In a particular electroplating process, the metal being plated has a +4 charge. If 861.8 C of charge pass through the cell, how many moles of metal should be plated? Useful information: F = 96,500 C/mol e- Provide your response to four digits after the decimal.

Answer:

0.238 M

Explanation:

A 17.00 mL sample of the dilute solution was found to contain 0.220 M ClO₃⁻(aq). The concentration is an intensive property, so the concentration in the 52.00 mL is also 0.220 M ClO₃⁻(aq). We can find the initial concentration of ClO₃⁻ using the dilution rule.

C₁.V₁ = C₂.V₂

C₁ × 24.00 mL = 0.220 M × 52.00 mL

C₁ = 0.477 M

The concentration of Pb(ClO₃)₂ is:

[tex]\frac{0.477molClO_{3}^{-} }{L} \times \frac{1molPb(ClO_{3})_{2}}{2molClO_{3}^{-}} =0.238M[/tex]

Explanation:

1) The valence electrons of group 4A elements are in the 5s subshell.

The given statement is false .

The valence electrons of group 4A elements are in the ns and np subshell .Where n is equal to 1 to 7.

2) Period 3 elements have an inner electron configuration of [Ar].

The given statement is false .

The noble gas which  comes before third period is neon.Period 3 elements have an inner electron configuration of [Ne].

3)The highest principal quantum number of period 4 elements is 4.

The given statement is true.

The period number of periodic table indicates the value of highest principal quantum number.

4) Period 5 elements have six 4p electrons.

The given statement is true.

This so because 4p subshell is filled before filling of shells of fifth shell. So, the 4p subshell will obviously have 6 electrons.

5) Period 5 elements have an inner electron configuration of (Kr).

The given statement is true.

The noble gas which comes before third period is krypton .

6) Group 8A elements have full outer principal s and p subshells.

The given statement is true.

This is because the the group mentioned in the statement if noble gases and noble gases are inert in nature due to fully filled electronic configurations.

7) The valence electrons of group 3A elements are in an s and p subshell.

The given statement is true.

The valence electrons of group 3A elements are in the ns and np subshell .Where n is equal to 1 to 7. there general electronic configuration is [tex]ns^2np^1[/tex].

8) The highest principal quantum number of period 4 elements is 5

The given statement is false.

The period number of periodic table indicates the value of highest principal quantum number.

The electron address in an atom can be given with the help of quantum mechanics. It helps to locate electrons with quantum numbers.

(1) The valence electrons of group 4A elements are in the 5s subshell.

The ns in group 4A can vary from 1 to 7. Thus the statement is false.

(2) Period 3 elements have an inner electron configuration of [Ar).

The nearest noble gas has been Neon, thus the inner electron configuration is of [Ne]. The statement is false.

(3) The highest principal quantum number of period 4 elements is 4.

The period number has been equal to the highest principal quantum number. Thus, the statement is true.

(4) Period 5 elements have six 4p electrons.

It has 5 subshells, and 4 subshells have been filled prior to 5s. Thus, the elements have a 4p orbital filled with 6 electrons. The statement is true.

(5) Period 5 elements have an inner electron configuration of (Kr).

The nearest noble gas has been Krypton, thus the inner electron configuration is of [Kr]. The statement is true.

(6) Group 8A elements have full outer principal s and p subshells.

Group 8A belongs to a noble gas. The subshells of noble gas are completely filled. The statement is true.

(7) The valence electrons of group 3A elements are in an s and p subshell.

The electrons of the 3A group have a general configuration [tex]\rm ns^2\;np^1[/tex]. The electrons are filled in s and p subshell. The statement is true.

(8) The highest principal quantum number of period 4 elements is 5.

The period number has been equal to the highest principal quantum number. Thus, the statement is false.

For more information about the quantum numbers, refer to the link:

https://brainly.com/question/18835321

Answer:   -2359 Joules

Explanation:

According to first law of thermodynamics:

[tex]\Delta E=q+w[/tex]

[tex]\Delta E[/tex]= Change in internal energy  = -2886 J

q = heat absorbed or released  = ?

w = work done or by the system  

w = work done by the system=[tex]-P\Delta V[/tex]  {Work is done by the system as the final volume is greater than initial volume and is negative}

w = -527 J

q = ?  

[tex]-2886=q+(-527)[/tex]

[tex]q=-2359J[/tex]  {Heat released by the system is negative}

Thus the amount of heat released by the system and carried away by the cooling system is -2359 Joules

According to first law of thermodynamics:

= Change in internal energy  = -2886 J

q = heat absorbed or released  = ?

w = work done or by the system  

w = work done by the system=  {Work is done by the system as the final volume is greater than initial volume and is negative}

w = -527 J

q = ?  

 {Heat released by the system is negative}

Thus the amount of heat released by the system and carried away by the cooling system is -2359 JoulesAnswer:

Explanation:

The final temperature of the mixture is approximately [tex]\(-57.97^\circ\text{C}\)[/tex].

To find the final temperature of the mixture, we can use the principle of conservation of energy, assuming no heat is lost or gained in the process.

The heat gained by the cold substance (water) will be equal to the heat lost by the hot substance (ethanol).

The heat gained or lost [tex](\(Q\))[/tex] can be calculated using the equation:

[tex]\[ Q = mc\Delta T \][/tex]

where:

- [tex]\(m\)[/tex] is the mass of the substance,

- [tex]\(c\)[/tex] is the specific heat capacity of the substance,

- [tex]\(\Delta T\)[/tex] is the change in temperature.

The sum of the heats gained and lost is zero:

[tex]\[ Q_{\text{water}} + Q_{\text{ethanol}} = 0 \][/tex]

Since the final temperature is the same for both substances, we can write the equation:

[tex]\[ m_{\text{water}}c_{\text{water}}\Delta T_{\text{water}} + m_{\text{ethanol}}c_{\text{ethanol}}\Delta T_{\text{ethanol}} = 0 \][/tex]

Rearrange the equation to solve for the final temperature [tex](\(T_{\text{final}}\))[/tex]:

[tex]\[ \Delta T_{\text{water}} + \Delta T_{\text{ethanol}} = 0 \][/tex]

[tex]\[ T_{\text{final}} - T_{\text{initial, water}} + T_{\text{final}} - T_{\text{initial, ethanol}} = 0 \][/tex]

[tex]\[ 2T_{\text{final}} = T_{\text{initial, water}} + T_{\text{initial, ethanol}} \][/tex]

[tex]\[ T_{\text{final}} = \frac{T_{\text{initial, water}} + T_{\text{initial, ethanol}}}{2} \][/tex]

Now, substitute the given values:

[tex]\[ T_{\text{final}} = \frac{(55.0 \ \text{mL} \times 1.0 \ \text{g/mL} \times 4.18 \ \text{J/g}^\circ\text{C} \times (28.6^\circ\text{C} - T_{\text{final}})) + (55.0 \ \text{mL} \times 0.789 \ \text{g/mL} \times 2.44 \ \text{J/g}^\circ\text{C} \times (T_{\text{final}} - 9.0^\circ\text{C}))}{2} \][/tex]

[tex]\[ T_{\text{final}} = \frac{(55.0 \ \text{mL} \times 4.18 \ \text{J/g}^\circ\text{C} \times (28.6^\circ\text{C} - T_{\text{final}})) + (55.0 \ \text{mL} \times 0.789 \ \text{g/mL} \times 2.44 \ \text{J/g}^\circ\text{C} \times (T_{\text{final}} - 9.0^\circ\text{C}))}{2} \][/tex]

Combine like terms:

[tex]\[ T_{\text{final}} = \frac{(55.0 \times 4.18 \times 28.6) + (55.0 \times 0.789 \times 2.44 \times (T_{\text{final}} - 9.0))}{2} \][/tex]

[tex]\[ T_{\text{final}} = \frac{(6132.86 + 107.79 \times (T_{\text{final}} - 9.0))}{2} \][/tex]

[tex]\[ T_{\text{final}} = \frac{6132.86 + 107.79T_{\text{final}} - 971.1}{2} \][/tex]

[tex]\[ 2T_{\text{final}} = 7101.76 + 107.79T_{\text{final}} - 971.1 \][/tex]

Combine like terms:

[tex]\[ 2T_{\text{final}} - 107.79T_{\text{final}} = 7101.76 - 971.1 \][/tex]

[tex]\[ -105.79T_{\text{final}} = 6130.66 \][/tex]

[tex]\[ T_{\text{final}} \approx -57.97^\circ\text{C} \][/tex]

Therefore, the final temperature of the mixture is approximately [tex]\(-57.97^\circ\text{C}\)[/tex]. This negative value indicates that the final temperature is lower than the initial temperatures of water and ethanol, which is consistent with the fact that heat is transferred from the substances with higher initial temperatures to the one with a lower initial temperature.

Answer:

a. 1,157x10⁻⁵M/s

b. 5,787x10⁻⁶M/s

Explanation:

For the reaction:

2H₂O₂(aq) → 2H₂O(l) + O₂(g).

a. The rate law of descomposition is:

[tex]rate=-\frac{1}{2} \frac{d[H_{2}O_{2}]}{dt}=\frac{d[O_{2}]}{dt}[/tex]

Where d[H₂O₂] is the change in concentration of H₂O₂ (between 0s and 2,16x10⁴s) is (1,000M-0,500M) and dt is (0s-2,16x10^4s). Replacing:

[tex]rate=-\frac{1}{2} \frac{0,500M}{-2,16x10^4s}[/tex]

[tex]rate=1,157x10^{-5}M/s[/tex]

As this rate is = d[O₂]/dt(Rate of production of O₂), the rate of production of O₂(g) is 1,157x10⁻⁵M/s

b. Between 2,16x10⁴s and 4,32x10⁴s, rate law is:

[tex]rate=-\frac{1}{2} \frac{0,500M-0,250M}{2,16x10^4s-4,32x10^4s}[/tex]

[tex]rate=5,787x10^{-6}M/s[/tex]

The rates are 5,787x10⁻⁶M/s

I hope it helps!

The study of chemicals and bonds is called chemistry. There are two types of elements these rare metals and nonmetals.

Thus the rate is,

[tex]rate=1157\times10^{-5}[/tex]

The rate law or rate equation for a chemical reaction is an equation that links the initial or forward reaction rate with the concentrations or pressures of the reactants and constant parameters.

The balanced reaction is:-

[tex]2H_2O_2(aq)----->2H_2O(l)+o_2(g)[/tex]

The rate law of decomposition:-

[tex]rate=-\dfrac{1}{2}\dfrac{d[H_2O_2]}{dt}[/tex]

Where d[H₂O₂] is the change in concentration of H₂O₂ (between 0s and 2,16x10⁴s) is (1,000M-0,500M) and dt is (0s-2,16x10^4s). Replacing:

[tex]rate=-\dfrac{1}{2}\dfrac{0.5}{2.16\times10^{4}}[/tex]

[tex]rate=1157\times10^{-5}[/tex]

As this rate is = d[O₂]/dt(Rate of production of O₂), the rate of production of O₂(g) is 1,157x10⁻⁵M/s

For more information about the rate of reaction, refer to the link:-

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Answer:

The true statements are given below.

Explanation:

1 In the electron transport chain a series of reactions moves electrons through carriers.

2 Coenzyme Q and cytochrome are the components of the electron transport chain.

   During electron transport chain electrons are transported from one electron carrier have high reduction potential to another electron carrier having comparatively low reduction potential than the previous one by a series of redox reactions.

      Coenzyme Q is a component of complex 2 whereas cytochrome is an important component of both complex 3 and complex 4.

The electron transport chain is a series of reactions that moves electrons through carriers and is located in the inner mitochondrial matrix in eukaryotic cells and the cytoplasmic membrane in prokaryotic cells. Its products are water and ATP. Coenzyme Q and cytochrome are components of the electron transport chain, but Coenzyme A is not. The electron transport chain operates independently of other metabolic processes.

The electron transport chain (ETC) is a series of reactions that moves electrons through carriers. It is located in the inner mitochondrial matrix in eukaryotic cells and the cytoplasmic membrane in prokaryotic cells. The products of the electron transport chain are water and ATP. Coenzyme Q and cytochrome are components of the electron transport chain, but Coenzyme A is not. The electron transport chain operates independently of other metabolic processes.

Answer:

b. ΔE rxn is a measure of heat

Explanation:

a. ΔHrxn is the heat of reaction. TRUE. ΔHrxn or change in enthalpy of reaction is per definition the change in heat that is involved in a chemical reaction.

b. ΔErxn is a measure of heat. FALSE. Is the change in internal energy of a reaction

c. An exothermic reaction gives heat off heat to the surroundings. TRUE. An exothermic reaction is a chemical reaction that releases heat.

d. Endothermic has a positive ΔH. TRUE. When a process is exothermic ΔH<0 and when the process is endothermic ΔH>0

e. Enthalpy is the sum of a system's internal energy and the product of pressure and volume. TRUE. Under constant pressure and volume the formula is ΔH = ΔE + PV

I hope it helps!

The statement that is FALSE is b. ΔErxn is a measure of heat.

The statement that is FALSE is b. ΔErxn is a measure of heat.

ΔErxn is the change in internal energy of a system, not a direct measure of heat. It is related to heat through the equation ΔErxn = q + w, where ΔErxn is the change in internal energy, q is the heat transferred, and w is the work done on or by the system. Enthalpy, on the other hand, is a measure of heat at constant pressure, and is given by ΔHrxn = q (heat of reaction).

Therefore, option b is the false statement.

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Answer:

The molarity of HI at the equilibrium is 2.8M

Explanation:

Step 1: Data given

Volume of the flask = 250.0 mL = 0.250L

Number of moles I2 = 1.3 mol

Number of moles HI = 1.0 mol

Kc = 0.983

Step 2: The balanced equation

H2(g) +I2(g) ⇆ 2HI(g)

For 1 mole I2 consumed, we need 1 mole H2 to produce 2 moles HI

Step 3: Calculate initial concentrations

Initial concentration I2 = 1.3mol / 0.25L

Initial concentration I2 = 5.2 M

Initial concentration HI = 1.0 mol / 0.25L

Initial concentration HI = 4.0 M

Step 4: Calculate concentrations at equilibrium

The concentration at equilibrium is:

[I2] = (5.2+x)M

[HI] = (4.0 - x)M

[H2] = xM

Kc = [HI]²/[H2][I2]

0.983 = (4-x)²/ (x*(5.2+x))

0.983 = (4-x)²/ (5.2x +x²)

5.1116x + 0.983 x² = 16 -8x +x²

-0.017x² +13.1116x -16 = 0

x = 1.222 = [H2]

[HI] = 4.0 - 1.222 = 2.778M ≈ 2.8 M

[I2] = 5.2 + 1.222 = 6.422 M ≈ 6.4 M

To control we can calculate:

[2.778]² / [1.222][6.422]  = 0.983 = Kc

The molarity of HI at the equilibrium is 2.8M

Answer:

a. strong interparticle attractions, low temperature, high pressure.

Explanation:

A gas behaves as an ideal gas when it fulfills the following conditions:

The gas deviates from the ideal behavior:

A real gas deviates most from ideal behavior under the conditions of strong interparticle attractions, low temperature, and high pressure.

A real gas deviates most from ideal behavior under the following set of conditions: strong interparticle attractions, low temperature, high pressure.

When a real gas has strong interparticle attractions, the gas particles are more likely to stick together and deviate from ideal behavior. Low temperature also contributes to the deviation as it slows down the gas particles and makes the attractive forces between them more prominent. High pressure further increases the deviation due to the decrease in empty space between the particles.

In summary, a real gas deviates most from ideal behavior when it has strong interparticle attractions, low temperature, and high pressure.

Answer:

Neutral nucleophile are: H2O, CH3OH, NH3, RNH2, R2NH, R3N, RCOOH, RSH and PR3. The products by nucleophilic substitution are diverse depending on the different nucleophiles, obtaining alcohol, eter, amines, ester and tioeter considering only the nucleophiles with a hydrogen available.

Explanation:

Please see the images attached.

Nucleophilic subtitution with water occurs under Sn1 mechanism. That's it because water as nucleophile is so weak.  With the other neutral nucleophiles, the reaction occur under Sn2 mechanism.

RSH + CH3I  -----> RSCH3  +  HI

Answer:

19,26 kJ

Explanation:

The work done when a gas expand with a constant atmospheric pressure is:

W = PΔV

Where P is pressure and ΔV is the change in volume of gas.

Assuming the initial volume is 0, the reaction of 500g of Zn with H⁺ (Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)) produce:

500,0g Zn(s)×[tex]\frac{1molZn}{65,38g}[/tex]×[tex]\frac{1molH_{2}(g)}{1molZn}[/tex] = 7,648 moles of H₂

At 1,00atm and 303,15K (30°C), the volume of these moles of gas is:

V = nRT/P

V = 7,648mol×0,082atmL/molK×303,15K / 1,00atm

V = 190,1L

That means that ΔV is:

190,1L - 0L = 190,1L

And the work done is:

W = 1atm×190,1L = 190,1atmL.

In joules:

190,1 atmL×[tex]\frac{101,325}{1atmL}[/tex] = 19,26 kJ

I hope it helps!

The amount of work done against an atmospheric pressure of 1.00 atm when 500.0 g of zinc dissolves in excess acid at 30.0°C is approximately 19601.44 joules.

The amount of work done against an atmospheric pressure when a gas is produced can be calculated using the ideal gas law, which is given by PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we need to determine the number of moles of hydrogen gas (H2) produced from the dissolution of 500.0 g of zinc. The balanced chemical equation is:

[tex]\[ \text{Zn}(s) + 2\text{H}^+(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{H}_2(g) \][/tex]

 From the stoichiometry of the reaction, 1 mole of zinc produces 1 mole of hydrogen gas. The molar mass of zinc is approximately 65.38 g/mol. Therefore, the number of moles of zinc (n_Zn) is:

[tex]\[ n_{\text{Zn}} = \frac{\text{mass of Zn}}{\text{molar mass of Zn}} = \frac{500.0 \text{ g}}{65.38 \text{ g/mol}} \approx 7.647 \text{ mol} \][/tex]

 Since the stoichiometry is 1:1, the number of moles of hydrogen gas produced (n_H2) is also approximately 7.647 moles.

 Next, we convert the temperature from degrees Celsius to Kelvin:

[tex]\[ T = 30.0^\circ\text{C} + 273.15 = 303.15 \text{ K} \][/tex]

 Now, we can use the ideal gas law to find the volume of hydrogen gas produced at 1.00 atm of pressure:

[tex]\[ PV = nRT \][/tex]

[tex]\[ V = \frac{nRT}{P} \][/tex]

 Plugging in the values:

[tex]\[ V = \frac{(7.647 \text{ mol})(0.0821 \text{ L·atm/mol·K})(303.15 \text{ K})}{1.00 \text{ atm}} \][/tex]

[tex]\[ V \approx 193.57 \text{ L} \][/tex]

 Finally, the work done (W) against the atmospheric pressure to produce this volume of hydrogen gas is given by:

[tex]\[ W = P \cdot V \][/tex]

[tex]\[ W = 1.00 \text{ atm} \cdot 193.57 \text{ L} \][/tex]

[tex]\[ W \approx 193.57 \text{ L·atm} \][/tex]

 To express this work in joules, we use the conversion factor[tex]1 L*atm[/tex] = [tex]101.325 J:[/tex]

[tex]\[ W \approx 193.57 \text{ L·atm} \times \frac{101.325 \text{ J}}{1 \text{ L·atm}} \][/tex]

[tex]\[ W \approx 19601.44 \text{ J} \][/tex]

Therefore, the amount of work done against an atmospheric pressure of 1.00 atm when 500.0 g of zinc dissolves in excess acid at 30.0°C is approximately 19601.44 joules.

 The answer is: [tex]19601.44 \text{ J}.[/tex]

Final answer:

The weight percent of NH3 in the aqueous waste was calculated from the titration data to be 0.2047%. The moles of NH3 were found to be 0.0029711 mol, corresponding to 0.050585g NH3 in the original waste sample.

Explanation:

The weight percent of NH3 in the aqueous waste is calculated using titration analysis. First, determine the moles of HCl that reacted:
# mol HCl = 0.02842 L × 0.1045 M = 0.0029711 mol HCl.

The reaction between HCl and NH3 is:
NH3(aq) + HCl(aq) → NH4Cl(aq).

Hence, 1 mole of HCl reacts with 1 mole of NH3. Therefore, the moles of NH3 in the aliquot are also 0.0029711 mol. To find the mass of NH3:
Mass of NH3 = moles × molar mass
              = 0.0029711 mol × 17.031 g/mol
              = 0.050585 g NH3.

Since the aliquot is part of the diluted waste, and assuming the ammonia is evenly distributed, we calculate the weight percent in the original waste sample:
   Weight percent = (mass of NH3 in aliquot / mass of waste sample) × 100
                 = (0.050585 g / 24.711 g) × 100
                 = 0.2047%.

The Ka of HX is mathematically given as

Ka = 9.11 *10^-8

Generally, the equation for the molarity of HX  is mathematically given as

M HX = moles HX / volume solution

Therefore

Molarity HX = 0.365 mol / 0.835 L

Molarity HX = 0.437 M

Therefore, ICE-chart

[H+] = [H3O+]

10^-3.70 = 10^-3.70 = 1.995 *10^-4

The concentration at the equilibrium is

[HX] = (0.437 - x)M

[H3O+] = 1.995*10^-4 M

x=1.995*10^-4

In conclusion

Ka = [X-]*[H3O+] / [HX]

Ka = ((1.995*10^-4)²)/ 0.437

Ka = 9.11 *10^-8

Read more about Solubility

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Final answer:

The Ka of HX, given its concentration and the pH of solution, can be determined through a series of calculations leading to a Ka value of 9.17×10-8.

Explanation:

To find the Ka of HX given a 0.365-mol sample dissolved in 835.0 mL of solution with a pH of 3.70, we start by calculating the concentration of HX, then use the pH to find the concentration of H+ ions, which will help determine the acid dissociation constant (Ka).

First, convert volume from mL to L: 835.0 mL = 0.835 L. The concentration of HX (Molarity, M) is moles of solute (HX) divided by volume of solution in liters, which is 0.365 mol / 0.835 L = 0.4371 M.

Given pH = 3.70, we calculate the concentration of H+ ions ([H+]) as 10-pH = 10-3.70 = 2.00×10-4 M.

Assuming all H+ comes from the dissociation of HX and ignoring autoionization of water because HX is a weak acid, the Ka expression for HX is Ka = [H+][X-] / [HX]. Assuming the concentration of [X-] is equal to [H+] because each molecule of HX donates one H+, Ka becomes 2.00×10-4 M * 2.00×10-4 M / 0.4371 M = 9.17×10-8, after considering the initial concentration of HX and the change in concentration due to dissociation.

Answer:

mass of O₂ = 800 g

Explanation:

Formula of Octane = C₈H₁₈

Formula of Oxygen = O₂

Balanced Chemical Reaction:

Octane react with oxygen and produce water and carbon dioxide.

                    2C₈H₁₈ + 25O₂ ---------> 16CO₂+ 18H₂O

Moles calculation:

From the number of moles of the reactant used in a chemical reaction specific number of product produced. we can find the amount of any reactant and product from the mole ratio of a chemical reaction.

