Answer:
b. 3.35
Explanation:
To calculate the pH of a solution containing both acid and its salt (produced as a result of titration) we need to use Henderson’s equation i.e.
pH = pKa + log ([salt]/[acid]) (Eq. 01)
Where
pKa = -log(Ka) (Eq. 02)
[salt] = Molar concentration of salt produced as a result of titration
[acid] = Molar concentration of acid left in the solution after titration
Let’s now calculate the molar concentration of HNO2 and KOH considering following chemical reaction:
HNO2 + KOH ⇆ H2O + KNO2 (Eq. 03)
This shows that 01 mole of HNO2 and 01 mole of KOH are required to produce 01 mole of KNO2 (salt). And if any one of them (HNO2 and KOH) is present in lower amount then that will be considered the limiting reactant and amount of salt produced will be in accordance to that reactant.
Moles of HNO2 in 50 mL of 0.01 M HNO2 solution = 50/1000x0.01 = 0.005 Moles
Moles of KOH in 25 mL of 0.01 M KOH solution = 25/1000x0.01 = 0.0025 Moles
As it can be seen that we have 0.0025 Moles of KOH therefore considering Eq. 03 we can see that 0.0025 Moles of KOH will react with only 0.0025 Moles of HNO2 and will produce 0.0025 Moles of KNO2.
Therefore
Amount of salt produced i.e [salt] = 0.0025 moles (Eq. 04)
Amount of acid left in the solution [acid] = 0.005 - 0.0025 = 0.0025 moles (Eq.05)
Putting the values in (Eq. 01) from (Eq.02), (Eq. 04) and (Eq. 05) we will get the following expression:
pH= -log(4.5x10 -4) + log (0.0025/0.0025)
Solving above we get
pH = 3.35
The pH value in the titration flask after 25.00 mL of the 0.10 M KOH solution is added to 50.00 mL of 0.10 M HNO2 solution is 3.35.
The subject of this question is titration, which is a method used in chemistry to measure the concentration of an unknown solution. Given 50.00 mL of 0.10 M HNO2 (nitrous acid, Ka = 4.5 × 10-4), titrated with 0.10 M KOH (potassium hydroxide), we need to calculate the pH after 25.00 mL of the KOH solution is added.
First, we need to find the moles of the HNO2 and the KOH. Moles equals Molarity times Volume. So, for HNO2, it is 0.10 M * 0.050 L which equals 0.005 moles. For KOH, it is 0.10 M * 0.025 L which equals 0.0025 moles.
Then, subtract the moles of OH- from the moles of HNO2 to determine the concentration of HNO2 left, which is 0.005 moles - 0.0025 moles = 0.0025 moles. Divide this by the total volume of the solution (50.00 mL + 25.00 mL = 75.00 mL or 0.075 L to determine the new concentration of HNO2, 0.0025 moles / 0.075 L = 0.033 M. Then use the given Ka value with the equation [H+] = sqrt(Ka * [HNO2]) to get [H+].
To find acids' pH, we use the formula pH = -log[H+]. Use the calculated [H+] to find the pH.
Upon performing these calculations, the resulting pH value should be approximately 3.35 after 25.00 mL of the KOH solution is added, so the answer is (b) 3.35.
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Answer:
use google and use the first link
Explanation:
Treatment of (S)-( )-5-methyl-2-cyclohexenone with lithium dimethylcuprate followed by protonolysis produces dextrorotatory ketone A and a trace of isomer B. Compound A can yield optically active, dextrorotatory hydrocarbon C when treated with zinc amalgam and HCl.
Treatment of (S)-( )-5-methyl-2-cyclohexenone with lithium dimethylcuprate followed by protonolysis yields a mixture containing primarily a dextrorotatory ketone A and a trace of an optically inactive isomer B. Compound A can be treated with zinc amalgam and HCl to produce an optically active, dextrorotatory hydrocarbon C.
To provide a specific identification and stereochemical configuration of compounds A, B, and C, I would need additional information or further context about the starting components and reaction conditions, as well as any other relevant data or observations.
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Answer: The mole fraction of calcium chloride and water in the solution is 0.057 and 0.943 respectively
Explanation:
We are given:
Molality of calcium chloride = 3.35 m
This means that 3.35 moles of calcium chloride are present in 1 kg or 1000 g of water
To calculate the number of moles, we use the equation:
Given mass of water = 1000 g
Molar mass of water = 18.02 g/mol
Putting values in above equation, we get:
Total moles of solution = [3.35 + 55.49] = 58.84 moles
Mole fraction of a substance is given by:
Hence, the mole fraction of calcium chloride and water in the solution is 0.057 and 0.943 respectively
Answer:
49.3% water
Explanation:
What is the binding energy of the electron in kJ/mol? [Note that KE = = mv2 and 1 electron volt (eV) = 1.602 x 10-19 J.]
Answer:
An X-ray photon of wave length 0.989 nm strikes a surface. The emitted electron has a kinetic energy of 969 eV. What is the binding energy of the electron in kJ/mol? [KE=
1
2
mv2;1 electron volt (eV)=1.602×10−19J][KE=
2
1
mv
2
;1 electron volt (eV)=1.602
The photoelectric effect equation is used to find the binding energy of an electron when an X-ray photon with a specific wavelength strikes a surface, taking into account the kinetic energy of the ejected electron and the energy of the photon.
To calculate the binding energy of an electron when an X-ray photon with a given wavelength strikes a surface, you use the photoelectric effect equation which connects the energy of the photon (E = hc/λ) with the kinetic energy (KE) of the ejected electron and the binding energy (BE) that keeps the electron attached to the atom.
The equation is: KE + BE = hc/λ, where h is Planck's constant (6.626 x 10-34 J s), c is the speed of light (3.00 x 108 m/s), and λ is the wavelength of the photon.
The given kinetic energy of the electron is 959 eV, which can be converted to joules (1 eV = 1.602 x 10-19 J). The energy of the photon can be calculated using the wavelength. Binding energy is then found by subtracting the electron's kinetic energy from the energy of the photon.
To find the binding energy per mole, you can use Avogadro's number (6.022 x 1023 mol-1) to calculate the total binding energy in a mole of such electrons and then convert it to kilojoules.
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1. Sodium 4-amino-1-naphthalenesulfonate makes it very soluble in water as it contains a hydrate salt sodium sulfate .
2. In the structure of this compound, sodium sulphate is polar in nature.
The molecular structure of sodium 4-amino-1-naphthalenesulfonate is .
The polar part of the structure sodium sulfate makes sodium 4-amino-1-naphthalenesulfonate a hydrate salt. Salt are polar and are usually soluble in water.
Find more information about Molecular formula here:
Answer: it contains a hydrate salt sodium sulfate NaO4S.
4-amino-1-naphthalenesulfonate is a sodium salt. Sodium sulfate is Polar.
Explanation:
The molecular structure of sodium 4-amino-1-naphthalenesulfonate is
C10H10NNaO4S
The polar part of the structure sodium sulfate NaO4S makes sodium 4-amino-1-naphthalenesulfonate a hydrate salt. Salt are polar and are usually soluble in water.
Answer:
The three subatomic particles are the particles contained in the iota. They are protons, neutrons, and electrons. Protons and electrons convey a positive and negative charge, individually, while neutrons don't convey any change
Explanation: