a chemist determines that a sample contains 0.15ug of U-235 and that the sample has undergone two half lives how much U-235 was there originally, before the same decayed?

The ratio of the actual value of a colligative property to the value calculated, assuming the substance to be a nonelectrolyte, is referred to as d. the van't Hoff factor

In comparison to the value anticipated for a nonelectrolyte, the Van't Hoff factor measures how much a solute dissociates or ionises in a solution. It is used to account for the impact of solute ionisation or dissociation on the computation of the collinear characteristics of solutions and is represented by the symbol "I".

Colligative qualities are those of a solution that, independent of the chemical composition of the solute particles, rely exclusively on their concentration. Collaborative qualities include things like reduced vapour pressure, increased boiling point, decreased freezing point, and reduced osmotic pressure. If the solute is a nonelectrolyte that doesn't dissolve or ionise, these qualities are utilised to compute the molar mass or molecular weight of the solute in a solution.

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1. The ratio of 12C to 13C in living organisms is generally constant at 1:1. This is because all living organisms take in carbon from their environment that is composed of 12C and 13C in a 1:1 ratio.

Through the process of photosynthesis, carbon is incorporated into organic molecules, and the same 1:1 ratio is maintained.

2. The ratio of 12C to 13C in non-living sources can vary greatly. This is because the two isotopes of carbon behave differently in the atmosphere, oceans, and other natural environments. For example, 12C is more soluble and can be taken up by plants and organisms more easily, while 13C is less soluble and tends to accumulate in the environment. As a result, the ratio of 12C to 13C in non-living sources can vary widely.

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Respiration breaks down dead organic matter for energy using oxygen and releasing carbon dioxide.

What is Respiration?

Respiration is the biological process by which living organisms exchange gases, usually oxygen and carbon dioxide, with the environment. In animals, respiration involves the absorption of oxygen from the air or water, the transportation of oxygen to the cells of the body, the use of oxygen by the cells to generate energy through cellular respiration, and the elimination of carbon dioxide, which is a waste product of this process.

In the case of dead organic matter, microorganisms such as bacteria and fungi carry out respiration as a means of obtaining energy. These microorganisms use the organic matter as a source of food, breaking it down into simpler compounds through a process called decomposition. During this process, oxygen is consumed and carbon dioxide is released as a byproduct.

Therefore, respiration breaks down dead organic matter for energy using oxygen and releasing carbon dioxide.

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The amount of element Y remaining is between 0.25 g and 0.4 g.

The half-life of a radioactive element is the time taken for half of the initial sample to decay. Using this information, we can determine the amount of element X remaining after a certain time has elapsed.

If the half-life of element X is 50 days, then after 50 days, half of the initial sample will remain. After another 50 days (i.e., a total of 100 days), half of that remaining amount will remain, which is 0.25 g. After another 50 days (i.e., a total of 150 days), half of that remaining amount will remain, which is 0.125 g. Therefore, after some time has passed and 0.2 g of element X remains, the time elapsed must be between 100 and 150 days.

Now, let's determine how much of element Y remains after the same amount of time has elapsed. If the half-life of element Y is 100 days, then after 100 days, half of the initial sample will remain. After another 100 days (i.e., a total of 200 days), half of that remaining amount will remain, which is 0.25 g. Therefore, after the same time has elapsed as for element X, which is between 100 and 150 days, the amount of element Y remaining should be between 0.25 g and 0.4 g.

To narrow down the range further, we can use the fact that the initial mass of both samples was 0.8 g. If 0.2 g of element X remains, then 0.6 g must have decayed. Therefore, the total mass of the two samples after some time has elapsed is 0.8 g - 0.6 g = 0.2 g.

If we assume that element Y has also decayed by the same amount, then the mass of element Y remaining is 0.2 g - 0.2 g = 0 g. This would mean that all of element Y has decayed, which is possible given that its half-life is longer than that of element X. However, if we assume that element Y has decayed less than element X, then the mass of element Y remaining must be between 0.25 g and 0.4 g.

Therefore, we can conclude that the amount of element Y remaining is between 0.25 g and 0.4 g.

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The one which is not the empirical formula is C₂H₆O₂. The correct option is D.