Mole formula =  

no. of moles = mass in grams / molecular mass ............... (1)

Given Data:

Molecular Weight of O₂  =  (16 x2 ) = 32 g/mol

Mass of O₂ = To be find

Calculations:

                             2C₈H₁₈  +  25O₂      --------->  16CO₂   +    18H₂O

                              2mole      25 mole                 16 mole     18 mole

From the above balanced chemical equation it is know that 2 mole of octane consume 25 mole of oxygen.

so we have to calculate the mass of oxygen that is consumed

by using mole formula  (1) we can fine the mass of oxygen

we know

Molecular Weight of O₂  =  32 g/mol

number of moles of Oxygen molecule = 25 mol

putting the value in the below formula

            no. of moles of O₂ = mass of O₂ / molecular mass of O₂

            25 mole = mass of O₂ / 32 g/mol ....... (2)

By rearragming the equation (2)

            mass of O₂ = 25 mole x 32 g/mol

            mass of O₂ = 25 mole x 32 g/mol

            mass of O₂ = 800 g

So in the octan reaction with oxygen 800g of oxygen will use.  

Answer:

To release 7563 kJ of heat, we need to burn 163.17 grams of propane

Explanation:

Step 1: Data given

C3H8 + 5O2 -----------> 3CO2 + 4H2O      ΔH° = –2044 kJ

This means every mole C3H8

Every mole of C3H8 produces 2044 kJ of heat when it burns (ΔH° is negative because it's an exothermic reaction)

Step 2: Calculate the number of moles to produce 7563 kJ of heat

1 mol = 2044 kJ

x mol = 7563 kJ

x = 7563/2044 =  3.70 moles

To produce 7563 kJ of heat we have to burn 3.70 moles of C3H8

Step 3: Calculate mass of propane

Mass propane = moles * Molar mass

Mass propane = 3.70 moles * 44.1 g/mol

Mass propane = 163.17 grams

To release 7563 kJ of heat, we need to burn 163.17 grams of propane

The mass of propane needed by the reaction to release 7563 KJ of heat energy is 162.8 g

C₃H₈ + 5O₂ —> 3CO₂ + 4H₂O ΔH° = –2044 KJ

Molar mass of propane C₃H₈ = (3×12) + (1×8) = 44 g/mol

Mass of propane C₃H₈ from the balanced equation = 1 × 44 = 44 g

Heat released = 2044 KJ

SUMMARY

From the balanced equation above,

2044 KJ of heat was released by the reaction of 44 g of propane

From the balanced equation above,

2044 KJ of heat was released by the reaction of 44 g of propane

Therefore,

7563 KJ of heat energy will be release by = (7563 × 44) / 2044 = 162.8 g of propane

Thus, 162.8 g of propane is needed for the reaction

Learn more enthalpy change:

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The predicted sign of the entropy change in each reaction are as follows;

To solve this question, we must understand the meaning of entropy.

Entropy of a system or chemical entity is the degree of disorderliness in the substance.

The change in entropy, ΔS∘ is a measure of the difference between the entropy of products and the reactants and is given mathematically as;

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In the first reaction entropy sign is negative while for the second reaction the sign on entropy is positive.

The Entropy is defined as degree of disorderliness in the system.

The entropy of the reaction can be calculated by,

[tex]\rm \bold{ \Delta S = S(products) - S(reactants)}[/tex]

Entropy for First reaction,

[tex]\rm \bold{ Ag(aq) + Br(aq) \rightarrow AgBr(s)}[/tex] will be negative

Entropy for second reaction,

[tex]\rm \bold{ CaCO_3(s) \rightarrow CaO(s)+CO_2(g) }[/tex] will be positive.

Hence, we can conclude that the the first reaction will be non- spontaneous while second reaction will spontaneous.

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Final answer:

The molar solubility of Mg(OH)2 in pure water is 2.41 x 10^-4 M. To calculate the molar solubility in a 0.130 M NaOH solution, we can use the common ion effect.

Explanation:

The molar solubility of Mg(OH)2 in pure water is 2.41 x 10^-4 M. The molar solubility of Mg(OH)2 in a 0.130 M NaOH solution can be determined using the common ion effect. Since NaOH dissociates into Na+ and OH-, the concentration of OH- in the NaOH solution is 0.130 M. To calculate the molar solubility, we can set up an ICE table:

Mg(OH)2 ⇌ Mg2+ + 2OH-

Initial: 0.130 M (from NaOH concentration)

Change: -x (due to dissolution of Mg(OH)2) and +2x (due to formation of OH- ions)

Equilibrium: 0.130 - x M

Ksp = (Mg2+)(OH-)²

Using the given value of the Ksp, 5.61x10^-11, we can substitute the equilibrium concentrations into the Ksp expression and solve for x. Then we can calculate the ratio of the solubility of Mg(OH)2 in pure water to the solubility in the NaOH solution by dividing the molar solubility in water by the molar solubility in the NaOH solution.

Mg(OH)₂ is approximately [tex]\( 0.472} \)[/tex] times more soluble in pure water compared to a 0.130 M NaOH solution. The solubility of Mg(OH)₂ in the presence of 0.130 M NaOH, is approximately [tex]\( 5.1 \times 10^{-4} \)[/tex] M.

To determine how many times more soluble Mg(OH)₂ is in pure water compared to a 0.130 M NaOH solution, we need to consider the effect of the common ion (OH-) from NaOH on the solubility of Mg(OH)₂.

Given:

- [tex]\( K_{sp} \)[/tex] for Mg(OH)₂ in pure water = [tex]\( 5.61 \times 10^{-11} \)[/tex]

- Molar solubility of Mg(OH)₂ in pure water = [tex]\( 2.41 \times 10^{-4} \)[/tex] M

- Molar solubility of NaOH in 0.130 M solution = [tex]\( 3.32 \times 10^{-9} \)[/tex] M

1. Calculate the concentration of OH- ions in the 0.130 M NaOH solution:

  The NaOH dissociates completely to produce Na+ and OH- ions.

