A cook prepare a solution for boiling by adding 12.5g of NaCl to a pot holding 0.750L of
water. At what temperature should the solution in the not boil? *

Answer:

0.458 M

Explanation:

Let's consider the following reaction at equilibrium.

2 NH₃(g) ⇄ N₂(g) + 3 H₂(g)

The concentrations at equilbrium are:

[NH₃] = 0.221 mol / 10.2 L = 0.0217 M

[N₂] = 0.315 mol / 10.2 L = 0.0309 M

[H₂] = ?

We can find the concentration of H₂ at equilibrium using the equilibrium constant (K).

K = [N₂] × [H₂]³ / [NH₃]²

[H₂] = ∛(K × [NH₃]²/ [N₂])

[H₂] = ∛(6.30 × 0.0217²/ 0.0309) = 0.458 M

Answer:

The equilibrium concentration of H2 is 0.4579 M

Explanation:

Step 1: Data given

The equilibrium constant for the following reaction is 6.30 at 723K

Volume = 10.2 L

Number of moles NH3 = 0.221 moles

Number of moles N2 = 0.315 moles

Step 2: The balanced equation

2NH3 <=> N2(g) + 3H2(g)

Step 3: Calculate concentration

Concentration = moles / volume

Concentration NH3= 0.221 moles / 10.2 L

Concentration NH3 = 0.0217 M

Concentration N2 = 0.315 moles / 10.2 L

Concentration N2 = 0.0309 M

Step 4: Define Kc

Kc = [H2]³[N2] / [NH3]²

6.30 = [H2]³ * (0.0309) / (0.0217)²

[H2]³ = 0.09601 M

[H2] = 0.4579 M

The equilibrium concentration of H2 is 0.4579 M

Answer:

C HI = 0.0968 M

Explanation:

∴ a: HI(g)

∴ order (α) = 1

∴ rate constant (K) = 7.21/s

∴ Initial concenttration HI(g) (Cao) = 0.440 M

∴ t = 0.210 s ⇒ Ca = ?

⇒ - δCa/δt = KCa

⇒ - ∫δCa/Ca = K*∫δt

⇒ Ln(Cao/Ca) = K*t

⇒ Cao/Ca = e∧(K*t)

⇒ Cao/e∧(K*t) = Ca

⇒ Ca = (0.440)/e∧((7.21*0,21))

⇒ Ca = 0.440/4.54533

⇒ Ca = 0.0968 M

Answer:

Explanation:

No reaction if there is no solid then the ions just stay there brah. All the ions remain in solution. We are all gonna make it!

Answer:

[tex]\large \boxed{\text{103 kPa}}[/tex]

Explanation:

We can use the Ideal Gas Law — pV = nRT

Data:

V = 66.8 L

m = 77.8 g

T = 25 °C

Calculations:

(a) Moles of N₂

[tex]\text{Moles of N}_{2} = \text{77.8 g N}_{2} \times \dfrac{\text{1 mol N}_{2}}{\text{28.01 g N}_{2}} = \text{2.778 mol N}_{2}[/tex]

(b) Convert the temperature to kelvins

T = (25 + 273.15) K = 298.15 K

(c) Calculate the pressure

[tex]\begin{array}{rcl}pV & =& nRT\\p \times \text{66.8 L} & = & \text{2.778 mol} \times \text{8.314 kPa$\cdot$ L$\cdot$K$^{-1}$mol$^{-1}\times$ 298.15 K}\\66.8p & = & \text{6886 kPa}\\p & = & \textbf{103 kPa}\end{array}\\\text{The pressure in the bag is $\large \boxed{\textbf{103 kPa}}$}[/tex]

Answer: 1) C; 2)D; 3)B; 4)B; 5) A

Explanation:Interpreting the following Arterial Blood gases, we have

1. pH 7.33 PaCO2 60 HCO3  34----Respiratory acidosis with partial compensation----C

2. pH 7.48 PaCO2 42 HCO3 30------. Metabolic alkalosis without compensation----D

3. pH 7.38 PaCO2 38 HCO3 24 ----- Normal---B

4. pH 7.21 PaCO2 60 HCO3 24------ Respiratory acidosis without compensation-----B

5. pH 7.48 PaCO2 28 HCO3 20 ----Respiratory alkalosis with partial compensation

The Arterial blood gas interpretation  from analysis shows  the pH and the partial pressures of oxygen and carbon dioxide in the arterial blood of an individual which can detect how well the lungs are functioning thereby making a physician make a diagnosis, estimate  the severity of a condition and profer treatment.

Higher concentration make a reaction faster due to faster collisions. The correct answer is "There are more collisions per second."

When the concentration of reactants is increased, it leads to a higher number of particles or molecules in the reaction mixture. As a result, there are more frequent collisions between the reactant particles per unit time.

In a chemical reaction, the reactant particles must collide with sufficient energy and proper orientation for a successful reaction to occur. By increasing the concentration, the chances of successful collisions increase because there are more particles available to collide with each other.

However, it's important to note that while a higher concentration increases the frequency of collisions, it does not necessarily guarantee that all collisions will result in a reaction. The energy of the collisions and proper orientation are also essential factors for a successful reaction.

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Answer:

what subject is this and how can u draw the structure :)

The reaction involves reduction of the given compound using NaBH4, followed by solvolysis with ethanol and protonation with H3O. Organic molecules structures are often simplified using the skeletal structure method. The stereochemistry of compounds can be represented using wedge and dash notation.

The given reaction involves the reduction of an organic compound using NaBH4, followed by the treatment with ethanol and then H3O. When a carbonyl compound reacts with NaBH4, the borohydride group of NaBH4 (it is a weak reducing agent with H-) adds to the carbonyl carbon. This process, known as reduction, transforms the carbonyl group to an alcohol. Following reduction, the compound is undergoing solvolysis in ethanol and then protonated with H3O. Due to the restriction of the platform, displaying the structural representation of the organic product is limited.

The wedge and dash notation can be used to represent the stereochemistry of the compounds. Solid wedges represent bonds coming up out of the plane whereas dashed lines represent bonds going into the plane. These techniques assist chemists to define molecular structures in three dimensions. Skeletal structure, a common way of simplifying complex organic structures in chemistry, represents carbons as each end of a line or bend in a line.

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The statement is true as all Ionic compounds are called Electrolytes because they  dissociate(break apart) in water to form a solution that conducts electric  current.

Explanation:

An ionic compound also called an electrolyte. Ionic compounds have capability to get dissociate into water to give ions(cation and anion)  which conduct electric current. This process is called conductivity.

Some insoluble compounds that are not dissolved in water still they form ions which conduct electricity.

The cations and anions released are responsible for carrying current.

Strong acids and strong bases are the strong electrolytes as they dissociate to give more ions as NaOH, HCl.

The cations move to the cathode and anions towards the anode. Cation and anion movement in the solution is electric current.

Ionic compounds are called electrolytes because they dissociate in water to form a solution that conducts electric current.

True. Ionic compounds are called electrolytes because they dissociate (break apart) in water to form a solution that conducts electric current.

When an ionic compound like NaCl is dissolved in water, the positive sodium ions and negative chloride ions separate from each other and are surrounded by water molecules. These separated ions can move freely in the solution and carry electric charge, allowing the solution to conduct electricity.

Other examples of electrolytes include compounds like potassium chloride (KCl) and magnesium sulfate (MgSO4). These compounds also dissociate in water to produce solutions that can conduct electric current.

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Answer:

b. It is when a change that indicates equivalence is observed in the analyte solution.

Explanation:

The end point of a titration is the point at which the indicator undergoes the change noticeable by our senses. Ideally, the equivalence point and end point coincide; but this does not usually happen in practice, because the indicator does not always change perceptibly at the same moment in which the equivalence point is reached and also for the change of the indicator, some of the reagent used in the evaluation is usually necessary .

The difference between the end point and the equivalence point of a valuation is called the valuation error or end point error.

At this point the pH changes the color of the indicator. nvhdurn

Final answer:

The end point of a titration is the moment observed (option b) when a change, due to an indicator or sensor, suggests that stoichiometric equivalence has been reached. It is an experimental determination and serves as the best estimate of the theoretical equivalence point.

Explanation:

The end point of a titration is not a synonym for the equivalence point. Rather, it represents the moment during a titration when a change indicating equivalence is observed in the analyte solution (option b). This point is typically detected through a change in color due to an indicator or through a change detected by a sensor. It is important to note that the end point is an experimental value, our best estimate of the equivalence point, and that any difference between the end point and the theoretical equivalence point is a source of determinate error.

In an acid-base titration, the end point is often observed when a pH indicator changes color, signifying that a stoichiometric amount of titrant has been added to the analyte. In redox and complexation titrations, it can be detected by changes in the solution conditions that are measurable by indicators and sensors.

The enthalpy change (ΔH°) for each reaction can be calculated using the given enthalpy of formation (ΔH°f) values. ΔH°1 = (4ΔH°f(NO) + 6ΔH°f(H2O)) - (4ΔH°f(NH3) + 5ΔH°f(O2)). ΔH°2 = (2ΔH°f(NO2)) - (2ΔH°f(NO) + ΔH°f(O2)). ΔH°3 = (2ΔH°f(HNO3) + ΔH°f(NO)) - (3ΔH°f(NO2) + ΔH°f(H2O)).

In order to determine the enthalpy change (ΔH°) for each reaction, we need to use the given enthalpy of formation (ΔH°f) values. The enthalpy change for each reaction can be calculated by subtracting the sum of the enthalpy of formation of the reactants from the sum of the enthalpy of formation of the products. Using this method, we can calculate the enthalpy change for each of the preceding reactions as follows:

ΔH°1 = (4ΔH°f(NO) + 6ΔH°f(H2O)) - (4ΔH°f(NH3) + 5ΔH°f(O2))

ΔH°2 = (2ΔH°f(NO2)) - (2ΔH°f(NO) + ΔH°f(O2))

ΔH°3 = (2ΔH°f(HNO3) + ΔH°f(NO)) - (3ΔH°f(NO2) + ΔH°f(H2O))

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The ΔH° for each reaction in the Ostwald process is calculated with a specific formula, and the overall ΔH° is found by summing the individual values. The final reaction is 4NH3(g) + 5O2(g) → 4HNO3(aq) with ΔH° = -1192 kJ.

In The Ostwald process, the ΔH° for each reaction is calculated with the formula: ΔH° = ΣΔH°f(products) - ΣΔH°f(reactants). So, for the first reaction, 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g), we calculate ΔH°1 = [4(90) + 6(-242)] - [4(-46) + 5(0)] = -906 kJ. The second reaction, 2NO(g) + O2(g) → 2NO2(g), gives ΔH°2 = [2(34)] - [2(90) + 0] = -112 kJ. Lastly, for 3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g), ΔH°3 = [2(-207) + 90] - [3(34) + -286] = -174 kJ.
Now, adding all three reactions, we get the overall reaction: 4NH3(g) + 5O2(g) → 4HNO3(aq), with the overall ΔH° = -1192 kJ.

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Answer:

0.625 mol

27.5 g

Explanation:

Given data

We can find the moles of dinitrogen monoxide gas using the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 0.500 atm × 30.0 L/0.0821 atm.L/mol.K × 292.2 K

n = 0.625 mol

Then, we can find the mass corresponding to 0.625 moles.

0.625 mol × 44.01 g/mol = 27.5 g

Answer:

0.626 moles of dinitrogen monoxide gas was collected, this is 27.6 grams

Explanation:

Step 1: Data given

Temperature of Dinitrogen monoxide gas = 19.0 °C = 292 K

Volume of the flask = 30.0 L

Pressure in the flask = 0.500 atm

Step 2: Calculate moles of Dinitrogen monoxide gas

p*V = n*R*T

⇒with p = the pressure of the gas = 0.500 atm

⇒with V = the volume of the flask = 30.0 L

⇒with n = the number of moles of N2O gas = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 292 K

n = (p*V) / (R*T)

n = (0.500 * 30.0) / (0.08206 * 292)

n = 0.626 moles

Step 3: Calculate mass N2O gas

Mass N2O gas = moles N2O * molar mass N2O

Mass N2O = 0.626 moles * 44.013 g/mol

Mass N2O = 27.55 grams ≈ 27.6 grams

0.626 moles of dinitrogen monoxide gas was collected, this is 27.6 grams

Answer:

742.2 L

Explanation:

First we must find the number of moles of nitroglycerine reacted.

Molar mass of nitroglycerine= 227.0865 g/mol

Mass of nitroglycerine involved = 1×10^3 g

Number of moles of nitroglycerine= 1×10^3g/227.0865 g/mol

n= 4.40361 moles

T= 1985°C + 273= 2258K

P= 1.100atm

R= 0.082atmLmol-1K-1

Using the ideal gas equation:

PV= nRT

V= nRT/P

V= 4.40361× 0.082× 2258/1.1

V= 742 L

Considering the reaction stoichiometry and ideal gas law, the volume of gas produced is 5375.626 L.

The balanced reaction is:

4 C₃H₅N₃O₉(s) → 12 CO₂(g) + 10 H₂O(g) + 6 N₂(g) + O₂(g)

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Then 4 moles of nitroglycerine C₃H₅N₃O₉(s) produce in total 29 moles of gas [12 moles of CO₂(g) + 10 moles of H₂O(g) + 6 moles of N₂(g) + 1 mole O₂(g)]

Being the molar mass of nitroglycerine C₃H₅N₃O₉(s) 227 g/mole, then the amount of moles that 1 kg (1000 g) of the compound contains can be calculated as:

[tex]1000 gramsx\frac{1 mole}{227 grams}= 4.405 moles[/tex]

Then you can apply the following rule of three: if by stoichiometry 4 moles of nitroglycerine C₃H₅N₃O₉(s) produce in total 29 moles of gas, 4.405 moles of C₃H₅N₃O₉(s) will produce how many moles of gas?

[tex]amount of moles of gas=\frac{4.405 moles of nitroglycerinex29 moles of gas}{4 moles of nitroglycerine}[/tex]

amount of moles of gas= 31.93625 moles

On the other hand, an ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P× V = n× R× T

In this case, you know:

Replacing:

1.1 atm× V= 31.93625 moles× 0.082 [tex]\frac{atmL}{molK}[/tex]× 2258 K

Solving:

V= (31.93625 moles× 0.082 [tex]\frac{atmL}{molK}[/tex]× 2258 K) ÷ 1.1 atm

V= 5375.625 L

Finally, the volume of gas produced is 5375.626 L.

Learn more:

Answer:

[tex]T = 34.493\,^{\textdegree}C[/tex]

Explanation:

The equilibrium temperature of the gas-container system is:

[tex]Q_{Cu} = -Q_{g}[/tex]

[tex](0.3\,kg)\cdot\left(386\,\frac{J}{kg\cdot ^{\textdegree}C} \right)\cdot (T-20\,^{\textdegree}C) = (1.5\,mol)\cdot \left(12.5\,\frac{J}{mol\cdot ^{\textdegree}C} \right)\cdot (124\,^{\textdegree}C-T)[/tex]

[tex]\left(115.8\,\frac{J}{^{\textdegree}C} \right)\cdot (T-20\,^{\textdegree}C) = \left(18.75\,\frac{J}{^{\textdegree}C} \right)\cdot (124\,^{\textdegree}C-T)[/tex]

[tex]134.55\cdot T = 4641[/tex]

[tex]T = 34.493\,^{\textdegree}C[/tex]

Answer:

pH  = 13.09

Explanation:

Zn(OH)2 --> Zn+2 + 2OH-   Ksp = 3X10^-15

Zn+2 + 4OH-   --> Zn(OH)4-2   Kf = 2X10^15

K = Ksp X Kf

  = 3*2*10^-15 * 10^15

  = 6

Concentration of OH⁻ = 2[Ba(OH)₂] = 2 * 0.15 = 3 M

                Zn(OH)₂ + 2OH⁻(aq)  --> Zn(OH)₄²⁻(aq)

Initial:           0             0.3                      0

Change:                      -2x                     +x

Equilibrium:               0.3 - 2x                 x

K = Zn(OH)₄²⁻/[OH⁻]²

6 = x/(0.3 - 2x)²  

6 = x/(0.3 -2x)(0.3 -2x)

6(0.09 -1.2x + 4x²) = x

0.54 - 7.2x + 24x² = x

24x² - 8.2x + 0.54 = 0

Upon solving as quadratic equation, we obtain;

x = 0.089

Therefore,

Concentration of (OH⁻) = 0.3 - 2x

                                    = 0.3 -(2*0.089)

                                  = 0.122

pOH = -log[OH⁻]

         = -log 0.122

          = 0.91

pH = 14-0.91

     = 13.09

Answer: a) 0.070 moles of oxygen were produced.

b) New pressure due to the oxygen gas is 2.4 atm

Explanation:

According to ideal gas equation:

[tex]PV=nRT[/tex]

P = pressure of gas = 2.7 atm

V = Volume of gas = 700 ml = 0.7 L

n = number of moles = ?

R = gas constant =[tex]0.0821Latm/Kmol[/tex]

T =temperature = 329 K

[tex]n=\frac{PV}{RT}[/tex]

[tex]n=\frac{2.7atm\times 0.7L}{0.0821 L atm/K mol\times 329K}=0.070moles[/tex]

Thus 0.070 moles of oxygen were produced.

When the 700 mL flask cools to a temperature of 293K.

[tex]PV=nRT[/tex]

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

[tex]P=\frac{0.070\times 0.0821\times 293}{0.7}[/tex]

[tex]P=2.4atm[/tex]

The new pressure due to the oxygen gas is 2.4 atm

The most likely substrates for the oligo-(α1→4)-(α1→4) glucanotransferase catalytic center of glycogen debranching enzyme are maltosyl and maltotriosyl residues.

The glycogen debranching enzyme has two activities; it acts as a glucosidase and a glucanotransferase. The (α‑6) glucosidase activity hydrolyzes α-1,6 glycosidic bonds, whereas the oligo-(α1→4)-(α1→4) glucanotransferase activity shifts α-1,4-linked glucose chains from one branch to another, often creating a substrate that the glucosidase can act upon. Based on the mechanism of glycogenolysis, the enzyme glycogen phosphorylase releases glucose units from the linear chain until a few are left near the branching point, and it is here that the glucan transferase action is relevant. The glucan transferase shifts the remaining α-1,4 linked glucose units, leaving a single α-1,6 linked glucose that the glucosidase can then release. Hence, the most likely substrates for the oligo-(α1→4)-(α1→4) glucanotransferase catalytic center are maltosyl and maltotriosyl residues, as these are the short α-1,4 linked glucose chains that are left after phosphorylase action.

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Answer:

celulares por que tienen una fuente de energiq

The final pressure of the ammonia gas, given that the the volume decreased by 50.0%, is 11 atm

How to calculate the final pressure f the ammonia gas?

To solve this question in the most simplified way, we shall begin by listing out the given data from the question. This is shown below:

The final pressure of the ammonia gas can be calculated as shown below:

[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\\\\\frac{4.6\ \times\ V}{252} = \frac{P_2\ \times\ 0.5V}{296}\\\\252\ \times\ P_2\ \times\ 0.5V = 4.6\ \times\ V\ \times\ 296\\\\P_2 = \frac{4.6\ \times\ V\ \times\ 296}{252\ \times\ 0.5V} \\\\P_2 = 11\ atm[/tex]

Thus, the final pressure is 11 atm

Answer:

See explanation below

Explanation:

First, we need to write the overall reaction which is:

CH₃COOH + NaOH <---------> CH₃COONa + H₂O

The acetic acid, is a weak acid, so it has a acid constant (Ka) which is 1.75x10⁻⁵. Now that we know this, we can solve the problem by parts:

a) Initial pH:

In this case, the base is not added yet, so the only thing we have is the acetic acid in solution. So we'll do an ICE chart for the dissociation of the acetic acid:

       CH₃COOH + H₂O <-------> CH₃COO⁻ + H₃O⁺     Ka  = 1.75x10⁻⁵

i)        0.105                                      0                0

e)       0.105 - x                                 x                x

Writting the Ka expression:    Ka = [CH₃COO⁻] [H₃O⁺] / [CH₃COOH]  replacing:

1.75x10⁻⁵ = x² / 0.105 - x

As Ka is a very small value, we can assume that x will be very small too, so we can assume that 0.105 - x = 0.105:

1.75x10⁻⁵ = x² / 0.105

1.75x10⁻⁵ * 0.105 = x²

x = [H₃O⁺] = 1.35x10⁻³ M

Then, the pH:

pH = -log[H₃O⁺]

pH = -log(1.35x10⁻³)

pH = 2.87

b) volume to reach the equivalence point

In this case, we use the following expression:

M₁V₁ = M₂V₂

And solve for the volume:

V₂ = 0.105 * 24 / 0.13

V₂ = 19.38 mL of NaOH

c) pH with 6 mL of added base

In this case we need to see the moles of each substance and then see how many moles of the acid remain:

moles Acid = 0.105 * 0.024 = 2.52x10⁻³  moles

moles base = 0.130 * 0.006 = 7.8x10⁻⁴ moles

remaining moles of acid: 2.52x10⁻³ - 7.8x10⁻⁴ = 1.74x10⁻³ moles

With these moles, we can calculate the concentration assuming the volume is the sum of the 24 mL and the 6 mL of added base (30 mL)

M = 1.74x10⁻³ / 0.030 = 0.058 M

Now we will do the same thing we did in part a) with an ICE chart to calculate the H₃O⁺ concentration and then, the pH.

