During an exothermic chemical reaction,
A. a system becomes warmer, and the chemical substances undergo a decrease in potential energy.
B. a system becomes warmer, and the chemical substances undergo an increase in potential energy.
C. a system becomes cooler, and the chemical substances undergo a decrease in potential energy.
D. a system becomes cooler, and the chemical substances undergo an increase in potential energy.
E. a system becomes warmer, and additional heat is gained from the surroundings.

Answer:

Explanation: AH, AG and AS for the reaction at 873 K

delta G =-RT ln Kp

G= -8.314 x 873 ln 0.9

= 764.7J/mol

AG = AH-TAS

AH = 764.7 - 8314(0.9)

Kp =pCO/pCo2

Substitute the values at different temp. Solve simultaneously

Since mole fraction =kp/pCo

Mf =1.5

Answer:

you looking beautiful

Answer:

The specific heat of zinc is 0.375 J/g°C

Explanation:

Step 1: Data given

Mass of zinc = 19.6 grams

Mass of water = 82.9 grams

Initial temperature of zinc T1= 98.37 °C

Initial temperature of water T1= 24.16 °C

Final temperature of water (and zinc) T2 = 25.70 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculate Specific heat of zinc

Q=m*c*ΔT

Qzinc = -Qwater

m(zinc)*C(zinc)*ΔT(zinc) = -m(water)*C(water)*ΔT(water)

⇒ with mass of water = 82.9 grams

⇒ with C(water) = 4.184 J/g°C

⇒ with ΔT(water) = T2 - T1 = 25.70 - 24.16 = 1.54

⇒ with mass of zinc = 19.6 grams

⇒ with C(zinc) = TO BE DETERMINED

⇒ with ΔT(zinc) = T2 -T1 = 25.70 - 98.37 = -72.67°C

Qzinc = -Qwater

m(zinc)*c(zinc)* ΔT(zinc) = - m(water)*c(water)* ΔT(water)

19.6g* C(zinc) * (-72.67°C) = - 82.9g* 4.184 J/g°C * 1.54 °C

-1424.332*C(zinc) = -534.155

C(zinc) = 0.375 J/g°C

The specific heat of zinc is 0.375 J/g°C

Answer:

V = 18.8 L

Explanation:

This question is solved by using the ideal gas law :

PV = nRT ∴  V= nRT/P

where n= 0.788 mmol = 0.788 mol (1 mmol is a thousandths of a mol)

R = R constant for ideal gases =  0.08206 Latm/Kmol

T = 17.0 ºC =  (17.0 + 273) K = 290 K

and V = volume in liters our unknown

Plugging our values and solving for V,

V = 0.788 mol x 0.08206 Latm/Kmol x 290 K / 1atm = 18.8 L

Applying the Ideal Gas Law with the given number of moles, temperature and pressure shows that the volume of dinitrogen difluoride gas collected is approximately 19.0 L.

The volume of a gas can be calculated using the Ideal Gas Law if the number of moles, temperature and pressure are known. In your question, we're given that 788mmol of dinitrogen difluoride gas is evolved at a temperature of 17.0 degree C and a pressure of 1 atm. Firstly, convert the temperature to Kelvins by adding 273.15 to the Celsius temperature, giving us 290.15K.

The Ideal Gas Law is expressed as PV=nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant (0.0821 L·atm/ K·mol for this situation) and T is temperature. We can rearrange the equation to solve for volume: V=nRT/P. Therefore, substituting in the known values: V = 0.788 mol * 0.0821 L·atm/ K·mol * 290.15K / 1 atm, we find that the volume of dinitrogen difluoride gas collected is approximately 19.0 L, given correct to the number of significant digits.

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Answer: 72.4 kJ/mol

Explanation:

The balanced chemical reaction is,

[tex]C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(l)[/tex]  [tex]\Delta H=-2220.1kJ/mol[/tex]

The expression for enthalpy change is,

[tex]\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]

[tex]\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_3H_8}\times \Delta H_{C_3H_8})][/tex]

where,

n = number of moles

[tex]\Delta H_{O_2}=0[/tex] (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

[tex]-2220.1=[(3\times -393.5)+(4\times -241.8)]-[(5\times 0)+(1\times \Delta H_{C_3H_8})][/tex]

[tex]\Delta H_{C_3H_8}=72.4kJ/mol[/tex]

Therefore, the heat of formation of propane is 72.4 kJ/mol

The heat of formation of propane can be found using Hess's Law and the given enthalpy formations of CO2 and H2O, along with the enthalpy combustion of propane. Calculations reveal a heat of formation for propane of -103.85 kJ/mol.

The heat of formation of propane can be calculated using Hess's Law, which states that the enthalpy change in a chemical reaction is independent of the pathway between the initial and final states. In this case, we know the enthalpy of combustion for propane (C3H8) is -2220.1kJ/mol, and that the standard enthalpy of formation (ΔHf°) for CO2 and H2O are -393.5kJ/mol and -241.8kJ/mol, respectively.