A) CHO = 1 : 1 : 1 = simplest form , this is an empirical formula.

B) CH₂O = 1 : 2 : 1 = simplest form , this is an empirical formula.

C) C₂H₄O = 2 : 4 : 1 = simplest form , this is an empirical formula.

D) C₂H₆O₂ = 2 : 6 : 2 = not the simplest form, this is not the empirical formula.

E ) C₃H₈O = 3 : 8 : 1  = simplest form , this is an empirical formula.

The empirical formula for the chemical compound is the formula that is the simplest whole number ratio of the atoms which is present in the compound.

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Hi! I'm happy to help with your question. The set in which all elements tend to form cations in binary ionic compounds is A) Li, Sn, O.

Step 1: Understand that cations are positively charged ions formed when an element loses electrons.
Step 2: Determine which elements in the given sets typically lose electrons to form cations.
Step 3: Li (lithium) loses one electron to form Li+, Sn (tin) loses two or four electrons to form Sn2+ or Sn4+, and O (oxygen) gains two electrons to form O2-. However, in the context of binary ionic compounds, Li and Sn will bond with other elements and form cations, while O will form an anion.

So, the correct answer is A) Li, Sn, O.

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The number of amino acid residues in a strand: Each protein strand typically contains anywhere from 50 to 2,000 amino acid residues. The exact number varies greatly depending on the specific protein and its function.

Depending on the function of the protein and the precise placement of the strand within the protein's structure, the number of amino acid residues in a protein strand might change. For instance, whereas some proteins may have strands with 20 or more amino acids, others may only have strands with a few amino acids.

A protein strand typically has between 5 and 30 amino acid residues. However, certain proteins also contain longer strands. For instance, the protein beta-sheet's beta-strands can be fairly lengthy and some of them can contain up to 100 amino acid residues.

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The percent composition of ZnO is approximately 80.50% Zn and 19.50% O.

To calculate the percent composition of a compound, we need to determine the mass of each element present in the compound and express it as a percentage of the total mass of the compound.

Given;

Mass of Zn = 1.63 g

Mass of oxygen (O₂) = 0.40 g

Balanced chemical equation for the reaction between Zn and oxygen to form ZnO is;

2Zn + O₂ → 2ZnO

From the balanced equation, we can see that 2 moles of Zn react with 1 mole of O₂  to produce 2 moles of ZnO.

1 mole of Zn has a molar mass of 65.38 g/mol, and 1 mole of O₂  has a molar mass of 32.00 g/mol.

Now we can calculate the moles of Zn and O₂  in the given masses:

Moles of Zn = Mass of Zn / Molar mass of Zn

Moles of Zn = 1.63 g / 65.38 g/mol ≈ 0.025 mol

Moles of O₂  = Mass of O₂  / Molar mass of O₂

Moles of O₂  = 0.40 g / 32.00 g/mol ≈ 0.0125 mol

The molar mass of ZnO is the sum of the molar masses of Zn and O, which are 65.38 g/mol and 16.00 g/mol, respectively;

Molar mass of ZnO = 65.38 g/mol + 16.00 g/mol = 81.38 g/mol

Now we can calculate the percent composition of ZnO;

Mass of Zn in ZnO = Moles of ZnO × Molar mass of Zn

Mass of Zn in ZnO = 0.0125 mol × 65.38 g/mol ≈ 0.81725 g

Mass of O in ZnO = Moles of ZnO × Molar mass of O

Mass of O in ZnO = 0.0125 mol × 16.00 g/mol ≈ 0.200 g

Total mass of ZnO = Mass of Zn in ZnO + Mass of O in ZnO

Total mass of ZnO = 0.81725 g + 0.200 g = 1.01725 g

Now we can calculate the percent composition of ZnO;

Percent composition of Zn in ZnO = (Mass of Zn in ZnO / Total mass of ZnO) × 100%

Percent composition of Zn in ZnO = (0.81725 g / 1.01725 g) × 100%

≈ 80.50%

Percent composition of O in ZnO = (Mass of O in ZnO / Total mass of ZnO) × 100%

Percent composition of O in ZnO = (0.200 g / 1.01725 g) × 100%

≈ 19.50%

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Lowering a solution's vapor pressure can result in an increased boiling point, a higher freezing point, and a reduced evaporation rate, all of which can have an impact on the solution's temperature.

