In a reduction half-reaction, electrons appear on the left side.
This is because reduction is a process that involves the gain of electrons. In other words, the species that is being reduced is gaining electrons and thus they need to appear on the side where this species is located.
A reduction half-reaction is a chemical reaction that describes the reduction of a species by the gain of electrons. It is typically written in the form of a chemical equation that shows the species being reduced along with the electrons that are being gained.
For example, the reduction half-reaction for the reduction of copper (II) ion to copper (I) ion can be written as follows:Cu2+ + e- → Cu+In this reaction, the copper (II) ion is being reduced to copper (I) ion by the gain of one electron. The electron is represented on the left-hand side of the equation.
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Write down the following:
1) the balanced molecular equation,
2) complete ionic equation and
3) net ionic equation
for complete neutralization of H2SO4 by KOH in aqueous solution.
The balanced molecular equation for the complete neutralization of H2SO4 by KOH in aqueous solution is H2SO4 + 2KOH → K2SO4 + 2H2O.
The balanced molecular equation:
H2SO4 + 2 KOH -> K2SO4 + 2 H2O
The complete ionic equation:
2 H+ + SO4^2- + 2 K+ + 2 OH- -> K2SO4 + 2 H2O The net ionic equation:
2 H+ + 2 OH- -> 2 H2O
The net ionic equation only includes the ions that actively participate in the reaction, excluding the spectator ions. In this case, the net ionic equation is
2H⁺ + 2OH⁻ → 2H₂OIn the net ionic equation, the sulfate ion (SO₄²⁻) and potassium ion (K⁺) are spectator ions and do not actively participate in the neutralization reaction.
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Consider the reversible reaction H2↔2H. Equilibrium considerations for this reaction provide insight into how much dissociation of molecular hydrogen (H2) into free hydrogen atoms (H) occurs at different temperatures and pressures.
a. Express the partial-pressure equilibrium constant Kp(T) for this equilibrium reaction in terms of the temperature and the Gibbs functions of formation of H2 and of H.
b. Develop a set of equations that can be used to solve for the equilibrium mole fractions of H2 and H in a system that contains only H2 molecules and H atoms at a specified temperature and pressure. Your result should be expressed in terms of the partial-pressure equilibrium constant Kp(T) for the given reversible reaction, and the mixture pressure P.
c. Will the dissociation of H2 molecules into free hydrogen atoms increase, decrease, or remain the same with increasing temperature, for a fixed pressure?
d. Will the dissociation of H2 molecules into free hydrogen atoms increase, decrease, or remain the same with increasing pressure, for a fixed temperature?
a. The fractional strain balance steady Kp(T) for this balance response can be communicated as far as the temperature and the Gibbs elements of development of H2 and of H as follows.
[tex]Kp(T) = exp(- ΔG°/RT)\\[/tex]
b. The balance mole parts of H2 and H in a framework that contains just H2 particles and H iotas at a predefined temperature and strain can be tackled utilizing the accompanying arrangement of conditions.
[tex]Kp(T) = PH^2/P(H2)[/tex]
[tex]PH2 + PH = P[/tex]
c. The separation of H2 particles into free hydrogen iotas will increment with expanding temperature for a decent tension in light of the fact that the worth of Kp(T) increments with expanding temperature.
d. For a fixed temperature, the dissociation of H2 molecules into free hydrogen atoms will increase with pressure because the value of Kp(T) decreases with pressure.
The physical concept of temperature expresses how hot or cold something is in numerical form. To measure temperature, use a thermometer. Thermometers are calibrated using several temperature scales that historically specified unique reference points and thermometric materials.
The most widely used scales are the Kelvin scale (K), which is mostly used for scientific purposes, the Fahrenheit scale (°F), and the Celsius scale, sometimes referred to as centigrade and denoted by the unit sign °C. The kelvin is one of the seven fundamental units that make up the SI. Absolute zero, often known as zero kelvin, or 273.15 °C, is the lowest temperature on the thermodynamic temperature scale.
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A 100 milligram (mg) sample of I-131 is used for a diagnostic procedure. The half-life of I-131 is 8.0 days. After 24 days have passed, how much of the I-131 remains? Assume no I-131 is lost due to metabolism or excretions.
The radioisotope iodine-131 (I-131) is used as a diagnostic tool in medicine. A 100 milligram sample of I-131 was taken for the diagnostic procedure.
The half-life of I-131 is 8.0 days. Given that the half-life of I-131 is 8.0 days and that after 24 days, we need to determine how much I-131 remains. The time elapsed is 24 days, or three half-lives. The following formula can be used to calculate the amount of radioactive substance remaining after a given number of half-lives: Amount remaining = initial amount × (1/2)number of half-lives. Substituting the values given:Amount remaining = 100 mg × (1/2)3
Amount remaining = 12.5 mg
Therefore, the amount of I-131 that remains is 12.5 mg.
After 24 days, the amount of I-131 that remains is 12.5 mg. I-131 has a half-life of 8.0 days. After 24 days, the elapsed time is three half-lives. The amount of the substance remaining is calculated using the formula Amount remaining = initial amount × (1/2)number of half-lives. The value of the initial amount is given as 100 mg. Substituting the values, we get Amount remaining = 100 mg × (1/2)3
= 12.5 mg.
Therefore, the amount of I-131 that remains after 24 days is 12.5 mg. This calculation assumes that no I-131 is lost due to metabolism or excretion.
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The rote constant of a certain reaction is konown to obcy the Arrtiersus equation, and to have an activation energy E
δ
=50.0 blanol, if the rate constant of this reaction is 37.M
−1
⋅S
−1
at. 78.0
∘
C, what will the rate constant be at 12.0
∘
C ? Round your answer to 2 sagnificant digts.
the rate constant at 12.0°C, we can use the Arrhenius equation:
Given:E = 50.0 kJ/mol,T1 = 78.0°C = 78.0 + 273.15 K,T2 = 12.0°C = 12.0 + 273.15 K
First, let's convert the activation energy from kJ/mol to J/mol:
E = 50.0 kJ/mol * 1000 J/1 kJ = 50000 J/mol Given that k1 = 37 M^-1 * s^-1, we can substitute the value of k1
Calculating this value, we find:k2 = 4.59 x 10^-12 M^-1 * s^-1Therefore, the rate constant at 12.0°C is approximately 4.59 x 10^-12 M^-1 * s^-1.
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The rate constant at 12.0°C is approximately [tex]5.38 M^-1 * s^-1[/tex]
The Arrhenius equation relates the rate constant of a reaction to its activation energy and temperature. It is given by k = A * exp(-Ea/RT), where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.
In this question, we are given that the rate constant at 78.0°C is [tex]37 M^-1 * s^-1.[/tex] We need to find the rate constant at 12.0°C.
To solve this, we need to convert the temperatures to Kelvin. Adding 273 to 78.0 gives us 351 K, and adding 273 to 12.0 gives us 285 K.
Now we can use the Arrhenius equation to find the rate constant at 12.0°C. We know the rate constant at 78.0°C, so we can rearrange the equation to solve for A: k1/k2 = exp((Ea/R)(1/T2 - 1/T1)). Plugging in the values, we have:
37 M^-1 * s^-1 / k2 = exp((50.0 kJ/mol) / (8.314 J/mol * K) * (1/285 K - 1/351 K))
Now we can solve for k2 by rearranging the equation:
k2 = 37 M^-1 * s^-1 / exp((50.0 kJ/mol) / (8.314 J/mol * K) * (1/285 K - 1/351 K))
Evaluating this expression, we find that the rate constant at 12.0°C is approximately 5.38 M^-1 * s^-1.
Therefore, the rate constant at 12.0°C is approximately [tex]5.38 M^-1 * s^-[/tex]1.
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Draw all possible Cis and Trans isomers of Cis and Trans 1,2-demethyl Cyclohexane. Give approximate Free Energy of each isomer.
There are four possible isomers of 1,2-demethyl cyclohexane, including two cis isomers and two trans isomers.
1,2-demethyl cyclohexane has methyl group removed from the cyclohexane ring, leaving two adjacent carbon atoms with a hydrogen atom on each. The two isomers can be either cis or trans depending on the relative orientation of the hydrogen atoms on the two carbon atoms.
The two cis isomers have the hydrogen atoms on the same side of the ring, while the two trans isomers have the hydrogen atoms on opposite sides of the ring. The four possible isomers are:
cis-1,2-demethyl cyclohexane (both hydrogen atoms on the same side)
- trans-1,2-demethyl cyclohexane (both hydrogen atoms on opposite sides)
- cis-1,2-dimethylcyclohexane (both methyl groups on the same side)
- trans-1,2-dimethylcyclohexane (both methyl groups on opposite sides)
The free energy of each isomer will depend on its specific molecular structure and the surrounding environment. Without additional information, it is not possible to provide approximate free energy values for each isomer.
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A neutron has (select all that apply)
Group of answer choices
A charge of +1
A charge of 0
A mass of 0 AMU
A mass of 1 AMU
A charge of -1
A neutron does not have a charge of +1, -1, or 0. The correct statement is that a neutron has a charge of 0. Therefore, option B is the correct answer.
A neutron is a subatomic particle that is present in the nucleus of an atom along with protons. Protons carry a positive charge, while electrons carry a negative charge. Neutrons, on the other hand, are electrically neutral, meaning they have no charge. This is why they are called neutrons.
Regarding the mass of a neutron, it is approximately equal to 1 atomic mass unit (AMU). An AMU is a unit used to express the mass of subatomic particles. Neutrons and protons have a mass of around 1 AMU each, while electrons have a much smaller mass (about 0.0005 AMU).
To summarize, a neutron has a charge of 0 and a mass of approximately 1 AMU. It plays a crucial role in determining the stability and properties of atomic nuclei, as it helps bind protons together through the strong nuclear force, counteracting their mutual electrostatic repulsion. Thus, option B is correct.
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Complete combustion of 4.20 g of a hydrocarbon produced 12.9 g of CO
2
and 5.96 g of H
2
O. What is the empirical formula for the bydrocarbon? Insert subscripts as necessary.
the empirical formula for the hydrocarbon is C4H9.
To determine the empirical formula of the hydrocarbon, we need to calculate the number of moles of carbon and hydrogen in the given compounds.
First, let's find the number of moles of CO2 and H2O:
Number of moles of CO2 = mass / molar mass
Number of moles of CO2 = 12.9 g / 44.01 g/mol ≈ 0.2935 mol
Number of moles of H2O = mass / molar mass
Number of moles of H2O = 5.96 g / 18.02 g/mol ≈ 0.3310 mol
Next, let's calculate the number of moles of carbon and hydrogen:
Number of moles of carbon = 0.2935 mol (since CO2 has one carbon atom)
Number of moles of hydrogen = 2 * 0.3310 mol (since H2O has two hydrogen atoms)
Number of moles of carbon = 0.2935 mol
Number of moles of hydrogen = 0.6620 mol
Now, let's find the simplest ratio of carbon to hydrogen by dividing the number of moles by the smallest value:
Carbon: Hydrogen ≈ 0.2935 mol / 0.2935 mol : 0.6620 mol / 0.2935 mol
Carbon: Hydrogen ≈ 1 : 2.255
To obtain whole-number ratios, we can multiply the ratio by 4:
Carbon: Hydrogen ≈ 4 : 9
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please solve Q4
4. Calculate the number of moles of carbon atoms in \( 5.00 \) grams of carbon. 5. Write the complete electron configuration for a carbon atom.
Number of moles of carbon atoms = 0.4167 mol. Electron configuration of carbon is 1s² 2s² 2p².
Given that the mass of carbon is 5.00 g.Molar mass of carbon is 12 g/mol. The number of moles of carbon can be calculated as:Number of moles of carbon = Mass of carbon/molar mass of carbon
= 5.00 g/12 g/mol
= 0.4167 mol. Number of moles of carbon atoms will be the same as number of moles of carbon because carbon is a non-metal and its atoms exist independently, so;Number of moles of carbon atoms = 0.4167 mol
Complete electron configuration of carbon is 1s² 2s² 2p².