Therefore, by rearranging the equation ΔH(combustion) = Σ(ΔHf° products) - Σ(ΔHf° reactants), we can solve for the ΔHf° of propane.

Inserting the given values into the equation, -2220.1 kJ/mol = [ 3(-393.5 kJ/mol) + 4(-241.8 kJ/mol)] - [x + 5(0)], where x represents the ΔHf° of propane, and '5(0)' is the contribution from the oxygen, which is zero because the change in enthalpy formation for an element in its stable form is zero. Thus, solving for x, you get a heat of formation for propane of -103.85 kJ/mol.

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Imatinib is a small molecule kinase inhibitor. The BCR-ABL kinase can phosphorylate a series of downstream substrates, leading to proliferation of mature granulocytes. Bcr-Abl kinase substrate is the tyrosine. The Protein Tyrosine Kinase activity is an important requirement for malignant transformation, and that it cannot be complemented by any downstream effector, though not all interactions of BCR-ABL with other proteins are phosphotyrosine dependent.

Answer:

a. 0.5 mol

b. 1.5 mol

c. 0.67

Explanation:

Fe3+ + SCN- -----> [FeSCN]2+

a. The ratio of the product to Fe3+ is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of Fe3+ was used. Leaving 0.5 mol remaining at equilibrium

b. The ratio of the product to SCN= is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of SCN- was used. Leaving 1.5 mol remaining at equilibrium

c. KC =  0.5/(0.5*1.5) =  0.67

Answer:

The solution boils at 102.76 °C

Explanation:

Δt = m* Kb x i  

i = 1 with 1 particle in solution

m = molality

Kb = 0.512 °C/molal

Step 2: Calculate moles C2H6O2

molar mass of  C2H6O2 = 62.068 g/mol

Calculate number of moles:

Moles = Mass / molar mass

Moles = 1400 grams / 62.068 g/mol

Moles C2H6O2 = 22.56 moles

Step 3: Calculate molality

these are in 4192 g of water:

22.56 moles / 4.192 kg water

⇒ moles / kg of water = 5.38 moles / Kg = m olal

Step 4: Calculate  ΔT

ΔT= Kb * molal = 5.38 molal* 0.512 °C/m

ΔT = 2.76 °C

Boiling point = 100°C + 2.75 °C = 102.76 °C

the solution boils at 102.76 °C

The boiling point of the ethylene glycol solution is calculated by first finding the molality of the solution and then using this to find the boiling point elevation. The normal boiling point of water is thus raised by this amount, giving a final boiling point of the solution as 102.75 degrees Celsius.

To solve this problem, we need to calculate the molality of the ethylene glycol solution and then use this to determine the boiling point elevation of the solution.

Firstly, we need to convert the mass of ethylene glycol (C2H6O2) and water into moles. The molar mass of ethylene glycol is approximately 62.07 g/mol, so 1.4 kg (or 1400 g) of ethylene glycol is equivalent to approximately 22.54 moles. The mass of water is 4192 g, or roughly 4.192 kg.

The molality (m), is thus given by the formula m = moles of solute / kg of solvent, therefore the molality of the ethylene glycol solution is 22.54 moles / 4.192 kg = 5.38 m.

Next, we use the formula for boiling point elevation: ΔTb = Kb * m, where Kb is the ebullioscopic constant for water (given as 0.512 °C/m), and m is the molality calculated earlier. Therefore, the boiling point of the solution rises by ΔTb = 0.512 × 5.38 = 2.75 °C.

The normal boiling point of water is 100 °C, thus, the boiling point of the ethylene glycol solution is 100 °C + 2.75 °C = 102.75 °C.

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The theoretical yield of Li3N is 1.24 mol or 138.16 g. The percent yield of the reaction is 4.23%.

To find the theoretical yield of Li3N, we first need to determine the limiting reactant. This is done by comparing the moles of each reactant to the stoichiometric ratio of the balanced equation. The molar mass of Li is 6.94 g/mol and the molar mass of N2 is 28.02 g/mol. First, convert the given masses of Li and N2 to moles using their respective molar masses:

12.7 g Li X (1 mol Li / 6.94 g Li) = 1.83 mol Li

34.7 g N2 X (1 mol N2 / 28.02 g N2) = 1.24 mol N2

Then, divide the number of moles of each reactant by its stoichiometric coefficient to find the mole ratio:

1.83 mol Li / 6 = 0.305 mol Li3N

1.24 mol N2 / 1 = 1.24 mol Li3N

Since N2 gives a smaller value, it is the limiting reactant. Therefore, the theoretical yield of Li3N is 1.24 mol or 138.16 g.

To calculate the percent yield, divide the actual yield by the theoretical yield, then multiply by 100:

Percent yield = (Actual yield / Theoretical yield) X 100

Percent yield = (5.85 g / 138.16 g) X 100 = 4.23%

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The product of the given reaction, an acid-catalyzed hydration of an alkene, would be 2-butanol, as per Markovnikov's Rule. This rule predicts the placement of the hydrogen and halide groups in a reaction.

The reaction given is an example of an acid-catalyzed hydration of an alkene. In such reactions, an alkene reacts with water in the presence of an acid (in this case, H3PO4) to form an alcohol. The prediction of the product involves recognizing the reaction type and the reagents involved.

For the given reaction: CH3CH=CHCH3 + H2OH3PO4⟶, the product would be 2-butanol.

But why does this process form 2-butanol? It all comes down to Markovnikov's rule, which predicts that in the addition of a protic acid HX to an alkene, the acid hydrogen (H) becomes attached to the carbon with fewer alkyl substituents, and the halide (X) group becomes attached to the carbon with more alkyl substituents.

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Answer:

-79.8 × 10⁴ J/mol

Explanation:

Arsine, AsH₃, is a highly toxic compound used in the electronics industry for the production of semiconductors. Its vapor pressure is 35 Torr at 111.95 °C and 253 Torr at 83.6 °C.

Then,

P₁ = 35 torr

T₁ = 111.95 + 273.15 = 385.10 K

P₂ = 253 torr

T₂ = 83.6 + 273.15 = 356.8 K

We can calculate the standard enthalpy of vaporization (ΔH°vap) using the two-point Clausius-Clapeyron equation.

[tex]ln(\frac{P_{2}}{P_{1}} )=\frac{-\Delta H\°_{vap}}{R} .(\frac{1}{T_{2}} -\frac{1}{T_{1}} )[/tex]

where,

R is the ideal gas constant

[tex]ln(\frac{253torr}{35torr})=\frac{-\Delta H\°_{vap}}{8.314J/K.mol} .(\frac{1}{356.8K}-\frac{1}{385.10K})\\ \Delta H\°_{vap}=-79.8 \times 10^{4} J/mol[/tex]

Answer: The [tex]\Delta H^o_{rxn}[/tex] for the reaction is -521.6 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical equation for the reaction of fluorine and water follows:

[tex]2F_2(g)+2H_2O(l)\rightarrow 4HF(g)+O_2(g)[/tex]   [tex]\Delta H^o_{rxn}=?[/tex]

The intermediate balanced chemical reaction are:

(1) [tex]H_2(g)+F_2(g)\rightarrow 2HF(g)[/tex]    [tex]\Delta H_1=-546.6kJ[/tex]    ( × 2)

(2) [tex]H_2(g)+O_2(g)\rightarrow 2H_2O(g)[/tex]    [tex]\Delta H_2=-571.6kJ[/tex]

The expression for enthalpy of reaction follows:

[tex]\Delta H^o_{rxn}=[2\times \Delta H_1]+[1\times (-\Delta H_2)][/tex]

Putting values in above equation, we get:

[tex]\Delta H^o_{rxn}=[(2\times (-546.6))+(1\times (571.6))]=-521.6kJ[/tex]

Hence, the [tex]\Delta H^o_{rxn}[/tex] for the reaction is -521.6 kJ.

Answer:

To prepare 50L of 32% solution you need: 11L of 30% solution, 22L of 50% solution and 17L of 10% solution.

Explanation:

A 32% solution of acid means 32L of acid per 100L of solution. As the chemist wants to make a solution using twice as much of the 50% solution as of the 30% solution it is possible to write:

2x*50% + x*30% + y*10% = 50L*32%

130x + 10y = 1600 (1)

-Where x are volume of 30% solution, 2x volume of 50% solution and y volume of 10% solution-

Also, it is possible to write a formula using the total volume (50L), thus:

2x + x +y = 50L

3x + y = 50L (2)

If you replace (2) in (1):

130x + 10(50-3x) = 1600

100x + 500 = 1600

100x = 1100

x = 11L -Volume of 30% solution-

2x = 22L -Volume of 50% solution-

50L - 22L - 11L = 17 L -Volume of 10% solution-

I hope it helps!

Final answer:

To create a 50-liter mixture with 32% acid concentration, the chemist should use 20 liters of the 10% acid solution, 10 liters of the 30% acid solution, and 20 liters of the 50% acid solution.

Explanation:

The question involves solving a system of linear equations to determine the volume of each acid solution required to create a 50-liter mixture with a 32% acid concentration. Let's denote the amount of the 10% solution as x liters, the 30% solution as y liters, and the 50% solution as 2y liters (since it's twice the amount of the 30% solution).

The total amount of acid should be 32% of the 50-liter mixture: 0.10x + 0.30y + 0.50(2y) = 0.32 × 50.