      CH₃COOH + H₂O <-------> CH₃COO⁻ + H₃O⁺     Ka  = 1.75x10⁻⁵

i)        0.058                                      0                0

e)       0.058- x                                  x                x

1.75x10⁻⁵ = x² / 0.058

1.75x10⁻⁵ * 0.058 = x²

x = 1.01x10⁻³ M

pH = -log(1.01x10⁻³)

pH = 3

d) at one half equivalence point

In this case, if the volume to reach equivalence is 19.38 mL, the half would be 9.69 mL, and now we will do the same thing as part c) but with these data:

moles Acid = 0.105 * 0.024 = 2.52x10⁻³  moles

moles base = 0.130 * 0.00969 = 1.26x10⁻³ moles

remaining moles of acid: 2.52x10⁻³ - 1.26x10⁻³ = 1.26x10⁻³ moles

With these moles, we can calculate the concentration assuming the volume is the sum of the 24 mL and the 9.69 mL of added base (33.69 mL)

M = 1.26x10⁻³ / 0.03369 = 0.037 M

          CH₃COOH + H₂O <-------> CH₃COO⁻ + H₃O⁺     Ka  = 1.75x10⁻⁵

i)        0.037                                      0                0

e)       0.037- x                                  x                x

1.75x10⁻⁵ = x² / 0.037

1.75x10⁻⁵ * 0.037 = x²

x = 8.05x10⁻⁴ M

pH = -log(8.05x10⁻⁴)

pH = 3.09

e) pH at equivalence point.

In this case the moles of the acid and the base are the same, therefore the pH would be higher than 7. so in this case we need to use the following expression:

pH = 7 + 1/2 pKa + 1/2log[CH₃COOH]

In this case the concentration of the acid in the equivalence point would be:

[C] = 2.52x10⁻³ / 0.04338 = 0.058 M

Applying the above expression:

pH = 7 + 1/2(-log(1.75x10⁻⁵) + 1/2log(0.058)

pH = 8.76

(a) The initial pH is approximately 2.87.

(b) The volume of added base required to reach the equivalence point is 18.46 mL.

(c) The pH at 6.00 mL of added base is approximately 4.74.

(d) The pH at one-half of the equivalence point is approximately 3.91.

(e) The pH at the equivalence point is approximately 8.39.

In the titration of acetic acid (CH3COOH) with sodium hydroxide (NaOH), the initial pH is determined by the concentration of the acetic acid before any base is added. Acetic acid is a weak acid, and its ionization in water results in the formation of hydronium ions (H3O+). The initial pH is calculated using the formula for weak acid dissociation, and for a 0.105 M solution of acetic acid, the initial pH is approximately 2.87.

The volume of added base required to reach the equivalence point is determined by stoichiometry, where the moles of base added equal the moles of acid initially present. Using the molarity and volume of the acid solution and the molarity of the base solution, the volume at the equivalence point is found to be 18.46 mL.

At 6.00 mL of added base, the pH is calculated by considering the remaining excess acid and the newly formed acetate ion. This involves solving the Henderson-Hasselbalch equation for a buffer system, resulting in a pH of approximately 4.74.

At one-half of the equivalence point, the pH is determined by finding the concentration of acetic acid and acetate ion in the solution. This is calculated using the initial concentration of acetic acid, the moles of base added, and the stoichiometry of the reaction. The pH at one-half of the equivalence point is approximately 3.91.

At the equivalence point, all the acetic acid has reacted with the sodium hydroxide, resulting in the formation of sodium acetate and water. The pH at the equivalence point is determined by the hydrolysis of the acetate ion. This results in a pH of approximately 8.39, indicating a basic solution due to the presence of excess hydroxide ions from the reaction.

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Answer:

Explanation:

Find attach the solution

The molar concentration of sodium ions in a 0.650M Na₃PO₄ solution is equal to 1.95M.

The concentration of the solution can be determined when we have the molecular formula and its molecular weight. We can easily determine the molar concentration of a substance in a solution.

The concentration of the solution calculated as the number of moles of a solute in a liter of a solution is called molarity or molar concentration.

The Molar concentration of the solution is determined in the following way.

Molarity (M) = Moles (n) of solute/Volume of the Solution ( in L)

Given, the molar concentration of the Na₃PO₄ solution = 0.650 M

The dissociation of Na₃PO₄ into ions can be represented as:

Na₃PO₄   →   3 Na⁺   +   PO₄³⁻

From the above equation, we can say that one mole of the  Na₃PO₄  gives 3 moles of sodium ions.

The molar concentration of sodium ions in  Na₃PO₄ solution = 3× 0.650

The molar concentration of Na⁺ions = 1.95M

Therefore, the molar concentration of Na⁺ions in Na₃PO₄ solution is 1.95M.

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n = 0.0989 moles

n = PV / RT

P = 2.09atm

V = 1.13L

R = 0.08206

T = 291K

Plug the numbers in the equation.

n = (2.09atm)(1.13L) / (0.08206)(291K)

n = 0.0989 moles

Final answer:

Using the ideal gas law PV = nRT, and rearranging for n (moles), we plug in the given values of pressure, volume, and temperature alongside the ideal gas constant to calculate the number of moles of gas present. There are approximately 0.0988 moles of gas present in 1.13 L of gas at [tex]\(2.09 \, \text{atm}\) and \(291 \, \text{K}\).[/tex]

Explanation:

To calculate the number of moles of gas present, we can use the ideal gas law equation:

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

First, we need to convert the given pressure to atm and the volume to liters if they are not already in those units.

Given:

- [tex]\( P = 2.09 \, \text{atm} \)[/tex]

-[tex]\( V = 1.13 \, \text{L} \[/tex]

- [tex]\( T = 291 \, \text{K} \)[/tex]

Now, we can rearrange the ideal gas law equation to solve for \( n \):

[tex]\[ n = \frac{PV}{RT} \][/tex]

[tex]\[ n = \frac{(2.09 \, \text{atm})(1.13 \, \text{L})}{(0.0821 \, \text{atm} \cdot \text{L} / \text{mol} \cdot \text{K})(291 \, \text{K})} \][/tex]

[tex]\[ n = \frac{2.3617}{23.9211} \][/tex]

[tex]\[ n \approx 0.0988 \, \text{mol} \][/tex]

Therefore, There are approximately 0.0988 moles of gas present in 1.13 L of gas at [tex]\(2.09 \, \text{atm}\) and \(291 \, \text{K}\).[/tex]

Answer:

15.1 seconds is the half life of the reaction when concentration of the substrate is 2.77 M.

Explanation:

A → B + C

The rate law of the reaction will be :

[tex]R=k[A]^x[/tex]

Initial rate of the reaction when concentration of the substrate was 0.4 M:

[tex]0.183 M/s=k[0.4 M]^x[/tex]..[1]

Initial rate of the reaction when concentration of the substrate was 0.8 M:

[tex]0.183 M/s=k[0.8 M]^x[/tex]...[2]

[1] ÷ [2] :

[tex]\frac{0.183 M/s}{0.183 M/s}=\frac{k[0.4 M]^x}{k[0.8 M]^x}[/tex]

x = 0

The order of the reaction is zero.

For the value of rate constant ,k:

[tex]0.183 M/s=k[0.4 M]^x[/tex]..[1]

x = 0

[tex]0.183 M/s=k[0.4 M]^0[/tex]

k= 0.183 M/s

The half life of the zero order kinetics is given by :

[tex]t_{1/2}=\frac{[A_o]}{2k}[/tex]

Where:

[tex][A_o][/tex] = Initial concentration of A

k = Rate constant of the reaction

So, the half-life for the decomposition of substrate 1 when the initial concentration of the substrate is 2.77 M:

[tex]t_{1/2}=\frac{2.77 M}{0.183 M/s}=15.1 s[/tex]

15.1 seconds is the half life of the reaction when concentration of the substrate is 2.77 M.

The half-life for the decomposition of substrate 1 when the initial concentration is 2.77 M is approximately 18 minutes.

The half-life of a reaction is the time required for one-half of a given amount of reactant to be consumed. In this case, we have initial rate data for the decomposition of substrate 1 with varying concentrations. To determine the half-life of substrate 1, we need to find the time it takes for the concentration to decrease to half its initial value.

Based on the given data, we can see that the initial rate of the reaction is constant at 0.183 M/s for different initial concentrations of substrate 1 (0.4 M, 0.8 M, and 2 M). Since the initial concentration of substrate 1 is given as 2.77 M, it will take the same amount of time as the other initial concentrations to reach half its initial concentration.

Therefore, the half-life for the decomposition of substrate 1 when the initial concentration of the substrate is 2.77 M is the same as the half-life observed when the initial concentration is 0.4 M, 0.8 M, or 2 M, which is approximately 18 minutes.

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Answer: The molarity of silver nitrate in the diluted solution is 0.015 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

According to the dilution law,

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1[/tex] = molarity of concentrated solution = 0.150 M

[tex]V_1[/tex] = volume of concentrated solution = 10.0 ml

[tex]M_2[/tex] = molarity of diluted solution = ?

[tex]V_2[/tex] = volume of diluted solution = 100.0 ml

[tex]0.150\times 10.0=M_2\times 100.0[/tex]

[tex]M_2=0.015M[/tex]

Thus the molarity of silver nitrate in the diluted solution is 0.015 M

Final answer:

The molarity of the diluted silver nitrate solution, prepared by diluting 10.0 mL of a 0.150 M solution to a final volume of 100.0 mL, is 0.015 M when calculated using the dilution equation.

Explanation:

The question is asking how to calculate the molarity of a diluted silver nitrate solution using the dilution equation. When 10.0 mL of a 0.150 M silver nitrate solution is diluted to 100.0 mL, the molarity of the diluted solution can be found using the equation M1V1 = M2V2, where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.

Using the dilution equation:

To find M2 (the molarity of the diluted solution), rearrange the equation to M2 = (M1V1) / V2.

The molarity of the diluted silver nitrate solution would be 0.015 M (calculated as (0.150 M * 10.0 mL) / 100.0 mL).

Answer:

It forms an octahedral complex

Rh[NH3]5I

Explanation:

Or see structure attached below

Answer:

[ Rhl(NH₃)₅]²⁺

Explanation:

Symbol of Rhodium atom = Rh

charge of Rhodium in complex = +3

charge on Iodine in complex = -1

Charge on NH₃= 0 (neutral)

Therefore, the total charge on the complex is given as follows;

Charge on complex= +3-1 + 0

                                        = +2

Therefore, the required formula of the complex = [ Rhl(NH₃)₅]²⁺

[Note: Inside the co-ordination complex,metal is written first and followed by negative and neutral ligands respectively and charge on complex is written on the right top of the complex. therefore, the chemical formular of the given complex is given as follows; [ Rhl(NH₃)₅]²⁺

Answer:

Approximately [tex]8.3 \times 10^{-2}\; \rm mol \cdot L^{-1}[/tex].

Explanation:

The [tex]K_a[/tex] in this question refers the dissociation equilibrium of [tex]\rm HC_6H_4ClO[/tex] as an acid:

[tex]\rm HC_6H_4ClO\, (aq) \rightleftharpoons H^{+} \, (aq) + C_6H_4ClO^{-}\, (aq)[/tex].

[tex]\displaystyle K_a\left(\mathrm{HC_6H_4ClO}\right) = \frac{\left[\mathrm{H^{+}}\right] \cdot \left[\mathrm{C_6H_4ClO^{-}}\right]}{\left[\mathrm{HC_6H_4ClO}\right]}[/tex].

However, the question also states that the solution here has a [tex]\rm pH[/tex] of [tex]11.05[/tex], which means that this solution is basic. In basic solutions at [tex]\rm 25\;^\circ C[/tex], the concentration of [tex]\rm H^{+}[/tex] ions is considerably small (typically less than [tex]10^{-7}\;\rm mol \cdot L^{-1}[/tex].) Therefore, it is likely not very appropriate to use an equilibrium involving the concentration of [tex]\rm H^{+}[/tex] ions.

Here's the workaround: note that [tex]\rm C_6H_4ClO^{-}\, (aq)[/tex] is the conjugate base of the weak acid [tex]\rm HC_6H_4ClO\, (aq)[/tex]. Therefore, when [tex]\rm C_6H_4ClO^{-}\, (aq)[/tex] dissociates in water as a base, its [tex]K_b[/tex] would be equal to [tex]\displaystyle \frac{K_w}{K_a} \approx \frac{10^{-14}}{K_a}[/tex]. ([tex]K_w[/tex] is the self-ionization constant of water. [tex]K_w \approx 10^{-14}[/tex] at [tex]\rm 25\;^\circ C[/tex].)

In other words,

[tex]\begin{aligned} & K_b\left(\mathrm{C_6H_4ClO^{-}}\right) \\ &= \frac{K_w}{K_a\left(\mathrm{HC_6H_4ClO}\right)} \\ &\approx \frac{10^{-14}}{6.6 \times 10^{-10}} \\ & \approx 1.51515 \times 10^{-5}\end{aligned}[/tex].

And that [tex]K_b[/tex] value corresponds to the equilibrium:

[tex]\rm C_6H_4ClO^{-}\, (aq) + H_2O\, (l) \rightleftharpoons HC_6H_4ClO\, (aq) + OH^{-}\, (aq)[/tex].

[tex]\displaystyle K_b\left(\mathrm{C_6H_4ClO^{-}}\right) = \frac{\left[\mathrm{HC_6H_4ClO}\right]\cdot \left[\mathrm{OH^{-}}\right]}{\left[\mathrm{C_6H_4ClO^{-}}\right]}[/tex].

The value of [tex]K_b[/tex] has already been found.  

The [tex]\rm OH^{-}[/tex] concentration of this solution can be found from its [tex]\rm pH[/tex] value:

[tex]\begin{aligned}& \left[\mathrm{OH^{-}}\right] \\ &= \frac{K_w}{\left[\mathrm{H}^{+}\right]} \\ & = \frac{K_w}{10^{-\mathrm{pH}}} \\ &\approx \frac{10^{-14}}{10^{-11.05}} \\ &\approx 1.1220 \times 10^{-3}\; \rm mol\cdot L^{-1} \end{aligned}[/tex].

To determine the concentration of [tex]\left[\mathrm{HC_6H_4ClO}\right][/tex], consider the following table:

[tex]\begin{array}{cccccc}\textbf{R} &\rm C_6H_4ClO^{-}\, (aq) & \rm + H_2O\, (l) \rightleftharpoons & \rm HC_6H_4ClO\, (aq) & + & \rm OH^{-}\, (aq) \\ \textbf{I} & (?) & \\ \textbf{C} & -x & & + x& & +x \\ \textbf{E} & (?) - x & & x & & x\end{array}[/tex]

Before hydrolysis, the concentration of both [tex]\mathrm{HC_6H_4ClO}[/tex] and [tex]\rm OH^{-}[/tex] are approximately zero. Refer to the chemical equation. The coefficient of [tex]\mathrm{HC_6H_4ClO}[/tex] and [tex]\mathrm{HC_6H_4ClO}[/tex] are the same. As a result, this equilibrium will produce [tex]\rm OH^{-}[/tex] and [tex]\mathrm{HC_6H_4ClO}[/tex] at the exact same rate. Therefore, at equilibrium, [tex]\left[\mathrm{HC_6H_4ClO}\right] \approx \left[\mathrm{OH^{-}}\right] \approx 1.1220 \times 10^{-3}\; \rm mol\cdot L^{-1}[/tex].

Calculate the equilibrium concentration of [tex]\left[\mathrm{C_6H_4ClO^{-}}\right][/tex] from [tex]K_b\left(\mathrm{C_6H_4ClO^{-}}\right)[/tex]:

[tex]\begin{aligned} & \left[\mathrm{C_6H_4ClO^{-}}\right] \\ &= \frac{\left[\mathrm{HC_6H_4ClO}\right]\cdot \left[\mathrm{OH^{-}}\right]}{K_b}\\&\approx \frac{\left(1.1220 \times 10^{-3}\right) \times \left(1.1220 \times 10^{-3}\right)}{1.51515\times 10^{-5}}\; \rm mol \cdot L^{-1} \\ &\approx 8.3 \times 10^{-2}\; \rm mol \cdot L^{-1}\end{aligned}[/tex].