The effects of lowering a solution's vapor pressure on temperatures are:

1. Lower boiling point: When the vapor pressure of a solution is lowered, it takes more energy (higher temperature) for the molecules to escape the liquid phase and become a gas. As a result, the boiling point of the solution increases.

2. Higher freezing point: Lowering the vapor pressure also affects the freezing point of a solution. With lower vapor pressure, the solution will have a higher freezing point, meaning it will solidify at a higher temperature than it would with higher vapor pressure.

3. Reduced evaporation rate: When the vapor pressure is lowered, it slows down the evaporation process. This means that the solution will take longer to evaporate at a given temperature, leading to a decrease in the cooling effect that evaporation usually provides.

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The maximum energy of the UV photons generated by this plasma pencil is approximately[tex]9.945 x 10^-19 Joules[/tex].

To find the maximum energy of the UV photons generated by this plasma pencil, you need to use the given wavelength range (lambda = 200-300 nm) and the provided constants (speed of light, c = 3.0 x 10^8 m/s and Planck's constant, [tex]h = 6.63 x 10^-34 J.s[/tex]).

Step 1: Convert the wavelength range to meters by multiplying with 10^-9 (since 1 nm = 10^-9 m). This gives you a range of [tex]200 x 10^-9 m to 300 x 10^-9 m.[/tex]
Step 2: To find the maximum energy, use the minimum wavelength (200 x 10^-9 m), since energy is inversely proportional to the wavelength.

Step 3: Use the energy formula:[tex]E = h * c / lambda[/tex].

In this case, E =[tex](6.63 x 10^-34 J.s) * (3.0 x 10^8 m/s) / (200 x 10^-9 m)[/tex].

Step 4: Calculate the energy:[tex]E ≈ 9.945 x 10^-19 J[/tex].

The maximum energy of the UV photons generated by this plasma pencil is approximately 9.945 x 10^-19 Joules.

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Alpha-D-glucopyranose in the chair conformation is a stable, six-membered ring structure with its hydroxyl groups and hydrogen atoms arranged in specific axial and equatorial positions to minimize steric hindrance and maximize stability.

The characteristics of alpha-D-glucopyranose in the chair conformation include the following:

1. Alpha-D-glucopyranose is a cyclic form of glucose, where the glucose molecule forms a six-membered ring structure with an oxygen atom as one of the members.
2. In the chair conformation, the six-membered ring adopts a stable and energetically favorable shape that resembles a chair.
3. The hydroxyl group on the anomeric carbon (C1) is in the axial position and points downward in the alpha-D-glucopyranose chair conformation.
4. The other hydroxyl groups and hydrogen atoms attached to the ring carbons are in alternating axial and equatorial positions, minimizing steric hindrance and maximizing stability.

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The number of moles of NaOH required to react completely with the acetic acid in the vinegar sample is also 0.00698 mol.

What is Moles?

Moles are a unit of measurement used in chemistry to express the amount of a substance. One mole of a substance is defined as the amount of that substance that contains the same number of entities (such as atoms, molecules, or ions) as there are in 12 grams of pure carbon-12.

Calculate the mass of acetic acid in the vinegar sample using its density and the percentage of acetic acid by mass. For example, if the density of the vinegar is 1.01 g/mL, then the mass of the vinegar sample is:

mass = density x volume = 1.01 g/mL x 8.3 mL = 8.383 g

The mass of acetic acid in the vinegar is:

mass of acetic acid = 5% x 8.383 g = 0.419 g

Convert the mass of acetic acid to moles using its molar mass, which is 60.05 g/mol:

moles of acetic acid = mass / molar mass = 0.419 g / 60.05 g/mol = 0.00698 mol

Use the molar ratio between NaOH and acetic acid, which is 1:1 according to the balanced chemical equation:

NaOH + [tex]CH_{3}[/tex]COOH → [tex]CH_{3}[/tex]COONa +[tex]H_{2}O[/tex]

This means that one mole of NaOH is required to react completely with one mole of acetic acid.