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Calculate the enthalpy of reaction for the decomposition of nitrogen triiodide atstandard conditions given ΔfHº (NI3) = +154.4 kJ/mol.
The enthalpy change for the decomposition of nitrogen triiodide at standard conditions is +217.2 kJ/mol.
The enthalpy change is the difference between the enthalpy of the products and the enthalpy of the reactants, both measured in the same standard conditions.
The decomposition of nitrogen triiodide at standard conditions is given as: N2I4 (g) ⟶ 2NI3 (g) ∆H = ?
We know that ΔfHº (NI3) = +154.4 kJ/mol.
Since the reaction as given is not balanced, we must first balance the equation.
The balanced equation is: N2I4 (g) ⟶ 2NI3 (g) + I2 (g)
The enthalpy change is equal to the enthalpy of formation of the products minus the enthalpy of formation of the reactants: = [2ΔfHº (NI3) + ΔfHº (I2)] - ΔfHº (N2I4)
Substitute the values from the table:
ΔH = [2 × (+154.4 kJ/mol) + 62.4 kJ/mol] - 205.2 kJ/mol= +217.2 kJ/mol
Therefore, the enthalpy change for the decomposition of nitrogen triiodide at standard conditions is +217.2 kJ/mol.
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Calculate the mole% of toluene in the vapor mixture at
equilibrium with a liquid mixture of 40% benzene and 60% toluene at
50 oC. Use DIPPR equation.
The mole percent of toluene in the vapor mixture can be obtained by multiplying y_toluene by 100. substitute the known values into the DIPPR equation, the mole fraction of toluene in the vapor phase (y_toluene) is 47.47%.
To calculate the mole percent of toluene in the vapor mixture at equilibrium with a liquid mixture of benzene and toluene, we can use the DIPPR equation. The DIPPR (Design Institute for Physical Property Data) equation is an empirical equation that relates the vapor-liquid equilibrium (VLE) composition to temperature and composition.
The DIPPR equation for calculating mole fraction in a vapor-liquid equilibrium is given as: y_i = x_i * P_i_sat(T) / P. where: y_i is the mole fraction of component i in the vapor phase, x_i is the mole fraction of component i in the liquid phase, P_i_sat(T) is the vapor pressure of pure component i at temperature T, and P is the total pressure of the system.
In this case, we have a liquid mixture of 40% benzene and 60% toluene. Let's assume a total pressure of P for the system and consider the vapor phase at equilibrium. We can use the DIPPR equation to calculate the mole fraction of toluene (y_toluene) in the vapor phase.
First, we need to determine the mole fraction of toluene in the liquid phase (x_toluene). Since the liquid mixture is composed of 40% benzene and 60% toluene, we have x_toluene = 0.60. Next, we need the vapor pressure of pure toluene at the given temperature of 50 °C. We can obtain this value from reliable sources or thermodynamic databases, such as the DIPPR database.
Finally, substitute the known values into the DIPPR equation to calculate the mole fraction of toluene in the vapor phase (y_toluene). The mole percent of toluene in the vapor mixture can be obtained by multiplying y_toluene by 100.
It's important to note that the DIPPR equation is an approximation, and for accurate calculations, it's advisable to consult more comprehensive thermodynamic models or databases specific to the system being analyzed.
We can calculate the mole fraction of toluene in the vapor phase: yT = 0.6 * 0.7911 = 0.4747 Therefore, the mole% of toluene in the vapor mixture at equilibrium with a liquid mixture of 40% benzene and 60% toluene at 50°C is 47.47%.
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Wrise the chrmical equatien between oopqur[iz walfate atad anc- 13. Write the chemical equation between copper(M) sulfate and rinc. Describe the reaction. Recond any observation. 14. Method 1: Determination of Limiting Reapent - The stoichiometric molar ratio is - The experimental molar ratio is 15. Method 2: Determination of Limiting Reagent - What is the molar amount of Cu expected to form if all the CuSOu is consumed? (Sim Calculations) - What is the molar amount ef Cu expected to form if all the Zn is consumed? (Shew Calculatiens) 21. Suggest possible sources of error in this experiment 22. Did the supernatant test (Procedure Part IV) agree with your prediction based on your calculations? Explain.
The chemical equation between copper(II) sulfate and zinc is:
CuSO4 + Zn → ZnSO4 + Cu
The reaction between copper(II) sulfate and zinc is a single replacement reaction. In this reaction, zinc displaces copper from copper(II) sulfate, resulting in the formation of zinc sulfate and copper. The zinc atoms donate electrons to the copper(II) ions, reducing them to copper atoms. This is an example of a redox (reduction-oxidation) reaction.
Observations during the reaction may include the color change of the solution from blue (copper(II) sulfate) to colorless (zinc sulfate), the formation of a reddish-brown precipitate of copper, and the evolution of hydrogen gas bubbles.
For method 1, the stoichiometric molar ratio between copper(II) sulfate and zinc can be determined from the balanced chemical equation. The experimental molar ratio can be obtained by measuring the amounts of reactants and products.
In method 2, the molar amount of Cu expected to form if all the CuSO4 is consumed can be calculated by converting the given mass of CuSO4 to moles and using the stoichiometric molar ratio from the balanced equation. Similarly, the molar amount of Cu expected to form if all the Zn is consumed can be calculated by converting the given mass of Zn to moles and using the stoichiometric molar ratio.
Possible sources of error in this experiment include incomplete reactions, side reactions, loss of reactants or products during transfer or filtration, and measurement errors in mass or volume.
The agreement between the supernatant test and the prediction based on calculations depends on the specific observation and prediction. It is important to evaluate the reasons for any discrepancies and consider factors such as experimental conditions, limitations of the procedure, and potential sources of error.
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The activation energy for the isomerization of cyclopropane is 274 kJ/mol. Find the rate constant at 250°C if it is equal to 8.23 x 10-3 at 280°C.
Using the formula:
student submitted image, transcription available below
I derived:
student submitted image, transcription available below
Ultimately, I came to the conclusion that:
student submitted image, transcription available below
Or the numerical answer of 0.250732.
This seems to be a larger number than what I started with and given that the temperature that I am solving for is lower, I believe this to be wrong. Can anyone guide me as to what I am doing wrong?
The new rate constant at the temperature that has been stated is 0.253
What is the rate constant?The proportionality constant known as the rate constant, or k, connects the rate of a chemical reaction to the concentrations of the reactants. It can be found in the rate equation or rate law, which describes how the concentrations of the reactants affect the rate of a reaction.
We know that;
[tex]ln(k_2/k_1) = -Ea/R(1/T_2 - 1/T_1)[/tex]
Where;
Ea = activation energy
R = gas constant
[tex]k_2[/tex] = final rate constant
[tex]k_1[/tex] = initial rate constant
[tex]T_1[/tex] = initial temperature
[tex]T_2[/tex] = Final temperature
If we then go on to substitute the values we have that;
ln([tex]k_2[/tex]/0.00823) = -274000/8.314(1/553- 1/523)
[tex]k_2[/tex] = [tex]e^{3.418}[/tex] * 0.00823
[tex]k_2[/tex] = 0.253
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Consider the following reaction. CO (g) + Cl2 (g) --> COCl2 (g)
The mechanism is believed to be,
(1) Cl2 <--> 2Cl (fast equilibrium)
(2) Cl + CO <--> COCl (fast equilibrium)
(3) COCl + Cl2 -> COCl2 + Cl (slow)
(4) 2 Cl -> Cl2 (fast)
Assuming that the mechanism is correct, derive the rate law for this reaction
The rate law for the given reaction is determined by examining the rate-determining step, which in this case is step (3) involving the reaction between COCl and Cl[tex]_{2}[/tex] to form COCl[tex]_{2}[/tex] and Cl.
The rate of the reaction is determined by the concentration of the reactants involved in the rate-determining step. From step (3), we can see that the rate is proportional to the concentrations of COCl and Cl[tex]_{2}[/tex]. Therefore, the rate law for this reaction can be expressed as:
Rate = k[COCl][Cl[tex]_{2}[/tex]]
Where k represents the rate constant, and [COCl] and [Cl[tex]_{2}[/tex]] are the concentrations of COCl and Cl[tex]_{2}[/tex], respectively.
This rate law suggests that the reaction rate depends on the concentrations of both COCl and Cl[tex]_{2}[/tex], with their respective exponents determined by the stoichiometric coefficients in the balanced equation. It is important to note that the rate law is determined based on the proposed mechanism and may require experimental confirmation to verify its accuracy.
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0.1M Ammonium Hydroxide (PKa = 9.26) G) 0.1 M Ammonium Chloride
Calculate the pH of:
a) 0.05 M HCl
b) 0.002 N NaOH
c) 0.001 M NaCl
d) 0.1 M acetic acid (pKa = 4.73)
e) 0.1 M sodium acetate
f) 0.1M ammonium hydroxide (pKa = 9.26)
g) 0.1 M ammonium chloride
The pH of the given solutions is as follows:
a) 0.05 M HCl: Very low (acidic)
b) 0.002 N NaOH: Very high (alkaline)
c) 0.001 M NaCl: Neutral
d) 0.1 M acetic acid (pKa = 4.73): Slightly acidic
e) 0.1 M sodium acetate: Slightly alkaline
f) 0.1 M ammonium hydroxide (pKa = 9.26): Alkaline
g) 0.1 M ammonium chloride: Slightly acidic
In order to calculate the pH of each solution, we need to consider the dissociation of relevant compounds and the equilibrium between the conjugate acid-base pairs.
a) 0.05 M HCl: HCl is a strong acid that completely dissociates in water, resulting in the formation of H+ ions. Since the concentration of HCl is relatively high, the solution will be highly acidic, with a low pH value.
b) 0.002 N NaOH: NaOH is a strong base that dissociates into Na+ and OH- ions in water. Since the concentration of NaOH is relatively low, the solution will be highly alkaline, with a high pH value.
c) 0.001 M NaCl: NaCl is a salt composed of a strong acid (HCl) and a strong base (NaOH), both of which fully dissociate in water. The resulting solution will be neutral, with a pH of 7.
d) 0.1 M acetic acid (pKa = 4.73): Acetic acid partially dissociates in water, releasing H+ ions. However, the concentration of acetic acid is relatively high, leading to a slightly acidic solution.
e) 0.1 M sodium acetate: Sodium acetate is the conjugate base of acetic acid. It hydrolyzes in water, releasing OH- ions and partially neutralizing the solution. Consequently, the solution becomes slightly alkaline.
f) 0.1 M ammonium hydroxide (pKa = 9.26): Ammonium hydroxide is a weak base that partially dissociates, yielding OH- ions. Since the concentration of ammonium hydroxide is relatively high, the solution will be alkaline.
g) 0.1 M ammonium chloride: Ammonium chloride is the salt of a weak base (NH4OH) and a strong acid (HCl). In water, it partially dissociates, leading to the formation of NH4+ and Cl- ions. The presence of NH4+ ions makes the solution slightly acidic.
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Consider the pentapeptide below:
Ala-Lys-Gly-Phe-Asp
Draw the structures of the products formed when a solution of it is treated with;
a. 6 M HCl and heat.
b. 1-fluoro-2,4-dinitrobenzene under mildly alkaline conditions.
c. Trypsin
d. Chymotrypsin
Consider the pentapeptide Ala-Lys-Gly-Phe-Asp.
The structures of the products formed when a solution of it is treated with the following reagents are given below:a. 6 M HCl and heatAla-Lys-Gly-Phe-Asp → Ala + Lys + Gly-Phe-Asp.When the pentapeptide Ala-Lys-Gly-Phe-Asp is treated with 6 M HCl and heat, it undergoes hydrolysis to form the products, alanine (Ala), lysine (Lys), and glycyl-phenylalanyl-aspartic acid (Gly-Phe-Asp).b. 1-fluoro-2,4-dinitrobenzene under mildly alkaline conditions.Ala-Lys-Gly-Phe-Asp + 1-fluoro-2,4-dinitrobenzene → Ala-Lys-Gly-Phe-Dnp + Asp.