  Concentration of OH- ions from NaOH = 0.130 M

2. Determine the solubility of Mg(OH)₂ in the presence of 0.130 M NaOH:

  According to the common ion effect, the presence of additional OH- ions from NaOH will shift the equilibrium of Mg(OH)2 dissolution. The solubility of Mg(OH)₂ will decrease due to the Le Chatelier's principle, which states that adding an ion that is involved in the equilibrium reaction will shift the equilibrium in the direction that reduces its effect.

Let's denote the solubility of Mg(OH)₂ in the presence of 0.130 M NaOH as [tex]\( S_{\text{NaOH}} \)[/tex].

Using the Ksp expression for Mg(OH)₂:

[tex]\[ K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 \][/tex]

In pure water, [tex]\( [\text{OH}^-] = S_{\text{water}} \)[/tex] (molar solubility of Mg(OH)2 in pure water).

In the presence of 0.130 M NaOH, [tex]\( [\text{OH}^-] = 3.32 \times 10^{-9} \)[/tex]M (concentration of OH- from NaOH).

So, [tex]\( S_{\text{NaOH}} \)[/tex] can be found from:

[tex]\[ K_{sp} = [\text{Mg}^{2+}](3.32 \times 10^{-9})^2 \][/tex]

Solve for [tex]\( [\text{Mg}^{2+}] \)[/tex]:

[tex]\[ [\text{Mg}^{2+}] = \frac{K_{sp}}{(3.32 \times 10^{-9})^2} \][/tex]

Calculate [tex]\( [\text{Mg}^{2+}] \)[/tex]:

[tex]\[ [\text{Mg}^{2+}] = \frac{5.61 \times 10^{-11}}{(3.32 \times 10^{-9})^2} \][/tex]

[tex]\[ [\text{Mg}^{2+}] = \frac{5.61 \times 10^{-11}}{1.1 \times 10^{-17}} \][/tex]

[tex]\[ [\text{Mg}^{2+}] \approx 5.1 \times 10^{-4} \, \text{M} \][/tex]

3. Calculate the ratio of solubility in pure water to solubility in 0.130 M NaOH:

[tex]\[ \text{Ratio} = \frac{S_{\text{water}}}{S_{\text{NaOH}}} = \frac{2.41 \times 10^{-4}}{5.1 \times 10^{-4}} \][/tex]

[tex]\[ \text{Ratio} \approx 0.472 \][/tex]

Answer:

Molarity of Unknown Acid = 0.1332 M

Explanation:

Data for solving problem:

Molarity of base in buret (M₁)= 0.1517 M

volume of the acid in Erlenmeyer flask (V₂)= 25.0 mL

Volume of the base in the buret (V₁) = final volume of buret - initial volume in buret

final volume of buret = 22.5 mL

initial volume in buret = 0.55 mL

So

Volume of the base in the buret (V₁) = 22.5 mL -0.55 mL = 21.95 mL

Volume of the base in the buret (V₁)  = 21.95 mL

Molarity of Unknown acid in the Erlenmeyer flask (M₂) = To be find

Explanation:

It is acid base titration and  formula for this titration is as follows:

Molarity of base x Volume of base = Molarity of acid x volume of acid

it can be written as

M₁V₁ = M₂V₂ -------------------- equation (1)

we have to find M₂

so by rearrangment the equation (1)

M₁V₁ / V₂ = M₂ ------------------ equation (2)

put the values in equation in equation (2)

M₂ = 0.1517 M x 21.95 mL / 25.0 mL

M₂ = 3.3298 /25.0

M₂ = 0.1332 M

so the Molarity of Unknown acid is 0.1332 M

The molarity of the unknown acid HA is calculated to be 0.1331 M by first determining the moles of NaOH that reacted and then using the volume of acid to find its concentration.

To determine the molarity of the unknown acid HA in a titration experiment, first, we calculate the volume of 0.1517 M NaOH used by subtracting the initial buret reading from the final buret reading: 22.50 mL - 0.55 mL = 21.95 mL. This volume is then converted to liters by dividing by 1000: 21.95 mL/1000 = 0.02195 L. Next, we use the molarity of the NaOH solution to find the moles of NaOH that have reacted: 0.1517 M × 0.02195 L = 0.003328665 moles.

Assuming a 1:1 mole ratio between NaOH and HA in the reaction, the moles of unknown acid HA that reacted are also 0.003328665. We then calculate the molarity of the unknown acid, by dividing moles of acid by the volume of acid in liters (25.0 mL = 0.025 L): 0.003328665 moles / 0.025 L = 0.1331466 M.

The molarity of the unknown acid HA is therefore 0.1331 M, rounded to four significant figures to match the precision of the given data.

Answer:

The weight percent in the sample is 17,16%

Explanation:

The dissolution of the Ce(IV) salt provides free Ce⁴⁺ that reacts, thus:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃²⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03247M Na₂S₂O₃ = 4,228x10⁻⁴ moles of S₂O₃²⁻.

As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,228x10⁻⁴ moles of S₂O₃⁻×[tex]\frac{1molI_{3}^-}{2molS_{2}O_{3}^{2-}}[/tex] = 2,114x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,114x10⁻⁴ moles of I₃⁻× [tex]\frac{2molCe^{4+}}{1molI_{3}^-}[/tex] =  4,228x10⁻⁴ moles of Ce(IV).

These moles are:

4,228x10⁻⁴ moles of Ce(IV)×[tex]\frac{140,116g}{1mol}[/tex] = 0,05924 g of Ce(IV)

As was taken an aliquot of 25,00mL from the solution of 250,0mL:

0,05924 g of Ce(IV)×[tex]\frac{250,0mL}{25,00mL}[/tex] =0,5924g of Ce(IV) in the sample

As the sample has 3,452g, the weight percent is:

0,5924g of Ce(IV) / 3,452g × 100 = 17,16 wt%

I hope it helps!

Answer:

For a liquid, it is 25°F

Explanation:

The standard state for a liquid is 25°C

The state which is not a standard state for solid is 25 °C and for liquid is 25 F.

Every matter will behave idealy at a particular set of items and that set is known as standard state of that matter.

