5 and 2.75 kg has been given as:

From the histogram, the number of cats weighing between 2.5 and 2.75 kg, [tex]C_2_._5_-_2_._7_5=20[/tex]

The % of cats weighing between 2.5 and 2.75 kg,  [tex]C_2_._5_-_2_._7_5\%[/tex], has been:

[tex]C_2_._5_-_2_._7_5\%=\dfrac{20}{144}\;\times\;100\\ C_2_._5_-_2_._7_5\%=13.88\%[/tex]

The approximately % of cats weighing between 2.5 and 2.75 kg has been 13.88 %.

The percentage of cats weigh between 2.75 and 3.5 kg has been given as:

From the histogram, the number of cats weighing between 2.75 and 3.5 kg, [tex]C_2_._7_5_-_3_._5=54[/tex]

The % of cats weighing between 2.75 and 3. 5 kg,  [tex]C_2_._7_5_-_3_._5\%[/tex], has been:

[tex]C_2_._7_5_-_3_._5\%=\dfrac{54}{144}\;\times\;100\\ C_2_._7_5_-_3_._5\%=37.5\%[/tex]

The approximately % of cats weighing between 2.75 and 3.5 kg has been 37.5 %.

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At 19.9 degrees Celsius, we dissolve a salt crystal.

The Red dot on the graph in the image above represents the time it took to dissolve this salt crystal and the temperature at which it dissolved.
What temperature is represented by the red dot?
Enter the number of degrees Celsius only; no units.

Answers

Answer:

The red dot represents the melting point of the element, which as stated is approximately 19.9 degrees Celsius and how long it took for the heat to properly completely dissolve it.

The question kind of answers itself however, is there a way to re-word it or is there a different answer you're looking for?

The temperature i,e represented by the red dot is 19.9 degrees.

Calculation of the temperature;

Since the Red dot on the graph shows for dissolve this salt crystal and the temperature at which it dissolved. So here the red dot shows the element melting point i.e. 19.9 degrees and also it should be take so much time for heating it properly for completely dissolving it.

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Using a spectrophotometer, and a cuvette with a path length of 1 cm you measure the absorbance (A275) of Guanosine to be 0.70. Calculate the concentration of guanosine in your sample

Answers

Answer : The concentration of guanosine in your sample is, [tex]8.33\times 10^{-5}M[/tex]

Explanation :

Using Beer-Lambert's law :

[tex]A=\epsilon \times C\times l[/tex]

where,

A = absorbance of solution  = 0.70

C = concentration of solution = ?

l = path length = 1.00 cm

[tex]\epsilon[/tex] = molar absorptivity coefficient guanosine  = [tex]8400M^{-1}cm^{-1}[/tex]

Now put all the given values in the above formula, we get:

[tex]0.70=8400M^{-1}cm^{-1}\times C\times 1.00cm[/tex]

[tex]C=8.33\times 10^{-5}M[/tex]

Thus, the concentration of guanosine in your sample is, [tex]8.33\times 10^{-5}M[/tex]

gas mixture with a total pressure of 765 mmHg contains each of the following gases at the indicated partial pressures: 131 mmHg CO2, 226 mmHg Ar, and 186 mmHg O2. The mixture also contains helium gas. what is the partial pressure of the helium gas

Answers

Answer:

322mmHg

Explanation:

To calculate the partial pressure of the helium gas, we use the Dalton’s law of partial pressure.

It states that for a mixture of gases which do not react, the total pressure is equal to the sum of the individual partial pressures.

This means that:

Total pressure = Partial pressure of CO2 + partial pressure of Ar + Partial pressure of O2 + Partial pressure of helium.

Hence, Partial pressure of the helium gas = 765-131-226-86 = 322 mmHg

The specific heat capacity of water is 4.18 J K⁻¹ g⁻¹. What is the enthalpy (heat) change when 10.00 g of water is heated from 285.0 to 300.0 K (Hint temperature change is the same in °C or Kelvin)?

Answers

Answer:627j

Explanation:

H=mcΔT

ΔH= 10*4.18*(300-285)

ΔH=10*4.18*15

ΔH=627j

The heat of the given water sample has been 0.627 kJ.