This reaction also releases the terminal amino acid, aspartic acid (Asp), as a separate product. The resulting compound, Ala-Lys-Gly-Phe-Dnp is an intermediate in the Edman degradation reaction.c. TrypsinAla-Lys-Gly-Phe-Asp → Ala-Lys-Gly + Phe-AspWhen the pentapeptide Ala-Lys-Gly-Phe-Asp is treated with trypsin, the enzyme hydrolyzes the peptide bond between lysine (Lys) and glycine (Gly), producing the products, Lys-Gly and Phe-Asp.d. ChymotrypsinAla-Lys-Gly-Phe-Asp → Ala-Lys + Gly-Phe-Asp.
When the pentapeptide Ala-Lys-Gly-Phe-Asp is treated with chymotrypsin, the enzyme hydrolyzes the peptide bond between phenylalanine (Phe) and aspartic acid (Asp), producing the products, Gly-Phe and Asp.
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Ammonium phosphate ((NH
4
)
3
PO
4
) is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid (H
3
PO)
4
) with ammonia (NH
3
What mass of ammonium phosphate is produced by the reaction of 2.83 g of phosphoric acid? Round your answer to 3 significant digits.
The mass of ammonium phosphate produced by the reaction of 2.83 g of phosphoric acid is 13 g (approximate to 3 significant digits).
Given mass of phosphoric acid is 2.83 g.
(NH₄)₃PO₄ can be made by reacting H₃PO₄ with NH₃.
The balanced equation for the given reaction is shown below:
H₃PO₄(aq) + 3NH₃(aq) → (NH₄)₃PO₄(aq)
Phosphoric acid reacts with ammonia to give ammonium phosphate, (NH₄)₃PO₄.
By looking at the balanced equation we can see that:
1 mol of H₃PO₄ reacts with 3 mol of NH₃ to give 1 mol of (NH₄)₃PO₄.
The molar mass of H₃PO₄ is 98 g/mol.
The number of moles of H₃PO₄ can be calculated as follows:
[tex]$$\text{Number of moles of H}_3\text{PO}_4=\frac{\text{Mass}}{\text{Molar mass}}$$[/tex]
[tex]$$=\frac{2.83\;g}{98\;g/mol}$$[/tex]
Number of moles of H3PO4= 0.0289 mol
As per the balanced equation, 1 mol of H₃PO₄ reacts with 1/3 mol of (NH₄)₃PO₄.
The number of moles of (NH₄)₃PO₄ produced can be calculated as follows:
[tex]$$\text{Number of moles of }(NH_4)_3\text{PO}_4=\frac{\text{Number of moles of } H_3\text{PO}_4}{1/3}$$[/tex]
[tex]$$= 3 × 0.0289\text{ mol}$$[/tex]
Number of moles of (NH₄)₃PO₄= 0.087 mol
The molar mass of (NH₄)₃PO₄ is 149.087 g/mol.
The mass of (NH₄)₃PO₄ produced can be calculated as follows:
[tex]$$\text{Mass of }(NH_4)_3\text{PO}_4 =\text{Number of moles}×\text{Molar mass}$$[/tex]
[tex]$$= 0.087\;mol × 149.087\;g/mol$$[/tex]Mass of (NH₄)₃PO₄ produced = 12.98 g ≈ 13 g
Therefore, the mass of ammonium phosphate produced by the reaction of 2.83 g of phosphoric acid is 13 g (approximate to 3 significant digits).
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Which of the following molecules has polar bonds but is a nonpolar molecule? PCl5, PCl3, NCl3, CO2
Carbon dioxide (CO₂) is a nonpolar molecule with polar bonds.
A molecule is polar if the electrons are distributed unevenly, which results in partial charges on the atoms. Whereas, a molecule is nonpolar if the electrons are distributed uniformly, and no part of the molecule has a positive or negative charge. CO₂ is a linear molecule with two identical polar bonds (C-O). Although the polar bonds make CO₂ a polar molecule, the bond polarities cancel out each other. This happens because the carbon atom is symmetrical, and the two O atoms are arranged symmetrically on opposite sides of the carbon atom.
As a result, the partial positive charges on one side of the molecule are canceled by the partial negative charges on the other side of the molecule, resulting in a net zero dipole moment. Thus, even though CO₂ has polar bonds, it is a nonpolar molecule. Among the other options, PCl₃ and NCl₃ have polar bonds as well as polar molecules, while PCl₅ has polar bonds and is polar too.
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20g of Orange fruits was blended with water to prepare 100 cm' of juice. If the result
of an lodometric titration of the juice indicates that the concentration of Ascorbic acid is
15260 mg/L, what will be the mass of Ascorbic acid in lg of the Orange?
The mass of ascorbic acid in the orange is 3.05 g.
To calculate the mass of ascorbic acid in the orange, we need to use the concentration of ascorbic acid in the juice and the volume of the juice.
Given that the concentration of ascorbic acid is 15260 mg/L and the volume of the juice is 100 cm³, we can convert the volume to liters by dividing it by 1000: 100 cm³ ÷ 1000 = 0.1 L.
Next, we multiply the concentration by the volume to obtain the total mass of ascorbic acid in the juice: 15260 mg/L × 0.1 L = 1526 mg.
Since the question asks for the mass of ascorbic acid in grams, we divide the result by 1000: 1526 mg ÷ 1000 = 1.526 g.
Therefore, the mass of ascorbic acid in the orange is 1.526 g.
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1.In the electrolysis of molten LiBr, which product forms at the anode? 1. Li(l) 2. Br2(g) 3. H2(g) 4. O2(g)
2.In the electrolysis of molten FeI3, which product forms at the anode? 1. Fe(l) 2. O2(g) 3. H2(g) 4. I2(g)
3.Hydrogen can be prepared by suitable electrolysis of aqueous calcium salts 1. True 2. False
4.Hydrogen can be prepared by suitable electrolysis of aqueous silver salts 1. False 2. True
5.What product(s) forms at the cathode in the electrolysis of an aqueous solution of NaCl? 1. O2 and H+ 2. Na 3. Cl2 4. H2 and OH-
[tex]Br_2[/tex](g) (option 2) is the byproduct created at the anode during the electrolysis of molten LiBr. [tex]I_2[/tex](g) (option 4) is the end result of the electrolysis of molten[tex]FeI_3[/tex] at the anode. True.