We've also established the relationship between the 30% and 50% solutions: 2y is twice the amount of y.

x + 3y = 50

0.10x + 0.30y + 1.00y = 16

Combining and rearranging gives us x = 20 liters, y = 10 liters, and thus 2y = 20 liters. Therefore, to obtain the desired mixture, the chemist should use 20 liters of the 10% solution, 10 liters of the 30% solution, and 20 liters of the 50% solution.

Answer:

The partial pressure of krypton in the flask is 0.59 atm and the total pressure in the flask is 1.39 atm

Explanation:

This must be solved with the Ideal Gas Law equation.

First of all we need the moles or Ar and Kr in the mixture

Moles = Mass / Molar mass

Molar mass Ar 39.95g/m

Moles Ar = 6.18 g/39.95 g/m → 0.154 moles

Molar mass Kr 83.8 g/m

Moles Kr = 9.66 g/ 83.8g/m → 0.115 moles

Total moles in the mixture: 0.154 moles + 0.115 moles = 0.269moles

Now, we have the total moles, we can calculate the total pressure.

P . V = n . R . T

(T° in K = T° in C + 273)

P. 5.38L = 0.269mol . 0.082 L.atm/mol.K . 340K

P = (0.269mol . 0.082 L.atm/mol.K . 340K) / 5.38 L

P = 1.39 atm

Now we have the total pressure, we can apply molar fraction so we can know the partial pressure of Kr.

Kr pressure / Total Pressure = Kr moles / Total moles

Kr pressure / 1.39 atm = 0.115 moles / 0.269 moles

Kr pressure = (0.115 moles / 0.269 moles) / 1.39atm

Kr pressure = 0.59 atm

Answer:

3 Fe₂O₃(s) + H₂(g) ⇒ 2 Fe₃O₄(s) + H₂O(g)    ΔH° = -6.00 kJ

Explanation:

Let's consider the following balanced equation.

3 Fe₂O₃(s) + H₂(g) ⇒ 2 Fe₃O₄(s) + H₂O(g)

When 1 mole of Fe₂O₃(s) reacts, 2.00 kJ of energy are evolved. Energy is an extensive property. In the balanced equation there are 3 moles of Fe₂O₃(s), so the evolved energy is:

[tex]3molFe_{2}O_{3}.\frac{2.00kJ}{1molFe_{2}O_{3}} =6.00kJ[/tex]

By convention, when energy is evolved it takes the negative sign. At constant pressure, the thermochemical equation is:

3 Fe₂O₃(s) + H₂(g) ⇒ 2 Fe₃O₄(s) + H₂O(g)    ΔH° = -6.00 kJ

where

ΔH° is the standard enthalpy of reaction (heat released at constant pressure)

Answer:

To calculate the molar mass add the atomic mass of each atom in the chemical formula.

Explanation:

For example Molar mass of NaHCO₃  calculated as

Molar mass is the sum of masses of all atom present in formula.

Atomic mass of sodium = 23 g/mol

Atomic mass of hydrogen = 1.008 g/mol

Atomic mass of carbon = 12 g/mol

Atomic mas of Oxygen = 16 g/mol

Molar mass of NaHCO₃ = 23 + 1.008 + 12 + 16× 3

Molar mass of NaHCO₃ = 23 + 1.008 + 12 + 48

Molar mass of NaHCO₃ = 84.008 g/mol

Explanation:

The heat gained by the solution = q

[tex]q=mc\times (T_{final}-T_{initial})[/tex]

where,

q = heat gained = ?

c = specific heat of solution= [tex]4.18 J/^oC[/tex]

Mass of the solution(m) = mass of water + mass of calcium chloride

Mass of water = ?

Volume of water = 50.00 mL

Density of water = 1.00 g/mL

Mass = Density × Volume

m = 1.00 g/mL × 50.00 mL = 50.00 g

Mass of the solution (m) = 50.00 g + 4.51 g =54.51 g

[tex]T_{final}[/tex] = final temperature = [tex]25.8 ^oC[/tex]

[tex]T_{initial}[/tex] = initial temperature = [tex]22.6 ^oC[/tex]

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

[tex]q=54.51 g\times 4.18 J/g^oC\times (25.8-22.6)^oC[/tex]

[tex]q=729.126 J[/tex]

The heat gained by the solution is 729.126 J.

Heat energy released during the reaction  = q'

q' = -q  ( law of conservation of energy)

q' = -729.126 J

The heat energy released during the reaction is -729.126 J.