The specific heat has been described as the amount of heat required to raise the temperature of 1 gram of substance by 1 degree Celsius. The expression for specific heat has been given as:

Heat = mass × specific heat × change in temperature

For the given water sample:

Specific heat = 4.18 J/K/g

Mass = 10 g

Initial temperature ([tex]T_i[/tex]) = 285 K

Final temperature ([tex]T_f[/tex]) = 300 K

Change in temperature ([tex]\rm \Delta T[/tex]) can be given as:

[tex]\Delta T =T_f\;-\;T_i\\\Delta T=300\;\text K\;-\;285\;\text K\\\Delta T=15\;\text K[/tex]

Substituting the values for heat:

[tex]\rm Heat=10\;g\;\times\;4.18\;J/K/g\;\times\;15\;K\\Heat\;=\;627.6\;J\\Specific\;heat\;=\;0.627\;kJ[/tex]

The heat of the given water sample has been 0.627 kJ.

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When a strong acid is titrated with a strong base using pheolphthalein as an indicator, the color changes abruptly at the endpoint of the titration and can be switched back and forth by the addition of only one drop of acid or base. The reason for the abruptness of this color change is that:

Answers

Final answer:

The abrupt color change in a titration using phenolphthalein occurs because this indicator exhibits a sharp transition at the equivalence point, which corresponds to a steep pH change when titrating strong acids with strong bases.

Explanation:

The abrupt color change observed when titrating a strong acid with a strong base using phenolphthalein as an indicator is due to the steep pH change that occurs at the equivalence point during the titration.

Phenolphthalein is a visual indicator that exhibits a clear and distinct color change—it turns from colorless to pink—around the equivalence point.

This sharp transition is suitable for accurately determining the end point of a titration between strong acids and bases, where the pH rises rapidly, allowing for the indicator to shift from its acidic form (colorless) to its basic form (pink) in a small volume interval of titrant addition.

The general chemistry of indicators can be represented by the equation: HIn(aq) → H+ (aq) + In¯¯(aq), where 'HIn' is the acid form of the indicator that is different in color compared to its ionized form 'In¯¯'.

Such distinct color changes of acid-base indicators are critical in titrations, since they provide a visual cue for the completion of the reaction without the need to continuously monitor the pH level.

A 10.0 mL 10.0 mL aliquot is removed from the described stock solution and diluted to a total volume of 100.0 mL. 100.0 mL. Calculate the molarity of the dilute solution.

Answers

The molarity of the dilute solution is 0.5 M.

The given parameters;

initial volume of the liquid, V₁ = 10 mLvolume of the diluted solution, V₂ = 100 mLConcentration of the initial stock, C = 5 M

The molarity of the dilute solution is calculated as follows;

[tex]C_{stock} = D.F \times c_{dilute}[/tex]

where;

D.F is dilute factor

The dilute factor of the given solution is calculated as follows;

[tex]D.F = \frac{V_{dilute}}{V_{concentrate}} \\\\D.F = \frac{100}{10} \\\\D.F = 10[/tex]

[tex]C_{stock} = D.F \times c_{dilute}[/tex]

[tex]5 = D.F \times c_{dilute}\\\\c_{dilute} = \frac{5}{DF} \\\\c_{dilute} = \frac{5}{10} \\\\c_{dilute} = 0.5 \ M[/tex]

Thus, the molarity of the dilute solution is 0.5 M.

"Your question is incomplete, it seems to be missing the following information":

A 10.0 mL of a liquid is removed from the described stock solution with molarity of 5M and diluted to a total volume of 100.0 mL. Calculate the molarity of the dilute solution.

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Predict the precipitate produced by mixing a(n) Al(NO3)3 solution with a(n) NaOH solution. Write the molecular equation for the reaction. (Use the lowest possible coefficients. Include states-of-matter under SATP conditions in your answer.)

Answers

Answer:

Al(NO₃)₃ (aq) + NaOH(aq) → Al(OH)₃ (s)↓ + 3NaNO₃ (aq)

Explanation:

Al³⁺ cation can generate a precipitate when it bonds to OH⁻ from a strong base but it is important to the base, not to be in excess.

When the OH⁻is in excess, the produced aluminium hydroxide will be soluble.