In the chemical process of electrolysis, a substance is broken down into its individual elements or ions. It involves causing chemical processes to take place at the electrodes by passing an electric current through an electrolyte, often a liquid or solution containing ions. Anode and cathode are the terms used to describe the electrodes linked to the positive and negative terminals of a power source, respectively.
1) [tex]Br_2[/tex](g) (option 2) is the byproduct created at the anode during the electrolysis of molten LiBr.
2) [tex]I_2[/tex](g) (option 4) is the end result of the electrolysis of molten[tex]FeI_3[/tex] at the anode.
3) True. Through the proper electrolysis of aqueous calcium salts, hydrogen can be produced.
4)False. The appropriate electrolysis of aqueous silver salts cannot produce hydrogen.
5) [tex]H_2[/tex] and [tex]OH^-[/tex] are the product(s) generated at the cathode during the electrolysis of a NaCl aqueous solution (option 4).
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Describe how to prepare 100ml of a 0.150M acetate buffer at a pH=5 from sodium acetate trihydrate and your standardized 0.200MHCl solution.
To prepare a 100 ml of a 0.150 M acetate buffer at a pH of 5 from sodium acetate trihydrate and your standardized 0.200 M HCl solution, the following procedures can be followed Calculate the amount of sodium acetate trihydrate required to prepare 100 ml of 0.15 M acetate buffer.
The molecular weight of sodium acetate trihydrate is 136.08 g/mol.The weight of sodium acetate trihydrate required can be calculated as follows:0.15 M = (weight of sodium acetate trihydrate / volume of solution in litres)Weight of sodium acetate trihydrate = 0.15 × 0.1 × 136.08 = 2.042 g (to 3 significant figures) Dissolve 2.042 g of sodium acetate trihydrate in 80 ml of distilled water and stir until the salt is completely dissolved. Use a volumetric flask for measuring the volume.
Adjust the pH of the solution to 5 using a pH meter. If the pH is too low, add a few drops of 0.2 M HCl to lower the pH. If the pH is too high, add a few drops of 0.2 M NaOH to raise the pH. Check the pH after each addition and adjust accordingly.Step 4: Bring the total volume of the solution to 100 ml by adding distilled water and mix well. The solution is now ready and can be used as an acetate buffer at pH 5.
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For the following reaction and its equilibrium constant, determine whether the position of equilibrium lies towards reactants, towards products, or somewhere reasonably close to the midd (intermediate). NH3(aq)+HBrO(aq)⇔NH4+(aq)+BrO (aq) Kc=5.0 Reactants Products Intermediate Question 4 For the following reaction and its equilibrium constant, determine whether the position of equilibrium lies towards reactants, towards products, or somewhere reasonably close to the middle (intermediate). H3O+(aq)+OH−(aq) ⇔2H2O(l)Kc=1.0×1014 Reactants Products Intermediate Question 5 For the following reaction and its equilibrium constant, determine whether the position of equilibrium lies towards reactants, towards products, or somewhere reasonably close to the middle (intermediate). Fe(OH)2( s)⇔Fe2+(aq)+2OH−(aq)Kc=4.9×10−17 Reactants Products Intermediate
For the following reaction and its equilibrium constant, the position of equilibrium lies towards the products as the value of Kc is 10^14.H3O+(aq)+OH−(aq) ⇔2H2O(l)
Kc=1.0×1014
Equilibrium constant (Kc) is defined as the ratio of the concentration of products raised to their stoichiometric coefficient to the concentration of reactants raised to their stoichiometric coefficient, with each concentration term raised to a power equal to the number of molecules or ions in the balanced equation. The value of Kc is used to determine the direction of the reaction. The position of equilibrium lies towards the reactants if the Kc is very small and the position of equilibrium lies towards the products if the Kc is very large.
The position of equilibrium lies in the middle of the reaction if the Kc is equal to 1. For the following reaction and its equilibrium constant, the position of equilibrium lies towards the reactants as the value of Kc is 4.9 × 10^-17.Fe(OH)2(s) ⇔Fe2+(aq) + 2OH-(aq)Kc=4.9×10−17. In this case, the value of Kc is very small (10^-17) which indicates that the reactants are favored. Therefore, the position of the equilibrium lies towards the reactants. The position of equilibrium lies towards the products as the value of Kc is 10^14.
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Express your answer as a chemical equation. Identify all of the phases in your answer. 5.5 (Pages 179 - 183) ; Watch KCV 5.5, IWE 5.6. Enter NOREACTION if no reaction occurs. Enter a molecular equation for the precipitation reaction that occurs (if any) when each pair of aqueous solutions is mixed. LiBr(aq)+Pb(C2H3O2)2( s)→PbBr+LiC2(aq)+H2O(aq) A chemical reaction does not occur for this question. X Incorrect; Try Again; 5 attempts remaining
The given chemical reaction is LiBr(aq) + Pb(C2H3O2)2(s) → PbBr + LiC2H3O2(aq) + H2O(aq). This chemical reaction does not occur.
Therefore, the answer is NOREACTION. Hence, "NOREACTION". chemical reaction can be identified by the formation of a precipitate, water, or gas. This chemical reaction does not occur because there is no formation of precipitate, water, or gas. Therefore, the answer is NOREACTION.
The given chemical reaction is LiBr(aq) + Pb(C2H3O2)2(s) → PbBr + LiC2H3O2(aq) + H2O(aq).
The aqueous solution of lithium bromide (LiBr) and solid lead (II) acetate [Pb(C2H3O2)2] are mixed together. There is no reaction as a result of the mixing. Therefore, there are no phases in the answer.