Moles of calcium chloride, n = [tex]\frac{4.51 g}{111 g/mol}=0.04063 mol[/tex]

[tex]\Delta H_{rxn}=\frac{q'}{n}=\frac{-729.126 J}{0.04063 mol}=-17,945.23 J/mol= -17.945 kJ/mol[/tex]

The ΔH of the reaction is -17.945 kJ/mol.

a. 2257.7 kW  b. 6.057kw/K

Steam existing at 100KPa with t = 20C, we can obtain values of enthalpy and entropy from table

h1 = hf = 417.4 kJ/kg

s1 = 1.302 KJ/kg.K

h2 = hg = 2675.1 kJ/kg

s2 = 7.359 kJ/K

a. Electrical Power consumption of system is given by Wel = mass x Change in Enthalpy

Wel = 1 x (h2 - h1) = 1 x (2675.1 - 417.4) = 2257.7 kw

b . Entropy Generation is given by Change in Entropy  = Entropy generation - heat dissipated/temperature, since we have a well insulated system with no losses, q =0, hence

ΔS = Sgen + 0/T = Sgen

m(S2 - S1) = Sgen

Sgen = 1 x (7.359 - 1.302) = 6.057 kW/K

Answer:

The standard reduction potential E°cell (Cr6+/Cr3+) is -0.13V

Explanation:

Step 1: Data given

3 Ni^2+(aq) + 2 Cr(OH)3(s) + 10 OH− (aq) → 3 Ni(s) + 2 CrO4^2−(aq) + 8 H2O(l) ΔG∘ = +87000 J/mol

Ni2+(aq) + 2 e− → Ni(s)    E∘red = -0.28 V

Step 2: The half reactions:

Cathode:  Ni2+(aq) + 2 e− → Ni(s)    E° = -0.28 V

Anode: CrO4^2-(aq) + 4H2O(l) +3e- → Cr(OH)3(s) + 5OH- (aq)   E°= unknown

Step 3: Calculate E°cell

ΔG° = -n*F*E°cell

⇒ with ΔG° = the gibbs free energy

⇒ n = the number of electrons in the net reaction = 6

⇒ F = the Faraday constant = 96485 C

⇒ E°cell= the standard cell potential

Step 4: Calculate E°(Cr6+/Cr3+

E°cell= ΔG°/(-n*F)

E°cell = 87000 /(-6*96485)

E°cell = -0.150 V

E°cell = E°(Ni2+/Ni) - E°(Cr6+/Cr3+)

E°(Cr6+/Cr3+) = -0.13V

The standard reduction potential E°cell (Cr6+/Cr3+) is -0.13V

Answer:

The change in entropy of gas is [tex]\Delta S= nC_{P}ln3[/tex]

Explanation:

n= Number of moles of gas

Change in entropy of gas = [tex]ds= \int \frac{dQ}{T}[/tex]

[tex]dQ= nC_{p}dT[/tex]

From the given,

[tex]V_{i}=V[/tex]

[tex]V_{f}=3V[/tex]

Let "T" be the initial temperature.

[tex]\frac {V_{i}}{T_{i}}=\frac {V_{f}}{T_{f}}[/tex]

[tex]\frac {V}{T}=\frac {3V}{T_{f}}[/tex]

[tex]{T_{f}} = 3T[/tex]

[tex]\int ds = \int ^{T_{f}}_{T_{i}} \frac{nC_{P}dT}{T}[/tex]

[tex]\Delta S = nC_{p}ln(\frac{T_{f}}{T_{i}})[/tex]

[tex]\Delta S = nC_{p}ln3[/tex]

Therefore, The change in entropy of gas is [tex]\Delta S= nC_{P}ln3[/tex]

Answer:

0.443 g

Explanation:

In the electrolysis of an aqueous solution of NaCl, the following half-reactions take place:

Reduction: Na⁺(aq) + 1 e⁻ ⇒ Na(s)

Oxidation: 2 Cl⁻(aq) ⇒ Cl₂(g) + 2 e⁻

Let's consider the following relations:

Suppose a current of 18.0 A is passed through an aqueous solution of NaCl for 67.0 seconds. The mass of chlorine produced is:

[tex]67.0s.\frac{18.0c}{s} .\frac{1mole^{-} }{96468c} .\frac{1molCl_{2}}{2mole^{-} } .\frac{70.9gCl_{2}}{1molCl_{2}} =0.443gCl_{2}[/tex]

Using Faraday's law of electrolysis, we can calculate the mass of chlorine produced by an 18.0 A current over 67.0 seconds in the chlor-alkali process. The result is approximately 0.44 grams.

To determine the mass of chlorine produced in the chlor-alkali process using a current of 18.0 A for a duration of 67.0 seconds, we need to use Faraday's law of electrolysis. This states that the amount of a substance produced at an electrode during electrolysis is directly proportional to the quantity of electricity that passes through the solution.

The amount of a substance produced can be calculated using the formula: It = zF, where I is the current in amperes, t is the time in seconds, z is the ionic charge(1 for Cl-), and F is Faraday’s constant (96500 coulombs per mole of electrons).

First, let’s find the quantity of electric charge (Q) that has passed through the solution using the formula Q = It. In this case, I = 18 A and t = 67.0 seconds, so Q = 18 * 67.0 = 1206 Coulombs.  

Now we can use this Q in the formula of m = QM/zF, where M is the molar mass of chlorine, which is approximately 35.5 g/mol. After plugging the numbers in, we find the mass (m) of the chlorine produced is 0.44 g.