Al³⁺(aq) + 3OH⁻(aq) ⇄ Al(OH)₃(s) ↓    Kps

In an ionic compound, the size of the ions affects the internuclear distance (the distance between the centers of adjacent ions), which affects lattice energy (a measure of the attractive force holding those ions together). Based on ion sizes, rank these compounds of their expected lattice energy..
Note: Many sources define lattice energies as negative values. Please rank by magnitude and ignore the sign. |Lattice energy| = absolute value of the lattice energy.
RbCl ,RbBr ,Rbl ,RbF

Answers

Answer:

RbF>RbCl>RbBr>RbI

Explanation:

The lattice energy is an indicator of the strength of an ionic bond. It is also a rough indicator of the probability that an ionic substance will dissolve in water. The higher the lattice energy, the more difficult it is for the substance to dissolve in water.

Lattice energy depends on the relative sizes of ions present in the substance. As already known, the order of increasing sizes of halogen atoms is F<Cl<Br<I. The lesser the size, the higher the lattice energy.

Since the cation size is constant, lattice energy is only affected by increasing anion size and follows the pattern highlighted in the paragraph above. Hence RbF has the highest lattice energy and RbI has the least lattice energy.

a 25% alcohol solution is to be mixed with a 40% alcohol solution to obtain 18 liters of a 30% alcohol solution. How many liters of each solution should be used

Answers

Answer : The volume of solution used should be, 12 L

Explanation :

Let the volume of solution be, x

Thus the equation will be:

[tex]25\%\times (x)+40\%\times (18-x)=30\%\times (18)[/tex]

Now solving the term 'x', we get:

[tex]\frac{25}{100}\times (x)+\frac{40}{100}\times (18-x)=\frac{30}{100}\times (18)[/tex]

[tex]0.25\times (x)+0.4\times (18-x)=0.3\times (18)[/tex]

[tex]0.25x+7.2-0.4x=5.4[/tex]

[tex]-0.15x=-1.8[/tex]

[tex]0.15x=1.8[/tex]

[tex]x=12[/tex]

Thus, the volume of solution used should be, 12 L

Write balanced chemical equations for the complete neutralization reactions that take place when the following acids are titrated with sodium hydroxide (a) hydrochloric acid, (b) acetic acid, and (c) phosphoric acid.

Answers

Answer:

(a) [tex]NaOH_{(aq)} + HCl_{(aq)} --> NaCl_{(aq)} + H_2O_{(l)}[/tex]

(b) [tex]NaOH_{(aq)} + CH_3COOH_{(aq)} --> CH_3COONa_{(aq)} + H_2O_{(l)}[/tex]

(c) [tex]H_3PO_4_{(aq)} + 3NaOH_{(aq)} --> 3H_2O{(l)} + Na_3PO_4_{(aq)}[/tex]

Explanation:

For a reaction involving sodium hydroxide and hydrochloric acid, the balanced equation of reaction is:

[tex]NaOH_{(aq)} + HCl_{(aq)} --> NaCl_{(aq)} + H_2O_{(l)}[/tex]

For a reaction involving sodium hydroxide and acetic acid, the balanced equation of reaction is:

[tex]NaOH_{(aq)} + CH_3COOH_{(aq)} --> CH_3COONa_{(aq)} + H_2O_{(l)}[/tex]

For a reaction involving sodium hydroxide and phosphoric acid, the balanced equation of reaction is:

[tex]H_3PO_4_{(aq)} + 3NaOH_{(aq)} --> 3H_2O{(l)} + Na_3PO_4_{(aq)}[/tex]

Final answer:

This answer provides balanced chemical equations for the neutralization reactions of hydrochloric acid, acetic acid, and phosphoric acid with sodium hydroxide.

Explanation:

Neutralization Reactions:

Hydrochloric Acid with Sodium Hydroxide: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

Acetic Acid with Sodium Hydroxide: HC2H3O2(aq) + NaOH(aq) → NaC2H3O2(aq) + H₂O(l)

Phosphoric Acid with Sodium Hydroxide: H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 3H₂O(l)

If a student performs an exothermic reaction in a calorimeter, how does the calculated value of ΔH (Hcalc) differ from the actual value (Hactual) if the heat exchanged with the calorimeter is not taken into account?

Answers

Answer:

The actual ∆H would be greater than the calculated value of ∆H with no calorimeter.

Explanation:

The amount of heat changed during this process at a fixed pressure is termed Enthalpy

enthalpy change ∆H = ∆E + P∆V

∆E = internal energy change

P = fixed pressure

∆V = change in volume

When energy is absorbed during reaction, it is called endothermic reaction.

Endothermic reaction carried out in the calorimeter and enthalpy change for the reaction. Since we have that

q(surrounding) = q(solution)+q(calorimeter)

Therefore, q(calorimeter) > 0(endothermic).

The actual ∆H would be greater than the calculated value of ∆H with no calorimeter.

The actual ∆H would be greater than the calculated value of ∆H with no calorimeter.

What is Enthalpy change?

The amount of heat changed during this process at a fixed pressure is termed Enthalpy.

Enthalpy change ∆H = ∆E + P∆V

∆E = internal energy change

P = fixed pressure

∆V = change in volume

When energy is absorbed during reaction, it is called endothermic reaction.

Endothermic reaction carried out in the calorimeter and enthalpy change for the reaction. Since we have that

q(surrounding) = q(solution)+q(calorimeter)

Thus, q(calorimeter) > 0(endothermic).

The actual ∆H would be greater than the calculated value of ∆H with no calorimeter.

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What condition must occur when a chemical reaction is at equilibrium?

Answers

The condition is that the amounts of reactants or products do not change when a chemical reaction is in state of equilibrium.

Amounts of reactants or products do not change.

Chemical Reaction

Chemical equilibrium is a dynamic process in which the rate of product formation by the forward reaction equals the rate of product re-formation by the reverse reaction.

A chemical reaction is a process that results in the chemical change of one set of chemical substances into another set of chemical substances. Rust is an example of iron and oxygen mixing. Sodium acetate, carbon dioxide, and water are formed when vinegar and baking soda are combined.

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At chemical equilibrium, the forward and reverse reactions are happening at equal rates, resulting in no net change in the concentrations of reactants and products. This state is known as dynamic equilibrium. The system must also be closed for equilibrium to be maintained.

The correct answer is: The forward and reverse reactions are happening at equal rates.

Chemical equilibrium is the state in which both reactants and products are present in concentrations that remain constant over time. This occurs when the rate of the forward reaction equals the rate of the reverse reaction, meaning that there are no net changes in the concentrations of the reactants and products. This is known as dynamic equilibrium, indicating that reactions continue to occur in both directions at equal rates.

A compound is composed of 13% carbon, 4.3% hydrogen, 30.4% nitrogen, and 52.2% oxygen. The mystery compound has a molar mass of 184 grams per mole. What is the molecular formula of the compound

Answers

Answer:

C₂H₈N₄O₆ is the molecular formula for the compound

Explanation:

Data from the problem:

13 g of C in 100 g of compound

4.3 g of H in 100 g of compound

30.4 g of N in 100 g of compound

52.2 g of O in 100 g of compound

Firstly we determine, the mass of each in 184 g of compound, which is 1 mol

(13 g / 100 g) . 184 g  = 24 g C

(4.3 g  / 100 g) . 184 g  = 7.91 g H ≅ 8 g H

(30.4 g / 100 g) . 184 g  = 56 g N

(52.2 g  / 100 g) . 184 g  = 96 g O

And now, we divide the mass by the molar mass of each to determine the moles:

24 g C / 12 g/mol = 2 mol C

8g H / 1 g/mol = 8 mol H

56 g N / 14 g/mol = 4 mol N

96 g O / 16 g/mol = 6 mol O

So the molecular formula of the compound is C₂H₈N₄O₆

Final answer:

To determine the molecular formula of the given compound, one must calculate the moles of each element from their percentages, derive the empirical formula, and scale it by the compound's molar mass. However, the provided data seems contradictory and does not accurately represent a real calculation process.

Explanation:

To determine the molecular formula of a compound with 13% carbon, 4.3% hydrogen, 30.4% nitrogen, and 52.2% oxygen with a molar mass of 184 grams per mole, we need to first calculate the empirical formula from the given percentages. Assuming we have 100 grams of this compound, the masses of each element would directly convert to grams (e.g. 13g C, 4.3g H, 30.4g N, and 52.2g O).