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The mass of

(a) 0.0146 mol KOH is 0.818 g

(b) 10.2 mol ethane, C2H6 is 307 g

(c) 1.6 × 10−3 mol Na2 SO4 is 0.227 g

(d) 6.854 × 103 mol glucose, C6 H12 O6 is 1.236 × 106 g

(e) 2.86 mol Co(NH3)6Cl3 is 884 g

To determine the mass of each substance, we need to use the molar mass of each compound and multiply it by the number of moles given.

(a) 0.0146 mol KOH:

The molar mass of KOH is 56.11 g/mol (39.10 g/mol for K + 16.00 g/mol for O + 1.01 g/mol for H). Therefore, the mass of 0.0146 mol KOH is:

0.0146 mol KOH x 56.11 g/mol = 0.818 g

(b) 10.2 mol ethane, C2H6:

The molar mass of C2H6 is 30.07 g/mol (2 x 12.01 g/mol for C + 6 x 1.01 g/mol for H). Therefore, the mass of 10.2 mol C2H6 is:

10.2 mol C2H6 x 30.07 g/mol = 307 g

(c) 1.6 × 10−3 mol Na2 SO4:

The molar mass of Na2SO4 is 142.04 g/mol (2 x 22.99 g/mol for Na + 32.06 g/mol for S + 4 x 16.00 g/mol for O). Therefore, the mass of 1.6 × 10−3 mol Na2SO4 is:

1.6 × 10−3 mol Na2SO4 x 142.04 g/mol = 0.227 g

(d) 6.854 × 103 mol glucose, C6H12O6:

The molar mass of C6H12O6 is 180.16 g/mol (6 x 12.01 g/mol for C + 12 x 1.01 g/mol for H + 6 x 16.00 g/mol for O). Therefore, the mass of 6.854 × 103 mol C6H12O6 is:

6.854 × 103 mol C6H12O6 x 180.16 g/mol = 1.236 × 106 g

(e) 2.86 mol Co(NH3)6Cl3:

The molar mass of Co(NH3)6Cl3 is 309.29 g/mol (58.93 g/mol for Co + 6 x 17.03 g/mol for NH3 + 3 x 35.45 g/mol for Cl). Therefore, the mass of 2.86 mol Co(NH3)6Cl3 is:

2.86 mol Co(NH3)6Cl3 x 309.29 g/mol = 884 g

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The empirical formula of the substance that contains the 2.64 g of C, the 0.444 g of H, and 7.04 g of O is CH₂O₂. The correct option is A.

The mass of the carbon, C = 2.64 g

The mass of the hydrogen, H = 0.444 g

The mass of the oxygen, O = 7.04 g

The number of the moles = mass / molar mass

The number of moles of carbon = 2.64 / 12

The number of  moles of carbon = 0.22 mol

The number of  moles of hydrogen = 0.444 / 1

The number of moles = 0.444 mol

The number of  moles of oxygen = 7.04 / 16

The number of moles of oxygen = 0.44 mol

Dividing by smallest, we get :

Moles of C = 1

Moles of H = 2

Moles of O = 2

The empirical formula of the compound is CH₂O₂. The option A is correct.

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True. In a concentration cell, the emf is generated by a concentration gradient between the two half-cells. The half-cell with lower concentration will act as the anode and the one with higher concentration will act as the cathode.

The emf of a concentration cell is derived from a concentration gradient wherein the less concentrated cell acts as the anode and the more concentrated cell acts as the cathode. The emf of a concentration cell is indeed derived from a concentration gradient. In this setup, the less concentrated cell acts as the anode, and the more concentrated cell acts as the cathode. This creates an electrochemical potential difference, which generates the electromotive force (emf) in the concentration cell.

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The buffer is made by adding 44.9 g of the (CH₃)₂NH₂I to the 250. 0 mL of the 1.42 M (CH₃)₂NH. The pH of this buffer is 10.74.

The number of the moles of (CH₃)₂NH₂I = (44.9 g) / (162.24 g/mol)

The number of the moles of (CH₃)₂NH₂I = 0.276 mol

The number of the moles of (CH₃)₂NH = (1.42 mol/L) x (0.250 L)

The number of the moles of (CH₃)₂NH = 0.355 mol

The number of the moles of (CH₃)₂NH = 0.355 - 0.276

The number of the moles of (CH₃)₂NH =  0.079 mol

The number of the moles of (CH₃)₂NH₂ = 0.329 mol

Kb = [CH₃)₂NH₂][OH⁻] / [(CH₃)₂NH] = 5.4 x 10⁻⁴

pKb = -log(Kb) = 3.26

pH = pKb + log([CH₃)₂NH] / [CH₃)₂NH₂])

pH = 10.74

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The statement "Classification of an acid or a base is useful when predicting how a chemical will react with another chemical" is true.\

Knowing whether a chemical is an acid or a base can help predict how it will react with other chemicals. Acids tend to donate protons (H+) while bases tend to accept protons, and these tendencies influence their reactions.

For example, acids react with bases to form salts and water, while bases react with acids to form salts and water as well. Additionally, acids can react with metals to form metal salts and hydrogen gas, while bases can react with certain organic compounds to form salts and water.

By understanding the properties and behavior of acids and bases, chemists can predict and control chemical reactions, which is important for many fields such as drug development, materials science, and environmental science.

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When moving from left to right across a period, the outer electrons feel a stronger attraction towards the nucleus, and the atomic radius decreases.

This is because, as we move from left to right, the number of protons in the nucleus increases, which results in a stronger positive charge. The increasing positive charge pulls the electrons closer to the nucleus, making the atomic radius smaller.

Additionally, the number of electrons remains the same across a period, so there are no extra shielding electrons to counteract the attraction towards the nucleus. This means that the electrons have a higher effective nuclear charge, which further increases the attraction towards the nucleus.

As a result, the electrons are held more tightly, and it becomes harder to remove them, which means that the element is less likely to form positive ions.

Therefore, when moving from left to right across a period, the outer electrons feel a stronger attraction towards the nucleus, and the atomic radius decreases.

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The steps to perform vacuum filtration is given below.

Vacuum filtration, in contrast to gravity filtering, is used to separate a solid suspended in a solvent when the desired component of the mixture is a solid. As in the process of recrystallization to obtain crystals. Because the air and dissolve are pushed through the filter paper by applying vacuum, vacuum filtration is faster than gravity filtration.

Here are the step-wise instructions to perform a vacuum filtration:

Note: It's essential to wear appropriate personal protective equipment (PPE) such as gloves, safety glasses, and a lab coat while performing vacuum filtration to ensure safety.

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Out of the given compounds, the ones that exhibit hydrogen bonding are H2O, NH3, CH3OH, and HF. Hydrogen bonding is a type of intermolecular force that occurs between a hydrogen atom bonded to a highly electronegative element (such as N, O, or F) and another highly electronegative element on a different molecule.

In H2O, the hydrogen atoms are bonded to an oxygen atom, which is highly electronegative, and thus, hydrogen bonding occurs between the molecules. Similarly, NH3 and CH3OH have highly electronegative nitrogen and oxygen atoms, respectively, that form hydrogen bonds with neighboring molecules.

Finally, HF has a highly electronegative fluorine atom bonded to a hydrogen atom, which forms strong hydrogen bonds with neighboring molecules.

On the other hand, HCl, CH4, H2Te, HI, CH2F2, AsH3, and (CH3)2O do not exhibit hydrogen bonding as they lack highly electronegative elements that can form hydrogen bonds.

Overall, the presence of hydrogen bonding in a compound affects its physical and chemical properties, such as boiling point, solubility, and reactivity.

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During a beta decay, a neutron in the nucleus changes into a proton and a beta (β-) electron. The proton remains, but the beta electron is ejected from the nucleus at high speed.

The beta decay process occurs when an unstable nucleus contains too many neutrons, or has too many protons compared to the number of neutrons, making it energetically favorable for a neutron to decay into a proton and a beta electron.

The beta electron is a high-energy electron that is ejected from the nucleus, along with a type of neutrino known as an electron antineutrino. This process results in the increase of one unit in atomic number (due to the creation of a proton) while the atomic mass number remains unchanged (because of the ejection of a low-mass beta electron).