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Answer:

[tex]K_{p}[/tex] of the reaction is  [tex]9.521 \times 10^{-3}[/tex].

[tex] K_{c}[/tex] of the reaction is  [tex]3.89 \times 10^{-4}[/tex].

Explanation:

Initial pressure of NOBr is "x" atm.

34% = 0.34

The equilibrium chemical reaction is a follows.

       [tex]2NOBr(g)\leftrightarrow 2NO(g)+Br_{2}(g)..............(1)[/tex]

Initial          x                                      0          0

Change    -0.34x                             +0.34x    +0.17x

Final        (x-0.34x)                          +0.34x     +0.17x

Total pressure = sum of the pressure at equilibrium.

[tex]0.25atm=x-0.34x+0.34x+0.17x[/tex]

[tex]0.25atm= 1.17x[/tex]

[tex]x= \frac{0.25}{1.17}=0.213[/tex]

Partial pressure of NOBr=  x-0.34x =x (1-0.34)

= 0.213(0.66) = 0.14 atm.

Hence, partial pressure of [tex][]P_{NOBr}][/tex] is 0.14 atm.

[tex]P_{NO}=0.34x = 0.34 \times 0.213 = 0.072[/tex]

[tex]P_{Br_{2}}=0.17x = 0.17 \times 0.213 = 0.036\,atm[/tex]

From equation (1)

[tex]K_{p}= \frac {P^{2}_{NO} P_{Br_{2}}}{P^{2}_{NOBr}}[/tex]

[tex]= \frac{(0.072)^{2} (0.036)}{(0.14)^{2}}= 9.521 \times 10^{-3}[/tex]

Therefore, [tex]K_{p}[/tex] of the reaction is  [tex]9.521 \times 10^{-3}[/tex].

[tex]K_{p}= K_{c}(RT)^{\Delta n}[/tex]

Rearrange the equation is as follows.

[tex]K_{c}= \frac{K_{p}}{(RT)^{\Delta n}}[/tex]

[tex]\Delta n[/tex] = number of moles of reactants - Number of moles of products

= 3-2 = 1

[tex]= \frac{9.521 \times 10^{-3}}{(0.0821 \times 298)^{1}}= 0.389 \times 10^{-3} = 3.89\times 10^{-4}[/tex]

Therefore,[tex] K_{c}[/tex] of the reaction is  [tex]3.89 \times 10^{-4}[/tex].

The [tex]K_p[/tex] and [tex]K_c[/tex] is " [tex]0.009649615 \ atm[/tex] and [tex]0.00039\ mol\ L^{-1}[/tex]"for the dissociation at this temperature.

[tex]2NOBr\rightleftharpoons 2NO+Br_2[/tex]

​When the initial pressure of [tex]NOBr[/tex] is p, then at equilibrium, therefore the dissociation percentage of [tex]NOBr \ \ is\ \ 34\%[/tex]

[tex]\to P(NOBr)=0.66\ p\\\\\to P(NO)=0.34\ p\\\\\to P(Br_2)= \frac{0.34p}{2}=0.17\ p\\\\[/tex]

Calculating the total pressure:

[tex]\to 0.66\ p+0.34\ p+0.17\ p=1.17\ p=0.25\ atm\\\\[/tex]

[tex]\to p=0.214 \ atm\\\\[/tex]

The equilibrium partial pressures:

[tex]\to P(NOBr)=0.66\ p=0.141\ atm\\\\\to P(NO)=0.34\ p=0.073\ atm\\\\\to P(Br_2)=0.17\ p=0.036\ atm\\\\[/tex]

Calculating the [tex]K_p[/tex]:

[tex]\to K_p =\frac{(P_{NO})^2 (P_{Br_2})}{(P_{NOBr})^2}\\\\[/tex]

          [tex]=\frac{(0.073)^2 \times (0.036)}{(0.141)^2}\\\\=\frac{(0.005329) \times (0.036)}{(0.019881)}\\\\=\frac{(0.000191844)}{(0.019881)}\\\\=0.009649615 \ atm[/tex]

Calculating the [tex]K_c[/tex]:

Using formula:

[tex]\to \Delta ng = \text{Number of moles of product - Number of moles of reactant}[/tex]

            [tex]= (2+1) - 2\\\\= 3 - 2\\\\=1 \\[/tex]

The relationship between [tex]K_p[/tex] and [tex]K_c[/tex] is:

[tex]\to K_p = K_C(RT)^{\Delta ng}[/tex]

Hence, by putting the values of  [tex]K_p[/tex], gas constant [tex]R, T,[/tex] and [tex]\Delta ng[/tex], we can calculate the value of [tex]K_c[/tex].