To find the moles of each element, we divide the mass of each by its atomic mass (C: 12.01, H: 1.01, N: 14.01, O: 16.00) and get the ratios of moles of each element. However, the provided dissolution involves an inaccurate representation of the compound's composition and seems to contradict the given percentages.

For an accurate molecular formula determination, we use the correct composition to find the lowest whole number ratio of elements and scale it up using the compound's molar mass. This approach accurately identifies the compound's molecular makeup by aligning the empirical formula mass to the given molar mass.

Which of the following is a legume?

wheat

clover

corn

oats

Answers

Answer:

The answer to your question is Clover

Explanation:

Legumes are plants or the seed of plants. Legumes are harvested for human consumption, for livestock forage and silage.

These plants are also important during the Nitrogen cycle due to they fix nitrogen.

Examples of legumes are:

Beans, alfalfa, clover, lentils, peas, mesquite, carob, tamarind, peanuts, soybeans, etc.

Clover

LegumeA legume is a plant or the fruit or seed of a plant of the Fabaceae family. The seed is also known as a pulse when utilized as a dry grain. Legumes are grown in agriculture for a variety of reasons, including human consumption, cattle fodder and silage, and soil-enhancing green manure.Clover is a legume.

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Based on Table F, which of these saturated solutions has the lowest concentration of dissolved ions? (Explain why shortly please)
1) NaCl (aq)
2) MgCl² (aq)
3) NiCl (aq)
4) AgCl (aq)

Answers

AgCl (aq)

Option: 4

Explanation:

AgCl has the lowest concentration of dissolved ions because it is insoluble. Hence, it does not allow any ions into the solution.

According to the Solubility rule, the salts that have Cl⁻ are usually soluble but Ag⁺ is an exception which shows that AgCl is insoluble. It is insoluble because the lattice structure of AgCl is very strong that it cannot be overcome by the forces that favor the formation of hydrated ions,  Ag⁺(aq) and Cl⁻(aq). Solubility of AgCl in water is very low but however, it can precipitate in water.

The saturated solution that contains the concentration of dissolved ions is AgCl (aq)

Solubility rule:

AgCl should contain a less concentration of dissolved ions due to insoluble. Due to this, it permits any ions into the solution. As per the above rule, the salts that have Cl⁻ are normally soluble however Ag⁺ should be an exception which represents that AgCl is insoluble. It is insoluble due to the lattice structure of AgCl should be very strong. The solubility of AgCl in water should be very low but it should be precipitated in water.

Hence, The saturated solution that contains the concentration of dissolved ions is AgCl (aq)

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If you combine 270.0 mL of water at 25.00 ∘ C and 140.0 mL of water at 95.00 ∘ C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of water.

Answers

The final temperature of the mixture of cold water and hot water is 48.9 ⁰C.

The given parameters;

volume of the cold water = 270 mltemperature of the cold water = 25⁰Cvolume of the hot water, = 140 mltemperature of the hot water, = 95⁰Cdensity of water, = 1 g/ml

The mass of the cold water is calculated as follows;

[tex]m = \rho \times V\\\\m_c = 1 \ g/ml \ \ \times \ \ 270 \ ml \\\\m_c = 270 \ g[/tex]

The mass of the hot water is calculated as follows;

[tex]m_h = 1 \ g/ml \ \ \times \ \ 140 \ ml\\\\m_h = 140 \ g[/tex]

The final temperature of the mixture is determined by applying the principle of conservation of energy;

[tex]m_c C \Delta \theta_c = m_h C \Delta \theta_h\\\\ m_c \Delta \theta_c = m_h \Delta \theta_h\\\\m_c (t - 25) = m_h(95 - t)\\\\270(t- 25) = 140(95-t)\\\\270t - 6750 = 13,300 - 140t\\\\410t = 20,050\\\\t = \frac{20,050}{410} \\\\t = 48.9 \ ^0C[/tex]

Thus, the final temperature of the mixture of cold water and hot water is 48.9 ⁰C.

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The final temperature of the mixture is 55.00 °C.

To calculate the final temperature of the mixture, we can use the following equation:

T_final = (m_1T_1 + m_2T_2) / (m_1 + m_2)

where:

* T_final is the final temperature of the mixture (in °C)

* m_1 is the mass of the first water sample (in grams)

* T_1 is the temperature of the first water sample (in °C)

* m_2 is the mass of the second water sample (in grams)

* T_2 is the temperature of the second water sample (in °C)

We are given the following information:

* m_1 = 270.0 mL * 1.00 g/mL = 270.0 g

* T_1 = 25.00 °C

* m_2 = 140.0 mL * 1.00 g/mL = 140.0 g

* T_2 = 95.00 °C

Substituting these values into the equation above, we get the following:

T_final = (270.0 g * 25.00 °C + 140.0 g * 95.00 °C) / (270.0 g + 140.0 g)

T_final = 55.00 °C

Therefore, the final temperature of the mixture is 55.00 °C.

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Given a diprotic acid, H 2 A , with two ionization constants of K a1 = 3.0 × 10 − 4 and K a2 = 4.0 × 10 − 11 , calculate the pH for a 0.117 M solution of NaHA.

Answers

Answer:

The pH for a 0.117 M solution of NaHA is 2.227

Explanation:

To solve the question we check the difference in the Ka values thus

Ka₁ / Ka₂ = 7500000 < 10⁸ so we are required to calculate each value as follows

We therefore have

H₂X→ H⁺¹+HX⁻¹ with Ka₁ = 3.0 × 10⁻⁴

Therefore

3.0 × 10⁻⁴ = (x²)/(0.117)

x² = 3.0 × 10⁻⁴ ×0.117 and x = 5.925 × 10⁻³ = [H⁺]

Similarly

Ka₂ =  4.0 × 10⁻¹¹

and

4.0 × 10⁻¹¹= (x²)/(0.117)

x²= 0.117× 4.0 × 10⁻¹¹

x= 2.16× 10⁻⁶

Total H⁺ =  5.925 × 10⁻³+2.16× 10⁻⁶ = 5.927 × 10⁻³

Since pH = -log of hydrogen ion concentration,

pH = - log 5.927 × 10⁻³ = 2.227

2' A mixture containing 2.75 gof ammonium chloride (NH4cl) in 5.0 g of water was heated to dissolve the solid and then allowed to cool in air. At 6f"C, the first crystals appeared in solution. What is the solubility of ammonium chloride (in g of NH4CIper 100 gof water) at 61 'C

Answers

Answer:

55g NH₄Cl / 100g Water

Explanation:

Solubility of a substance define the amount of solute per solvent in a saturated solution. The solution can dissolve additional solute if heated.

In the problem, as the first crystal appears at 61°C the solubility in this temperature is the concentration of the solution, that is:

2,75g NH₄Cl / 5,0g water ₓ 100 = 55g NH₄Cl / 100g Water

I hope it helps!

The solubility of ammonium chloride at 61°C is 55 grams per 100 grams of water.

The solubility of ammonium chloride at 61°C is 2.75 grams per 5 grams of water, which can be expressed as a ratio to find the solubility per 100 grams of water.

To find the solubility per 100 grams of water, we can set up a proportion:

[tex]\[\frac{2.75 \text{ g NH}_4\text{Cl}}{5 \text{ g water}} = \frac{x \text{ g NH}_4\text{Cl}}{100 \text{ g water}}\][/tex]

Now, we solve for [tex]\(x\)[/tex]:

[tex]\[x = \frac{2.75 \text{ g NH}_4\text{Cl}}{5 \text{ g water}} \times 100 \text{ g water}\] \[x = \frac{2.75}{5} \times 100\] \[x = 0.55 \times 100\] \[x = 55\][/tex]

The answer is: 55.

In the 1 H NMR spectrum of chloroethane the methylene group is split into a quartet by the α and β nuclear spins of the protons on the neighboring methyl group. If the external magnetic field, Bo, directs upward, which sequence of nuclear spins contributes to the second farthest peak downfield within the spin-spin splitting pattern?

Answers

Answer:+1/2 and -1/2.

Explanation:the presence of an external magnetic field (B0), two spin states exist, +1/2 and -1/2.

The magnetic moment of the lower energy +1/2 state is aligned with the external field, but that of the higher energy -1/2 spin state is opposed to the external field.