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Compressed air in a scuba tank belongs to the class of matter known as gases.

Gases are one of the three fundamental states of matter, alongside solids and liquids. In a gaseous state, molecules are in constant motion, and they have enough kinetic energy to overcome the attractive forces between them, allowing them to move freely and fill the container they are in.

In the case of compressed air, the gas is primarily composed of nitrogen (78%), oxygen (21%), and trace amounts of other gases such as carbon dioxide and argon. Scuba tanks store this air under high pressure, typically around 200-300 bar, which is 200-300 times the atmospheric pressure at sea level.

This high-pressure storage allows a large volume of air to be compressed into a smaller space, enabling scuba divers to carry a sufficient air supply for their underwater excursions.

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According to ideal gas equation, work done is equal to pressure of gas during this process which is 12886.7 atmospheres.

The ideal gas equation is a equation which is applicable in a hypothetical state of an ideal gas.It is a combination of Boyle's law, Charle's law,Avogadro's law and Gay-Lussac's law . It is given as, PV=nRT where R= gas constant whose value is 8.314.The law has several limitations. Substitution of values in the equation gives P= 2.5×8.314×310/0.5=12886.7 atmospheres.

The pressure of gas is equal to work done which is 12886.7 atmospheres.

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The half life of a radioisotope is the amount of time it takes for half of the original sample to decay. If the half life of the radioisotope is 3 years, then after 3 years, half of the sample will remain (45g) and the other half will have decayed.

After another 3 years (6 years total), half of the remaining sample will decay leaving only 22.5g remaining. After a total of 9 years (3 half lives), only one eighth (1/2 x 1/2 x 1/2) of the original sample will remain, which is 90g/8 = 11.25g. Therefore, after 9 years, only 11.25g of the original 90g sample will remain.

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The pH of a solution is defined as the negative logarithm to the base 10 of the value of the hydronium ion concentration in moles per litre. If the pH is less than 7, then it is acidic and if it is greater than 7, then it will be basic.

1 cm³ = 0.001 L

a) Moles of HCl = 0.144 M × 0.025 L =  3.6 × 10⁻³

Moles of NaOH = 0.125 × 0.025 = 3.125 × 10⁻³

The balanced chemical equation is:

HCl + NaOH → NaCl + H₂O

Moles of HCl remain unreacted is:

3.6 × 10⁻³ - 3.125 × 10⁻³ = 4.75 × 10⁻⁴

Molar concentration of HCl = 4.75 × 10⁻⁴ / 0.05 = 9.5 × 10⁻³ M

pH = - log [9.5 × 10⁻³] = 2.02

b) Moles of HCl = 0.00375

Moles of NaOH  = 0.00525

Moles of NaOH remain unreacted is:

0.00525 - 0.00375 = 0.0015

Molar concentration of NaOH = 0.0015 / 0.025 = 0.06 M

pOH = - log [OH⁻] = -log [ 0.06] = 1.22

pH = 14 -  1.22 = 12.78

[H₃O⁺] = 10⁻pH = 1.65 × 10⁻¹³

c) Moles of HCl = 0.22 × 0.0212 = 0.0046

Moles of NaOH = 0.30 × 0.01 = 0.003

Moles of HCl remain unreacted is:

0.0046 - 0.003 = 0.0016

Molar concentration of HCl =  0.0016 / 0.0312 = 0.0512

pH = - log [0.0512] =1.290

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To equalize the vapor pressure of a solution with a lower vapor pressure than its pure solvent to the ambient pressure, you must heat the solution until its boiling point is reached. This process will cause an increase in vapor pressure due to the increase in kinetic energy of the solvent molecules, eventually equalizing it with the ambient pressure.

It is clear that the vapor pressure of the solution is lower than that of the pure solvent. In order for the vapor pressure of the solution to equal the ambient pressure, the following steps need to occur:

1. First, the solution must be heated. As the temperature of the solution increases, so does its vapor pressure.

2. The increase in vapor pressure is caused by an increase in the kinetic energy of the solvent molecules within the solution. This allows more molecules to escape from the liquid phase into the vapor phase.