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For a steel alloy it has been determined that a carburizing heat treatment of 13 h duration at 832∘C will raise the carbon concentration to 0.54wt% at a point 3.9 mm from the surface. Estimate the time necessary to achieve the same concentration at a 5.9 mm position for an identical steel and at a carburizing temperature of 1030∘C. Assume that D₀ is 4.9×10⁻⁵ m²/s and Q is 119 kJ/mol.
i ..... h
Therefore, the time necessary to achieve the same carbon concentration of 0.54wt% at a distance of 5.9 mm from the surface for an identical steel and at a carburizing temperature of 1030°C is about 507 hours (or 21 days).
To calculate the time necessary to achieve the same carbon concentration of 0.54wt% at a 5.9 mm position for an identical steel and at a carburizing temperature of 1030°C, we can use Fick's second law of diffusion.
According to Fick's second law of diffusion, the change in concentration with time at a distance x from the surface can be given by:
[tex]$$\frac{\partial C}{\partial t}=\frac{D}{x^2}\frac{\partial}{\partial x}(x^2\frac{\partial C}{\partial x})$$[/tex]
Where C is the concentration of carbon, D is the diffusion coefficient, x is the distance from the surface, and t is the time of carburizing.
The equation can be simplified for one-dimensional diffusion (i.e., diffusion in one direction) and for constant diffusion coefficient D as follows:
[tex]$$C-C_0=erfc(\frac{x}{2\sqrt{Dt}})$$[/tex]
where C0 is the initial carbon concentration, erfc is the complementary error function, and t is the time.
To estimate the time necessary to achieve the same carbon concentration at a 5.9 mm position, we can use the following steps:1. Calculate the diffusion coefficient (D) for the given temperature of 1030°C. The diffusion coefficient is given by the Arrhenius equation as follows:
D=D₀exp(-Q/RT)
Where D₀ is the pre-exponential factor, Q is the activation energy, R is the universal gas constant, and T is the absolute temperature. Plugging in the values given:
D₀ = 4.9×10⁻⁵ m²/sQ = 119 kJ/mol
R = 8.314 J/mol K (universal gas constant)
T = 1030 + 273 = 1303 K
D = 4.9×10⁻⁵ exp(-119×10³/(8.314×1303)) = 4.09×10⁻⁶ m²/s2.
Use the simplified equation of Fick's second law to find the time necessary for the carbon concentration to reach 0.54wt% at a distance of 5.9 mm from the surface:
0.54-0.00=erfc(5.9/(2√(4.09×10⁻⁶t)))
erfc(5.9/(2√(4.09×10⁻⁶t))) = 0.54
erfc⁻¹(0.54) = 2.043√(4.09×10⁻⁶t) = 5.9/(2×2.043) = 1.44
t = (1.44)²/(4.09×10⁻⁶) = 507 h
Therefore, the time necessary to achieve the same carbon concentration of 0.54wt% at a distance of 5.9 mm from the surface for an identical steel and at a carburizing temperature of 1030°C is about 507 hours (or 21 days).
The answer is approximately 507 hours.
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Lquild hexane (CH
3
(CH
2
)
4
CH
3
) will react with gaseous oxygen (O
2
) to produce gaseous carbon dioxide (CO
2
) and gaseous water (H
2
O). Suppose 2.6 g of hexane is mixed with 5.849 of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction, Round your answer to 3 significant digits.
We need to determine the limiting reagent and the stoichiometric ratio between the reactants. The maximum mass of carbon dioxide that could be produced by the chemical reaction is approximately 7.955 grams (rounded to three significant digits).
To calculate the maximum mass of carbon dioxide (CO2) that could be produced by the chemical reaction between hexane (C6H14) and oxygen (O2), we need to determine the limiting reagent and the stoichiometric ratio between the reactants.
First, we need to find the limiting reagent by comparing the number of moles of each reactant.
Molar mass of hexane (C6H14) = 6 * 12.01 g/mol (carbon) + 14 * 1.01 g/mol (hydrogen) = 86.18 g/mol
Molar mass of oxygen (O2) = 2 * 16.00 g/mol = 32.00 g/mol
Moles of hexane = 2.6 g / 86.18 g/mol ≈ 0.0301 moles
Moles of oxygen = 5.849 g / 32.00 g/mol ≈ 0.1828 moles
From the balanced equation: C6H14 + 19/2 O2 -> 6 CO2 + 7 H2O, we can see that the stoichiometric ratio between hexane and carbon dioxide is 1:6.
Since the number of moles of hexane (0.0301 moles) is less than 1/6 times the number of moles of oxygen (0.1828 moles), hexane is the limiting reagent.
To determine the maximum mass of carbon dioxide produced, we use the stoichiometric ratio between hexane and carbon dioxide.
Moles of carbon dioxide produced = Moles of hexane * 6
= 0.0301 moles * 6
= 0.1806 moles
Mass of carbon dioxide produced = Moles of carbon dioxide produced * Molar mass of CO2
= 0.1806 moles * 44.01 g/mol
≈ 7.955 g
Therefore, the maximum mass of carbon dioxide that could be produced by the chemical reaction is approximately 7.955 grams (rounded to three significant digits).
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(NH₄)₂SO₄ has a van't Hoff factor of i = 2.31. What is the concentration of particles in a 1.251 M solution of (NH₄)₂SO₄?
The concentration of particles in a 1.251 M solution of (NH₄)₂SO₄ is approximately 2.897 M. The van't Hoff factor (i) represents the number of particles that a solute dissociates into in a solution.
In the case of (NH₄)₂SO₄, the van't Hoff factor is given as i = 2.31.
To calculate the concentration of particles in a solution, we multiply the van't Hoff factor by the initial molarity of the solute.
Given that the initial molarity of (NH₄)₂SO₄ is 1.251 M, we can calculate the concentration of particles as follows:
Concentration of particles = van't Hoff factor × initial molarity
Concentration of particles = 2.31 × 1.251 M
Concentration of particles ≈ 2.897 M
Therefore, the concentration of particles in a 1.251 M solution of (NH₄)₂SO₄ is approximately 2.897 M.
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Wet air enters a dryer at 98.5 kPa and 60oC. The partial
pressure of water vapor in the entering air is 8.7 kPa. Calculate
the kg-mol water present per 100 kg-mol dry air.
Answer:
The mass of water present per 100 kg-mol of dry air is 3.17 kg-mol.
