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2 months ago

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NiO is to be reduced to nickel metal in an industrial process by use of the reaction:NiO(s) + CO(g) <-----> Ni(s) + CO2(g)

At 1600K the equilibrium constant for the reaction is kp = 6.0 x 10^2. If a CO pressure of 150 torr is to be employed in the furnace and total pressure never exceeds 760 torr, will reduction occur?

1 answer:

alisha [2.9K]2 months ago

6 0

Answer:

The reaction will proceed in the forward direction, resulting in the reduction of NiO to Ni

Explanation:

Step 1: Provided data

Kp = 6.0 * 10²

Pressure of CO = 150 torr

Total pressure remains under 760 torr

Step 2: The balanced reaction equation:

NiO(s) + CO(g) ⇆ Ni(s) + CO2(g)

Step 3: Determine P(CO) and P(CO2)

For this equilibrium, the Kp expression is:

Kp = P(CO2)/P(CO) = 6.0*10^2

Given that NiO and Ni are solids, they do not affect the Kp

⇒ with P(CO) = 150 torr

⇒ P(CO) = 760 - 150 = 610 torr

Step 4: Calculate the reaction quotient

Q = 610/150 = 4.1

Since Q is significantly less than Kp, there are more reactants than products. Some reactants will convert into products, driving the reaction rightward.

Thus, the reaction will proceed forward, leading to the reduction of NiO to Ni.

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For each set of dilutions (bleach and mouthwash), the first tube contained a 1:11 dilution (0.5 mL of agent was added to 5.0 mL

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Response:

Tube 2: 8.26 * 10^-3; Tube 4: 6.83 * 10^-5

Explanation:

For the MIC test's serial dilutions, each tube should contain an equal volume of nutrient broth: 5.0 mL, while the agent's volume per dilution must also match: 0.5 mL.

The serial dilution process followed was:

Tube 1: 0.5/5.5
Tube 2: 0.5 mL from tube 1 was diluted with 5.0 mL of broth, resulting in a dilution of tube 2 as (1:11) * (1:11) = (0.5/5.5) * (0.5/5.5) = 1:121 = 8.26 * 10^-3
Tube 3: similar calculations yield 1:1331 = 7.51 * 10^-4
Tube 4: yields 1:14641 = 6.83 * 10^-5.

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2 months ago

If you were given a sample of a cotton ball and a glass stirring rod with identical mass (ex: 5.0 g), which sample would contain

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Answer:

The sample with more oxygen atoms is a glass stirring rod.

Explanation:

Referring to the periodic table,

the glass stirring rod is composed of Silicon dioxide whereas the cotton ball is made of cellulose.

The molar mass of the glass stirring rod $SiO_{2}$ = 60.08 grams/mole.

The molar mass of the cotton ball $C_{6} H_{10}O_{5}$ = 162.09 grams/mole.

Considering the total number of molecules for oxygen is 32,

therefore,

In the glass stirring rod,

oxygen atoms contained in 5g = $\frac{32}{60.07} \times 5$

= 2.66 g of oxygen

= $\frac{1}{16} \times 2.66$

= 0.16625 moles

= 0.16625 x $6.023\times 10^{23}$

= $1.001 \times10^{23}$ atoms

In the cotton ball,

oxygen atoms contained in 5g = $\frac{80}{162.09} \times 5$

= 2.467 g of oxygen

= $\frac{1}{16} \times 2.467$

= 0.15418 moles

= 0.15418 $\times 6.023 \times 10^{23}$

= 0.928 $\\ \times10^{23}$

Thus, the glass stirring rod has a greater quantity of oxygen atoms.

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3 months ago

A sample of solid naphthalene is introduced into an evacuated flask. Use the data below to calculate the equilibrium vapor press

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Answer: The vapor pressure of naphthalene within the flask remains at $2.906\times 10^{-4}$ atm.

Explanation:

The transformation from solid naphthalene to its gaseous form follows the equilibrium reaction:

$C_{10}H_8(s)\rightleftharpoons C_{10}H_8(g)$

The formula employed to determine the enthalpy change for the reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The formula for calculating the enthalpy change regarding the aforementioned reaction is:

$\Delta H^o_{rxn}=(1\times \Delta H^o_f_{(C_{10}H_8(g))})-(1\times \Delta H^o_f_{(C_{10}H_8(s))})$

The provided information includes:

$\Delta H^o_f_{(C_{10}H_8(s))}=78.5kJ/mol\\\Delta H^o_f_{(C_{10}H_8(g))}=150.6kJ/mol$

Substituting the values into the previous equation produces:

$\Delta H^o_{rxn}=(1\times 150.6)-(1\times 78.5)=72.1kJ/mol$

The formula utilized to compute Gibbs free energy change is of a reaction:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f(product)]-\sum [n\times \Delta G^o_f(reactant)]$

The equation for the enthalpy change for the reaction is:

$\Delta G^o_{rxn}=(1\times \Delta G^o_f_{(C_{10}H_8(g))})-(1\times \Delta G^o_f_{(C_{10}H_8(s))})$

The given factors include:

$\Delta G^o_f_{(C_{10}H_8(s))}=201.6kJ/mol\\\Delta G^o_f_{(C_{10}H_8(g))}=224.1kJ/mol$

By inserting values from the above equation, we arrive at:

$\Delta G^o_{rxn}=(1\times 224.1)-(1\times 201.6)=22.5kJ/mol$

For the calculation of $K_1$ (at 25°C) regarding the provided value of Gibbs free energy, the following relationship is applied:

$\Delta G^o=-RT\ln K_1$

where,

$\Delta G^o$ = Gibbs free energy = 22.5 kJ/mol = 22500 J/mol (Conversion factor: 1kJ = 1000J)

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

$K_1$ = equilibrium constant at 25°C =?

Inserting values into the above equation yields:

$22500J/mol=-(8.314J/Kmol)\times 298K\times \ln K_1\\\\K_1=1.14\times 10^{-4}$

To determine the equilibrium constant at 35°C, we refer to the equation proposed by Arrhenius, which states:

$\ln(\frac{K_2}{K_1})=\frac{\Delta H}{T}(\frac{1}{T_1}-\frac{1}{T_2})$

where,

$K_2$ = Equilibrium constant at 35°C =?

$K_1$ = Equilibrium constant at 25°C = $1.14\times 10^{-4}$

$\Delta H$ = Enthalpy change of the reaction = 72.1 kJ/mol = 72100 J

R = Gas constant = $8.314J/K mol$

$T_1$ = Initial temperature = $25^oC=[273+25]K=298K$

$T_2$ = Final temperature = $35^oC=[273+35]K=308K$

By plugging values into the equation above, we obtain:

$\ln(\frac{K_2}{1.14\times 10^{-4}})=\frac{72100J/mol}{8.314J/K.mol}(\frac{1}{298}-\frac{1}{308})\\\\K_2=2.906\times 10^{-4}$

In order to calculate the partial pressure of naphthalene at 35°C, we utilize the equation for $K_p$ , which is:

$K_p=\frac{p_{C_{10}H_8(g)}}{p_{C_{10}H_8(g)}}=p_{C_{10}H_8(g)$

The partial pressure of the solid phase is considered to be 1 at equilibrium.

Therefore, the value for $K_2$ will equal $K_p$

$p_{C_{10}H_8}=2.906\times 10^{-4}$

Consequently, the partial pressure of naphthalene at 35°C is $2.906\times 10^{-4}$ atm.

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1 month ago

The two-slit diffraction experiment shows how light can be treated as particles and how light waves carry the statistical inform

alisha [2963]

The double-slit experiment serves as a renowned method to exemplify concepts in quantum mechanics. Specifically, it highlights the idea of wave-particle duality. Employing a light wave shows diffraction and interference, which are typical characteristics of wave behavior. Unexpectedly, using an electron beam produces an interference pattern as well, indicating that electrons can exhibit wave-like properties.

Explanation:

The optical phenomenon would nearly resemble, yet be entirely distinct from, that involved with the exploitation of light. Interference and diffraction are the characteristics distinguishing waves from particles: waves can interfere and disperse, whereas particles cannot.

Light curves around obstacles akin to waves, and this bending results in the single-slit diffraction pattern.

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2 months ago

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lions [2927]

Answer:

The rate law for the decomposition reaction is:

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The unit for the rate constant will be $M^{-1}s^{-1}$

Explanation:

$D\rightarrow Product$

The rate law can be expressed as:

$R=k[D]^x$ ..[1]

When the drug concentration is tripled, the decomposition rate rises by a factor of nine.

$[D]'=3[D]$

$R'=9\times R$

$R'=k[D]'^x$ ...[2]

[1] ÷ [2]

$\frac{R}{R'}=\frac{k[D]^x}{k[D']^x}$

$\frac{R}{9R}=\frac{k[D]^x}{k[3D]^x}$

$9=3^x$

Solving for x results in:

x = 2.

This indicates a second-order reaction.

The decomposition reaction's rate law is:

$R=k[D]^2$

The unit for the rate constant will be:

$k=\frac{R}{[D]^2}=\frac{M/s}{(M)^2}=M^{-1}s^{-1}$

The unit for the rate constant will be $M^{-1}s^{-1}$ .

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