3. Continue heating the solution until its vapor pressure matches the ambient pressure. At this point, the solution has reached its boiling point.

4. Once the boiling point is reached, the solution will begin to change from a liquid state to a gaseous state, as more and more solvent molecules escape into the vapor phase.

5. As long as the temperature and ambient pressure are maintained, the vapor pressure of the solution will continue to equal the ambient pressure.

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To determine the concentration of the new solution, we need to use the equation:

C1V1 = C2V2

where C1 is the concentration of the concentrated solution, V1 is the volume of the concentrated solution used, C2 is the concentration of the new solution, and V2 is the final volume of the new solution.

We know that the concentrated solution has a concentration of 0.50m and we are diluting 250mL of it to a final volume of 2.5L (which is 2500mL). So:

C1 = 0.50m
V1 = 250mL = 0.25L
V2 = 2.5L

Now we can solve for C2:

C1V1 = C2V2
0.50m x 0.25L = C2 x 2.5L
0.125 = 2.5C2
C2 = 0.05m

where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.

In this case, you have:

C1 = 0.50 M (initial concentration of sodium thiosulfate)
V1 = 250 mL (initial volume)
V2 = 2.5 L (final volume)

First, we need to convert V1 to liters: V1 = 250 mL / 1000 = 0.25 L

Now we can solve for C2 (final concentration):

(0.50 M)(0.25 L) = C2(2.5 L)

Divide both sides by 2.5 L:

C2 = (0.50 M)(0.25 L) / 2.5 L

C2 = 0.1 M

So, the concentration of the new, diluted sodium thiosulfate solution is 0.1 M.

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The given statement " radioactivity of an element can be removed only by applying highly sophisticated chemical treatments." is false because the radioactivity of an element cannot be removed by any chemical or physical treatment as it is an intrinsic property of the atom's nucleus.

Radioactivity refers to the spontaneous decay of unstable nuclei, resulting in the emission of ionizing radiation in the form of alpha particles, beta particles, or gamma rays. While it is possible to shield oneself from this radiation, it cannot be removed by chemical or physical means. However, some radioactive isotopes can decay naturally over time, and their radioactivity can diminish or disappear completely as a result.

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Weaker reducing agents can only reduce aldehydes and ketones because these carbonyl compounds have a more electrophilic carbonyl carbon compared to other functional groups, such as carboxylic acids and esters.

Weaker reducing agents, such as NaBH4 (sodium borohydride) and LiAlH4 (lithium aluminum hydride), can only reduce aldehydes and ketones because these functional groups are not as strongly electron-withdrawing as carboxylic acids or esters. The electrophilic carbonyl carbon makes aldehydes and ketones more susceptible to nucleophilic attack by weaker reducing agents. In contrast, stronger reducing agents are required to reduce less electrophilic functional groups like carboxylic acids and esters.

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HBr is a strong acid and NaOH is a strong base. Therefore, we can assume that the reaction between HBr and NaOH goes to completion and produces water and NaBr.

The balanced chemical equation for the reaction is:

HBr + NaOH → NaBr + H2O

The moles of HBr present in the initial solution are:

moles HBr = Molarity x Volume in liters

moles HBr = 0.3000 mol/L x 0.02000 L

moles HBr = 0.00600 mol

Since the reaction between HBr and NaOH is a 1:1 stoichiometry, the moles of NaOH required to neutralize all the HBr in the solution are also 0.00600 mol.

The moles of NaOH added to the solution are:

moles NaOH = Molarity x Volume in liters

moles NaOH = 0.15 mol/L x 0.0403 L

moles NaOH = 0.006045 mol

Since the moles of NaOH added is slightly more than the moles of HBr in the solution, the excess moles of NaOH that have not reacted can be calculated as follows:

Excess moles NaOH = moles NaOH added - moles HBr

Excess moles NaOH = 0.006045 mol - 0.00600 mol

Excess moles NaOH = 0.000045 mol

The total volume of the solution after the addition of NaOH is:

Vtotal = Vinitial + VNaOH added

Vtotal = 20.00 mL + 40.3 mL.

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