The intermediate deshielding, when two methyl protons align with Bo and one opposes, breaking methylene protons into a quartet, is indicated by the second furthest peak in the ¹H NMR spectrum of chloroethane.

Understanding the spin-spin splitting in the ¹H NMR spectrum of chloroethane requires recognizing that the methylene (-CH₂-) protons are split into a quartet by the neighboring methyl (-CH₃) protons. These four peaks arise due to the interactions of the methylene protons with the three methyl protons, leading to different possible alignments.

The quartet results from the following combinations where the external magnetic field (Bo) is considered:

All three methyl protons aligned with Bo.Two aligned with Bo and one opposed.One aligned with Bo and two opposed.All three opposed to Bo.

Focusing on the second farthest peak downfield, this corresponds to the scenario where two of the methyl protons are aligned with Bo, and one is opposed. This configuration causes intermediate deshielding, placing this peak slightly less downfield than the most deshielded single proton state.

Which of the following statements about buffers is true? Group of answer choices A buffer composed of a weak acid of pKa = 5 is stronger at pH 4 than at pH 6. At pH values lower than the pKa, the salt concentration is higher than that of the acid. The strongest buffers are those composed of strong acids and strong bases. The pH of a buffered solution remains constant (exactly the same) no matter how much acid or base is added to the solution. When pH = pKa, the weak acid and salt concentrations in a buffer are equal.

Answers

The statement "When pH = pKa, the weak acid and salt concentrations in a buffer are equal." is true.

A solution or system that resists pH changes when small amounts of acid or base are introduced to it is called a buffer. A weak acid and its conjugate base, or a weak base and its conjugate acid, make up the substance. In order to maintain a steady pH environment for various chemical reactions or biological processes, buffers are frequently utilised in chemistry and biological sciences. They are essential for preserving homeostasis in biological systems like blood, where a steady pH is necessary for proper operation.

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Final answer:

The true statement about buffers is that they have their maximum buffering capacity when pH equals pKa, meaning the concentrations of the weak acid and its salt (conjugate base) are equal. Buffer solutions resist changes in pH effectively within a range of ± 1 pH unit from their pKa, and their buffering capacity depends on the buffer concentration.

Explanation:

The correct statement about buffers is: When pH = pKa, the weak acid and salt concentrations in a buffer are equal. This is because buffers consist of a weak acid and its conjugate base and exhibit maximum buffering capacity when the pH is numerically equal to the weak acid's pKa. At this pH, the buffer effectively resists changes in pH when small amounts of acid or base are added.

Buffer solutions are most effective within a pH range of ± 1 unit from their pKa. They are not invincible; their ability to maintain the pH level depends on the buffer concentration and the amount of strong acid or base added. The higher the concentration of the buffer components, the greater the buffer capacity, meaning more acid or base can be added before a significant change in pH occurs.

Buffers composed of weak acids work best for pH less than 7, while those composed of weak bases are more suitable for pH greater than 7. Moreover, buffers created from strong acids or bases are not effective, as they do not establish an equilibrium that is necessary for buffering action.

The requirements for one type of atom to substitute for another in a solid solution are:_______.A. All substitutions must be limited to the same element. B. An atom must be identical in size. C. An atom must be similar in size. D. The substituting atom must be from the same period. E. The substituting atom must be from the same group.

Answers

Answer:

E. The substituting atom must be from the same group.

Explanation:

Usually members higher-up in the group can replace or substitute lower members of a group

A chemistry student needs of isopropenylbenzene for an experiment. He has available of a w/w solution of isopropenylbenzene in acetone. Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button.

Answers

Question:

A chemistry student needs of 10 g isopropenylbenzene for an experiment. He has available 120 g of a 42.7% w/w solution of isopropenylbenzene in acetone. Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button.

Answer:

The answer to the question is as follows

The mass of solution the student should use is 23.42 g.

Explanation:

To solve the question we note the following

A solution containing 42.7 % w/w of isopropenylbenzene in acetone  has 42.7 g of isopropenylbenzene in 100 grams of the solution

Therefore we have 10 g of isopropenylbenzene contained in

100 g * 10 g/ 42.7 g = 23.42 g of solution

Available solution = 120 g

Therefore the quantity to used from the available solution = 23.42 g of the isopropenylbenzene in acetone solution.

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