Answer:

The phosphorylation of glucose to glucose-6-phosphate is endergonic reaction that is coupled to the exergonic hydrolysis of ATP.

Explanation:

In glycosis, the first reaction that takes place is the phosphorylation of glucose to glucose-6-phosphate by the enzyme hexokinase. This is an exergenic reaction. This is a coupled reaction in which phosphorylation of glucose is coupled to ATP hydrolysis. The free energy of ATP hydrolysis fuels glucose phosphorylation.

Answer:

The specific heat of copper is  [tex]C= 392 J/kg\cdot ^o K[/tex]

Explanation:

From the question we are told that

The amount of energy contributed by each oscillating lattice site  is  [tex]E =3 kT[/tex]

       The atomic mass of copper  is  [tex]M = 63.6 g/mol[/tex]

        The atomic mass of aluminum is  [tex]m_a = 27.0g/mol[/tex]

        The specific heat of aluminum is  [tex]c_a = 900 J/kg-K[/tex]

 The objective of this solution is to obtain the specific heat of copper

       Now specific heat can be  defined as the heat required to raise the temperature of  1 kg of a substance by  [tex]1 ^o K[/tex]

  The general equation for specific heat is  

                    [tex]C = \frac{dU}{dT}[/tex]

Where [tex]dT[/tex] is the change in temperature

             [tex]dU[/tex] is the change in internal energy

The internal energy is mathematically evaluated as

                       [tex]U = 3nk_BT[/tex]

      Where  [tex]k_B[/tex] is the Boltzmann constant with a value of [tex]1.38*10^{-23} kg \cdot m^2 /s^2 \cdot ^o K[/tex]

                    T is the room temperature

                      n is the number of atoms in a substance

Generally number of  atoms in mass of an element can be obtained using the mathematical operation

                      [tex]n = \frac{m}{M} * N_A[/tex]

Where [tex]N_A[/tex] is the Avogadro's number with a constant value of  [tex]6.022*10^{23} / mol[/tex]

          M is the atomic mass of the element

           m actual mass of the element

  So the number of atoms in 1 kg of copper is evaluated as  

             [tex]m = 1 kg = 1 kg * \frac{10000 g}{1kg } = 1000g[/tex]

The number of atom is  

                       [tex]n = \frac{1000}{63.6} * (6.0*0^{23})[/tex]

                          [tex]= 9.46*10^{24} \ atoms[/tex]

Now substituting the equation for internal energy into the equation for specific heat

          [tex]C = \frac{d}{dT} (3 n k_B T)[/tex]

              [tex]=3nk_B[/tex]

Substituting values

         [tex]C = 3 (9.46*10^{24} )(1.38 *10^{-23})[/tex]

            [tex]C= 392 J/kg\cdot ^o K[/tex]

Answer:

986.9 °C

Explanation:

To find the temperature we can use the Arrhenius equation:

[tex] k = Ae^{(-E_{a}/RT)} [/tex]   (1)

Where:

k: is the reaction rate coefficient

A: is the pre-exponential factor

Ea: is the activation energy

R: is the gas constant

T: is the absolute temperature

With the values given:

T = 590 °C = 863 K; Ea = 117 kJ/mol; k = 0.00289 s⁻¹,

we can find the pre-exponential factor, A:

[tex]A = \frac{k}{e^{(-E_{a}/RT)}} = \frac{0.00289 s^{-1}}{e^{(-117 \cdot 10^{3} J*mol^{-1}/(8.314 J*K^{-1}*mol^{-1}*863 K))}} = 3.49 \cdot 10^{4} s^{-1}[/tex]

Now, we can find the temperature by solving equation (1) for T:

[tex] ln(\frac{k}{A}) = -\frac{E_{a}}{RT} [/tex]

[tex] T = -\frac{E_{a}}{R*ln(\frac{k}{A})} = -\frac{117 \cdot 10^{3} J*mol^{-1}}{8.314 J*K^{-1}*mol^{-1}*ln(\frac{0.492 s^{-1}}{3.49 \cdot 10^{4} s^{-1}})} = 1259.9 K = 986.9 ^{\circ} C [/tex]

Therefore, at 986.9 °C, the rate constant will be 0.492 s⁻¹.

I hope it helps you!  

Answer:

The temperature is 1259.4 K (986.4°C)

Explanation:

Step 1: Data given

The activation energy Ea, = 117 kJ/mol

the rate constant for the reaction is 0.00289 s −1 at 590 °C

The new rate constant = 0.492 s −1

Step 2: Calculate the temperature

ln(kX / k 590°C)  =  Ea/R * (1/T1 - 1/T2)

⇒with kX is the rate constant at the new temperature = 0.492 /s

⇒with k 590 °C = the rate constant at 590 °C = 0.00289 /s

⇒with Ea = the activation energy = 117000 J/mol

⇒with R = the gas constant = 8.314 J/mol*K

⇒with T1 = 590 °C = 863 K

⇒ with T2 = the new temperature

ln (0.492 / 0.00289) = 117000/8.314 *(1/863 - 1/T2)

5.137 =14072.6 * (1/863 - 1/T2)

3.65*10^-4 = (1/863 - 1/T2)

3.65*10^-4 = 0.001159 - 1/T2

0,000794‬ = 1/T2

T2 = 1259.4 K

The temperature is 1259.4 K or 986.4 °C