Solids will remain at the standard state at 1 atm pressure, but 25°C (298K) is not the standard temperature for solids as they will melt in this temperature.

Standard state of liquids is 1 atm pressure and 25°C (298K) temperature, but in the question it is given 25F which is not true so.

Solution of 1 molariy will consider as the standard solution and it is define as the number of moles of solute present in per liter of the solution.

So, for solids and liquids 25°C and 25F respectively is not a standard state.

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Answer:

V= 0.544

Explanation:

Given that

σ=812 MPa

σ₁=1240 MPa

σ₂=300 MPa

Lets take volume fraction V

The strength of composite fiber given as

σ = σ₁ V + σ₂(1-V)

By putting the values

812 = 1240 V + 300 (1-V)

812 = 1240 V +300 - 300 V

812 - 300 = 1240 V- 300 V

512 = 940 V

V= 0.544

The percentage volume = 54.4 %

Final answer:

Leucine, being nonpolar, should travel farther than the polar amino acid aspartic acid on a TLC plate because nonpolar substances have a higher affinity for the stationary phase and a lower affinity for the polar mobile phase.

Explanation:

On a thin-layer chromatography (TLC) plate, the distance that a compound travels is closely related to its polarity. In this lab experiment, leucine is a nonpolar amino acid, while aspartic acid is polar due to its acidic side chain. Therefore, leucine should travel farther on the TLC plate than aspartic acid because the nonpolar amino acids have a higher affinity for the nonpolar stationary phase and a lower affinity for the polar mobile phase, so they're carried less distance by the solvent front.

The correct answer to this question is A. Leucine is nonpolar, thus it should travel farther than aspartic acid on my TLC plate.

Answer:

The valid set of quantum numbers are:

a. 3,1,1,-1/2

b. 4,3,1,-1/2

c. 2,0,0,-1/2

e. 3,2,-1,-1/2

j. 3,0,0,1/2

Explanation:

Quantum numbers (n, l, mℓ, ms) are the set of numbers that describe the state of an electron in an atom.

The four quantum numbers and their rules are:

Therefore,

a. 3,1,1,-1/2: Valid

b. 4,3,1,-1/2: Valid

c. 2,0,0,-1/2: Valid

d. 1,3,0,1/2: NOT Valid

Reason: ℓ ≤ (n-1). Therefore, ℓ can not be greater than n, .

e. 3,2,-1,-1/2: Valid

f. 3,3,-1,1/2: NOT Valid

Reason: ℓ ≤ (n-1). Therefore, ℓ can not be equal to n. (ℓ ≠n)

g. 3,2,1,-1: NOT Valid

Reason: the only allowed values of ms = (- s), (+ s) = (- 1/2), (+ 1/2)

h. 1,-1,-1,-1/2: NOT Valid

Reason: ℓ ≤ (n-1) and n ≥ 1. Therefore, ℓ can not be equal to n and also ℓ can't be negative.

i. 3,3,1,1/2: NOT Valid

Reason: ℓ ≤ (n-1). Therefore, ℓ can not be equal to n. (ℓ ≠n)

j. 3,0,0,1/2: Valid

k. 4,3,4,-1/2: NOT Valid

Reason: mℓ = (- ℓ) to (+ ℓ). Therefore, mℓ can't be greater than ℓ.

l. 0,2,1,1/2: NOT Valid

Reason: n ≥ 1. Therefore, value of n can't be 0 (n≠0)

Answer:

The correct option is: B. 14 H⁺ + 6 e⁻ + Cr₂O₇²⁻ ⟶ 2 Cr³⁺ + 7 H₂O

Explanation:

Redox reactions is an reaction in which the oxidation and reduction reactions occur simultaneously due to the simultaneous movement of electrons from one chemical species to another.

The reduction of a chemical species is represented in a reduction half- reaction and the oxidation of a chemical species is represented in a oxidation half- reaction.

To balance the reduction half-reaction for the reduction of Cr₂O₇²⁻ to Cr³⁺:

Cr₂O₇²⁻ ⟶ Cr³⁺

First the number of Cr atoms on the reactant and product side is balanced

Cr₂O₇²⁻ ⟶ 2 Cr³⁺

Now, Cr is preset in +6 oxidation state in Cr₂O₇²⁻ and +3 oxidation state in Cr³⁺. So each Cr gains 3 electrons to get reduced.

Therefore, 6 electrons are gained by 2 Cr atoms of Cr₂O₇²⁻ to get reduced.

Cr₂O₇²⁻ + 6 e⁻ ⟶ 2 Cr³⁺

Now the total charge on the reactant side is (-8) and the total charge on the product side is (+6).

From the given options it is evident that the reaction must be balanced in acidic conditions.

Therefore, to balance the total charge on the reactant and product side, 14 H⁺ is added on the reactant side.

Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺

Now to balance the number of hydrogen and oxygen atoms, 7 H₂O is added on the product side.

Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺ + 7 H₂O

Therefore, the correct balanced reduction half-reaction is:

Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺ + 7 H₂O

Final answer:

To calculate the mass of mineral fluoroapatite (Ca5(PO4)3F) required each year, we need to use stoichiometry. Given that approximately 2.00 x 10^9 gallons of concentrated phosphoric acid solution is produced annually containing 14.8 M concentration, we can convert this volume to moles of H3PO4 and then use stoichiometry to calculate the moles of Ca5(PO4)3F required. Finally, we can calculate the mass of Ca5(PO4)3F required, which is approximately 1.88 x 10^13 grams.

Explanation:

To calculate the mass of mineral fluoroapatite (Ca5(PO4)3F) required each year, we need to use stoichiometry.

Given that 2.00 x 109 gallons of concentrated phosphoric acid solution are produced annually, we need to convert this volume to moles of H3PO4 and then use the stoichiometric coefficients to calculate the moles of Ca5(PO4)3F required.

A balanced equation shows that for every 3 moles of H3PO4 produced, 1 mole of Ca5(PO4)3F is required.

Therefore, we can set up a ratio:

3 moles H3PO4 / 1 mole Ca5(PO4)3F

Now, we can calculate the moles of H3PO4:

14.8 M = 14.8 mol/L

2.00 x 109 gallons * 3.785 L/gallon = 7.57 x 109 L

7.57 x 109 L * 14.8 mol/L = 1.12 x 1011 mol H3PO4

Finally, we can use the ratio to calculate the moles of Ca5(PO4)3F:

1.12 x 1011 mol H3PO4 * (1 mol Ca5(PO4)3F / 3 mol H3PO4) = 3.73 x 1010 mol Ca5(PO4)3F

Since the molar mass of Ca5(PO4)3F is 504.05 g/mol, we can calculate the mass of Ca5(PO4)3F required:

3.73 x 1010 mol * 504.05 g/mol = 1.88 x 1013 g Ca5(PO4)3F

Therefore, approximately 1.88 x 1013 grams of mineral fluoroapatite (Ca5(PO4)3F) would be required each year.

Answer:

The charges on the given complexes are: +2, (-3), (-1)

Explanation:

A coordination complex is composed of a central metal atom or metal ion and ligands, bonded by coordinate covalent bonds.

The total charge of the complex is equal to the sum of the charge on central metal and the total charge of the ligands.

1. hexaaquairon(II), [Fe(H₂O)₆]

The oxidation state or charge on the central metal ion, Fe = +2

Charge on the ligand water molecule H₂O = 0

Therefore, the total charge on the complex = +2 + (0 × 6) = +2

2. tris(carbonato)aluminate(III), [Al(CO₃)₃]

The oxidation state or charge on the central metal ion, Al = +3

Charge on the ligand carbonate CO₃²⁻ = (-2)

Therefore, the total charge on the complex = +3 + (-2 × 3) = (-3)

3. diaquatetrachlorovanadate(III), [V(H₂O)₂Cl₄]

The oxidation state or charge on the central metal ion, V = +3

Charge on the ligand water molecule H₂O = 0

Charge on the ligand chloride Cl = (-1)

Therefore, the total charge on the complex = +3 + [(0 × 2)+(-1 × 4)] = +3 + (-4) = (-1)                  

Therefore, the charge on 1. hexaaquairon(II), [Fe(H₂O)₆] complex is +2

2. tris(carbonato)aluminate(III), [Al(CO₃)₃] complex is (-3)  

3. diaquatetrachlorovanadate(III), [V(H₂O)₂Cl₄] complex is (-1)                  

Answer:

The molarity of the Cl- anions is 0.0033 M

Explanation:

Step 1: Data given

Mass of MgCl2 = 48 mg = 48 *10^-3 grams

volume = 300 mL = 0.3L

Molar mas of MgCl2 = 95.21 g/mol

Step 2: The balanced equation

MgCl2 → Mg2+ + 2Cl-

This means for 1 mol MgCl2, we'll have 2 moles of Cl-

Step 3: Calculate moles MgCl2

Number of moles of MgCl2 = mass of MgCl2 / Molar mass of MgCl2

Moles MgCl2 = 48*10^-3 grams / 95.21 g/mol

Moles MgCl2 = 5.04 *10-4 moles

Step 4: Calculate moles of Cl-

This means for 1 mol MgCl2, we'll have 2 moles of Cl-

For  5.04 *10-4 moles MgCl2, we will have 2* 5.04 *10-4 moles = 0.001 moles Cl-

Step 5: Calculate molarity of the Cl- anion

Molarity = Moles / volume

Molarity Cl- = 0.001 moles / 0.3 L

Molarity Cl- = 0.0033 M

The molarity of the Cl- anions is 0.0033 M

Answer: The half reactions and net cell reaction of the cell is written below.

Explanation:

The given cell is:

[tex]Cu(s)/Cu^{2+}(aq.)||Ag^{+}(aq.)/Ag(s)[/tex]

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

Half reactions for the given cell follows:

Oxidation half reaction (anode): [tex]Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-[/tex]

Reduction half reaction (cathode): [tex]Ag^{+}(aq.)+e^-\rightarrow Ag(s)[/tex]   ( × 2)

Net cell reaction:  [tex]Cu(s)+2Ag^+(aq.)\rightarrow Cu^{2+}(aq.)+2Ag(s)[/tex]

Hence, the half reactions and net cell reaction of the cell is written above.

The half-reactions at each electrode and the net cell reaction for the electrochemical cell containing copper and silver are provided.

The half-reactions at each electrode in this electrochemical cell are:

Anode: Cu (s) → Cu2+ (aq) + 2e−

Cathode: 2Ag+ (aq) + 2e− → Ag (s)

The net cell reaction for this electrochemical cell is:

Cu (s) + 2Ag+ (aq) → Cu2+ (aq) + 2Ag (s)

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Answer:

only chlorine can expand its octet.

Explanation:

An atom can expand its octet is it has empty d orbital

the electronic configuration of given elements will be:

B : 1s2 2s2 2p1 [Valence shell n =2 no d orbital]

O :1s2 2s2 2p4 [Valence shell n =2 no d orbital]

F : 1s2 2s2 2p5 [Valence shell n =2 no d orbital]

Cl :1s2 2s2 2p6 3s2 3p5 3d0 [Valence shell n =2 no d orbital]

Out of given elements only chlorine has empty d orbitals in its valence shell

Thus only chlorine can expand its octet.

Answer:

The concentration of the chemist's silver(I) nitrate  solution is  0.897 mmol/L

Explanation:

Step 1: Data given

Number of moles AgNO3 = 269 micromol = 269 * 10^-6 mol

Volume of AgNO3 = 300 mL = 0.3 L

Molar mass of AgNO3 = 169.87 g/mol

Step 2: Calculate molarity of AgNO3

Molarity = number of moles per volume (in Liters)

Molarity AgNO3 = 269 *10^-6 mol / 0.3 L

Molarity AgNO3 = 8.97 * 10^-4 M

8.97 *10^-4 mol/L = 897 micromol/L = 0.897 mmol/L

The concentration of the chemist's silver(I) nitrate  solution is  0.897 mmol/L