[tex]\to 0.0096\ atm = - K_c(0.0821 \ atm \ mol L^{-1}K^{-1} \times 298 K)^1[/tex]

[tex]\to K_c = \frac{0.0096 }{0.0821\times 298}[/tex]

         [tex]= \frac{0.0096}{ 24.46}\\\\ = 0.00039\ mol\ L^{-1}[/tex]

Therefore, the [tex]K_p[/tex] and [tex]K_c[/tex] is " [tex]0.009649615 \ atm[/tex] and [tex]0.00039\ mol\ L^{-1}[/tex]"for the dissociation at this temperature.

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Answer:

Anode: Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻

Explanation:

In the anode takes place the oxidation, in which the reducing agent loses electrons.

In the cathode takes place the reduction, in which the oxidizing agent gains electrons.

The corresponding half-reactions are:

Anode: Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻

Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)

Answer:

[H3O+] = 0.0117 M

[OH-] = 8.5 * 10^-13 M

Explanation:

Step 1: Data given

Concentration of HCl = 0.0117 M

Step 2:

HCl is a strong acid

pH of a strong acid = -log[H+] = - log[H3O+]

[H3O+] = 0.0117 M

pH = -log(0.0117)

pH = 1.93

pOH =14 - 1.93 = 12.07

pOH = -log[OH+] = 12.07

[OH-] = 10^-12.07 = 8.5 * 10^-13

Or

Kw / [H3O+] = [ OH-]

10^-14 / 0.0117 = 8.5*10^-13

Final answer:

The hydronium ion concentration [H3O+] in a 0.0117 M HCl solution is 0.0117 M, and the hydroxide ion concentration [OH-] is calculated to be 8.55 × 10^-13 M using the water dissociation constant.

Explanation:

To calculate the hydronium ion concentration, [H3O+], of a 0.0117 M HCl solution, we first need to understand that HCl is a strong acid which dissociates completely in water. This means that for every mole of HCl dissolved, there will be one mole of H3O+ ions in solution. Hence, the hydronium ion concentration in a 0.0117 M HCl solution is also 0.0117 M.

As for the hydroxide ion concentration, [OH-], we use the water dissociation constant (Kw), which is 1.0 × 10-14 at 25 °C. The product of the concentrations of the hydronium and hydroxide ions in any aqueous solution is equal to Kw. We can find [OH-] by rearranging the expression Kw = [H3O+][OH-] to solve for [OH-], giving us [OH-] = Kw / [H3O+]. Substituting in the values we get [OH-] = 1.0 × 10-14 M / 0.0117 M = 8.55 × 10-13 M.

NO2Cl is a polar molecule because the electronegativity difference between atoms of the molecule, being the chloride tho most electronegative. This fact and the asymmetry of the molecule makes a dipole moment.

In order to don’t include formal charges in the drawing you can use the image number 1. Nevertheless is more usual to use a lewis structure like image number 2.

Answer:

The system does work on the surroundings.

Explanation:

The work (w) exerted in a chemical reaction depends on the change in the number of gaseous moles (Δn(g)), where

Δn(g) = n(gas, products) - n(gas, reactants)

These variables are linked through the following expression.

w = -R.T.Δn(g)

where,

R is the ideal gas constant

T is the absolute temperature

At constant pressure, which of these systems do work on the surroundings?

a. 2A(g) + 3B (g) ------> 4C (g)

Δn(g) = 4 -5 = -1. The surroundings do work on the system.

b. 2A(g) + 2B (g) -------> 5C(g)

Δn(g) = 5 -4 = 1. The system does work on the surroundings.

c. 2A (g) + B (g) -----> C(g)

Δn(g) = 1 - 3 = -2. The surroundings do work on the system.

d. A(s) + B(g) -------> 2C (g)

Δn(g) = 2 -2 = 0. No work is done.

Answer:

The correct answer is 1 the total number of proton plus neutrons in the products and reactants must be same.

Explanation:

During a radioactive reaction the nucleus of the radioactive atom undergo disintegration or breakdown to form new nucleus.

   Radioactive disintegration deals with the emission of α particle or β particle from a radioactive nucleus resulting in the formation of new nucleus.Although new nucleus is formed the total number of proton plus neutrons in product and reactant must be same.

A  rule for balancing a nuclear equation is: I. total number of protons + neutrons in the products and reactants must be the same.

A nuclear equation for a radioactive reaction, shows the reactants and product, where the proton number and atomic mass are conserved, unlike in chemical equations.

In a radioactive decay, there is emission of α particle or β particle from a radioactive nucleus to form a new nucleus.

Therefore, a rule for balancing a nuclear equation is: I. total number of protons + neutrons in the products and reactants must be the same.

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Answer:

11.3 L

Explanation:

Initially, the cylinder had n₁ = 0.120 mol of gas in an initial volume V₁ = 2.18 L. At the end, it had n₂ = 0.621 mol in an unknown volume V₂. According to the Avogadro's law, in the same conditions of pressure and temperature, the volume is directly proportional to the number of moles.

[tex]\frac{V_{1}}{n_{1}} =\frac{V_{2}}{n_{2}} \\V_{2}=\frac{V_{1}\times n_{2} }{n_{1}} =\frac{2.18L \times 0.621mol}{0.120mol} =11.3L[/tex]

Answer:

The final volume of the cylinder is 1.67 L

Explanation:

Step 1: Data given

Initial volume = 0.250 L

external pressure = 2.00 atm

Expansion does 288 J of work on the surroundings

Step 2: Definition of reversible work:

Wrev = -P(V2-V1) = -288 J

The gas did work, so V2>V1  (volume expands) and the work has a negative sign.(Wrev<0)

V2 = (-Wrev/P)  + V1

⇒ with Wrev = reverse work (in J)

⇒ with P = the external pressure (in atm)

⇒ with V1 = the initial volume

We can see that your pressure is in  atm  and energy in J

To convert from J to L * atm we should use a convenient conversion unit using the universal gas constants :

R = 8.314472 J/mol *K and R= 0.08206 L*atm/K*mol

V2 =- (-288 J * (0.08206 L*atm/K*mol  /8.314 J/mol *K))/2.00 atm  + 0.250L

V2 = 1.67 L

The final volume of the cylinder is 1.67 L

The final volume of the cylinder after the expansion is 1.673 liters. This was found by calculating the change in volume using the work done on the surroundings and the external pressure, and adding it to the initial volume.

To calculate the final volume of the cylinder after expansion, we need to use the work done on the surroundings and the external pressure. Work (W) is related to pressure (P) and volume change (ΔV) by the equation W = -PΔV, where the pressure is constant and work done on the surroundings is negative.

In this case, we know that the work done on the surroundings is 288 J and the external pressure is 2.00 atm. Since 1 L·atm is equivalent to 101.3 J, we convert the external pressure to joules by multiplying by the volume change ΔV in liters:

288 J = -(2.00 atm) · ΔV · 101.3 J/L·atm

Therefore, to find ΔV we divide 288 J by the product of 2.00 atm and 101.3 J/L·atm:

ΔV = -288 J / (2.00 atm · 101.3 J/L·atm)

ΔV = -1.423 L (taking the absolute value, since volume change is positive upon expansion)

The initial volume was 0.250 L, thus the final volume is:

V_final = V_initial + ΔV

V_final = 0.250 L + 1.423 L

V_final = 1.673 L

Answer:

The IUPAC of 1-pentanone is not correct.

Explanation:

The structures of each of the given compounds with correct IUPAC names are shown in the figure.

In case of ketones there must be two alkyl groups attached to the carbonyl carbon. In case of aldehydes there must be atleast one hydrogen attached to carbonyl carbon.

Now if we are saying that 1-pentanone, it means we are actually naming the compound with wrong functional group. This is infact an aldehyde with the correct name as "pentanal".

Rest of the given IUPAC names are correct.

All solutions A through E, being strong acids at the same concentration, will have virtually the same pH value under the same conditions. None will have a significantly higher pH. In general, the solution with the highest pH would be the one derived from the weakest acid or strongest base.

The pH scale is used to determine the acidity or basicity of a solution. The solutions given are all acidic, being solutions of hydrobromic acid (HBr), hydroiodic acid (HI), hydrofluoric acid (HF), hydrochloric acid (HCl), and perchloric acid (HClO4), respectively. Each of these acids dissociates in water to a different extent, affecting the pH of the solution. However, common to all of these acids is that they are strong acids and will fully dissociate in water.

In this case, all solutions have the same molarity (concentration) of 0.10 M. Because all the acids are strong acids and will fully dissociate, the pH of these solutions will be virtually the same. Therefore, none of the solutions will have a significantly higher pH than the others if measured under the same conditions.

In general, to identify the solution with the highest pH (or lowest acidity), you would look for the weakest acid or the strongest base out of the options provided.

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Among the given solutions, C) 0.10 M HF(aq) has the highest pH because HF is a weak acid and only partially dissociates in water, resulting in a higher pH compared to the strong acids HCl, HI, HBr, and HClO₄.

To determine which solution has the highest pH, we need to understand that pH is a measure of the hydrogen ion concentration in a solution. A lower pH signifies a higher hydrogen ion concentration. Since all the given solutions are acids, we must determine which one produces the least amount of hydrogen ions (H⁺).

HCl, HI, HBr, and HClO₄ are strong acids, meaning they fully dissociate in water. This full dissociation results in a high concentration of H⁺ ions, giving them all a pH of 1 for a 0.10 M solution:

HCl: pH = 1.0

HI: pH = 1.0

HBr: pH = 1.0

HClO4: pH = 1.0

HF, however, is a weak acid and only partially dissociates in water. This partial dissociation results in fewer H+ ions compared to the strong acids listed. Consequently, the pH of a 0.10 M HF solution will be higher (less acidic) than the pH of the other solutions.

Given this information, the solution that has the highest pH is:

C) 0.10 M HF(aq)