Explanation:
Given parameters:
The partial pressure of water vapor in the entering air = 8.7 kPa
The temperature of the entering air = 60°C
The pressure of entering air = 98.5 kPa
We are to calculate the number of kg-mol of water present per 100 kg-mol dry air
Solution: The mole fraction of water vapor in the air can be calculated as:
Y = Pv / PaPv = Partial pressure of water vapor in the entering air = 8.7 kPaPa = Total pressure of entering air = 98.5 kPaY = 8.7 / 98.5Y = 0.08817
Therefore, the mole fraction of water vapor in the air is 0.08817.
The total number of moles of air present can be calculated as:nA = PA * VA / RT... (i)
where, PA = Total pressure of entering air = 98.5 kPaVA = Volume of entering air = 1 kg-mol dry air
R = Universal gas constant = 8.314 J / mol K... (ii)
T = Temperature of entering air = 60 + 273 = 333 K
T = 333 K
Substituting the given values in equation (i)
nA = 98.5 × 1000 / (8.314 × 333) = 35.89 kg-mol
Therefore, the total number of moles of air present is 35.89 kg-mol. The mass of water present per 100 kg-mol of dry air can be calculated as:
nw = nA × Y
nw = 35.89 × 0.08817
nw = 3.17 kg-mol
Therefore, the mass of water present per 100 kg-mol of dry air is 3.17 kg-mol.
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What is the chemical formula for the compound formed between chromium(III) and the chlorate ion? chemical formuln: What is the chemical formola for the compound formed between chromium(III) and the carbonate ion? chemical formula:
The chemical formula for the compound formed between chromium(III) and the chlorate ion is Cr(ClO₃)₃.
Chromium (III) is a cation with a charge of +3. On the other hand, the chlorate ion has a charge of -1. So, to balance the charges, three chlorate ions are required for each chromium ion. The chemical formula for the compound formed between chromium(III) and the chlorate ion is Cr(ClO₃)₃.
The chemical formula for the compound formed between chromium(III) and the carbonate ion is Cr₂(CO₃)₃.
Chromium (III) is a cation with a charge of +3. Carbonate ion has a charge of -2. In order to balance the charges, two chromium ions are required for every three carbonate ions. Therefore, the chemical formula for the compound formed between chromium(III) and the carbonate ion is Cr₂(CO₃)₃.
Thus, the chemical formula for the compound formed between chromium(III) and the chlorate ion is Cr(ClO₃)₃, whereas the chemical formula for the compound formed between chromium(III) and the carbonate ion is Cr₂(CO₃)₃.
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Convert the following measurement. \[ 6.3 \times 10^{2} \frac{\mathrm{mg}}{\mathrm{dL}}=\frac{\mathrm{g}}{\mathrm{L}} \]
The converted value is 6.3 g/L. Conversion factors are ratios or relationships between different units of measurement that allow for the conversion of quantities from one unit to another.
To convert the given measurement of 6.3 x 10^2 mg/dL to g/L, we can use the following conversion factors:
1 g = 1000 mg (since there are 1000 milligrams in a gram)
1 L = 10 dL (since there are 10 deciliters in a liter)
Now, let's proceed with the conversion:
6.3 x 10^2 mg/dL x (1 g / 1000 mg) x (10 dL / 1 L)
First, we convert milligrams (mg) to grams (g) by multiplying by the conversion factor 1 g / 1000 mg. Then, we convert deciliters (dL) to liters (L) by multiplying by the conversion factor 10 dL / 1 L.
Simplifying the units and performing the calculation:
6.3 x 10^2 x 1 / 1000 x 10 g/L
This simplifies to:
6.3 g/L
Therefore, the converted value is 6.3 g/L.
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Each ""20-mg"" capsule of Prozac contains 22.5 mg of fluoxetine HCL (m.w. 345.63) equivalent to 20 mg of fluoxetine base. Calculate the molecular weight of the fluoxetine base. Round your answer to the HUNDRETHS decimal place.
The given formula is incorrect for determining the molecular weight of fluoxetine base. The molecular weight of fluoxetine base is approximately 309.17 g/mol.
The molecular weight of fluoxetine base can be calculated by subtracting the molar mass of the hydrochloride (HCl) portion from the molar mass of fluoxetine HCl.
Given:
Molar mass of fluoxetine HCl (MW) = 345.63 g/mol
Amount of fluoxetine base = 20 mg
To calculate the molecular weight of fluoxetine base, we need to subtract the molar mass of the HCl portion from the molar mass of fluoxetine HCl:
Molecular weight of fluoxetine base = MW - Molar mass of HCl
The molar mass of HCl is approximately 36.46 g/mol.
Molecular weight of fluoxetine base = 345.63 g/mol - 36.46 g/mol
= 309.17 g/mol
Therefore, the molecular weight of fluoxetine base is approximately 309.17 g/mol.
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You pipet 25.0 mL of the solution prepared in question 4 into a 250 mL Erlenmeyer flask and, following the procedure outlined in your lab manual, titrate to its endpoint with newly made EDTA titrant. The starting buret volume was 4.35 mL. The ending buret volume was 30.04 mL. Calculate the exact concentration of the EDTA titrant and report your answer in molarity.
EDTA is a hexadentate ligand that forms stable complexes with most metal ions. The metal-EDTA complex can be titrated with EDTA, and the endpoint of the titration can be detected with a metal ion indicator.
Accordingn to the given information:The starting buret volume is 4.35 mL
The ending buret volume is 30.04 mL
The volume of the EDTA titrant used in the titration was (30.04 - 4.35) mL
= 25.69 mL
Given that the mass of EDTA disodium dihydrate is 372.24 g/mol and the EDTA titrant was made using 0.8000 g of the disodium salt, then:
moles of EDTA = mass / molar mass
moles of EDTA = 0.8000 g / 372.24 g/mol = 0.002148 mol
Molarity = moles of EDTA / volume of EDTA titrant
Molarity = 0.002148 mol/25.69 mL
= 0.0835 M
Answer: The exact concentration of the EDTA titrant is 0.0835 M.
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