A 0.750g of commercial lye, impure NaOH, is dissolved in water and titrate with 32.00 mL of 0.500 M HCl. What is the percent purity of the lye sample (i.e., the %NaOH in the lye) ?

Density is the measure of how tightly packed the molecules of a substance are. In organic molecules, density refers to the mass of the molecule per unit volume.

Organic molecules are composed of carbon atoms and other elements such as hydrogen, oxygen, and nitrogen. These atoms are bonded together through covalent bonds, which determine the shape and size of the molecule. The density of an organic molecule is affected by the mass and volume of its constituent atoms.

The density of an organic molecule can also be affected by the functional groups attached to the carbon backbone. For example, if a molecule has a hydroxyl group (-OH), it will be more dense than a molecule without this group because of the added mass of the oxygen atom.

In addition, the density of an organic molecule can have important implications for its behavior in different environments. For example, if a molecule is less dense than water, it will float on the surface of water, while a more dense molecule will sink.

Overall, density is an important property of organic molecules that can provide valuable insights into their physical and chemical behavior.

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In the electrochemical cell: Zn(s) | Zn²⁺ (aq) || Ag+(aq) | Ag(s), the anode is Zn(s). This is because the anode is the electrode where oxidation occurs, and in this cell, Zn(s) is oxidized to Zn²⁺(aq).

Let us discuss this in detail.

1. In an electrochemical cell, the anode is the electrode where oxidation occurs.
2. Oxidation involves the loss of electrons.
3. In this cell, Zn(s) is oxidized to Zn²⁺(aq) by losing 2 electrons: Zn(s) → Zn²⁺(aq) + 2e⁻.
4. Since Zn(s) is undergoing oxidation, it is the anode in this electrochemical cell.

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A chemical reaction accompanied by a release of energy is called an exothermic reaction.

In an exothermic reaction, the energy stored in chemical bonds is converted into other forms of energy, such as heat or light, which are released into the surroundings. An exothermic reaction is one in which the overall standard enthalpy change (H) is negative, according to thermochemistry. Exothermic processes typically produce heat. Exergonic reaction, which the IUPAC defines as "... a reaction for which the overall standard Gibbs energy change G is negative," is frequently mistaken with the phrase. Because "H" contributes significantly to "G," a strongly exothermic process is typically also exergonic. Exothermic and exergonic chemical reactions make up the majority of the impressive demonstrations in schools. An endothermic reaction, on the other hand, frequently produces heat and is fueled by a rise in system entropy.

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The concentration of the H2SO4 solution is 1.00 mol/dm³ or 0.08 cm³/mol. Option D.

The balanced chemical equation for the reaction between Na2CO3 and H2SO4 is:

Na2CO3 + H2SO4 → Na2SO4 + H2O + CO2

From the equation, we can see that 1 mole of Na2CO3 reacts with 1 mole of H2SO4. Therefore, the number of moles of H2SO4 used in the reaction is:

moles of H2SO4 = 25.00/1000 × C

where C is the concentration of the H2SO4 solution in mol/dm³.

The number of moles of Na2CO3 in the solution is:

Since 20.00 cm³ of the solution contains the above amount of Na2CO3, then 100 cm³ of the solution contains:

Since the reaction between Na2CO3 and H2SO4 is a neutralization reaction, the number of moles of H2SO4 used in the reaction is equal to the number of moles of Na2CO3 in the solution. Therefore:

0.025 mol H2SO4 = 0.005 mol Na2CO3

Substituting the expression for moles of H2SO4 above, we get:

Therefore, the concentration of the H2SO4 solution is 1.00 mol/dm³ or 0.08 cm³/mol.

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Spreading of positive or negative charge over two or more atoms in a compound is called resonance. Resonance is a concept used to describe the delocalization of electrons within a molecule or ion.

Resonance occurs when a molecule can be represented by more than one Lewis structure, and the actual electronic structure of the molecule is a combination, or hybrid, of these different structures.

In resonance, the electrons in a molecule are not localized on a single atom, but instead are delocalized over several atoms, resulting in a more stable overall structure. This is because the delocalization of electrons allows for the formation of multiple covalent bonds, which are stronger than a single covalent bond.

Resonance is commonly observed in organic molecules, such as benzene, where the electrons in the carbon-carbon double bonds are delocalized over the entire ring structure. It is also observed in other molecules, such as ozone (O3), where the electrons are delocalized over all three oxygen atoms, resulting in a more stable structure than a single Lewis structure could provide.

Overall, resonance is an important concept in chemistry as it helps explain the stability and reactivity of many molecules and ions.

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The pH at the equivalence point is 9.36.

At the equivalence point, all of the HNO₂ will have reacted with an equal amount of NaOH, forming NaNO₂ and H₂O. Since NaNO₂ is a salt of a weak base (NO₂⁻) and a strong alkali metal cation (Na⁺), the solution will be basic. The pH at the equivalence point can be calculated using the equation:

pH = pKb + log([NaNO₂]/[HNO₂])

The pKb of NO₂⁻ is 4.64, so:

pH = 14 - pOH = 14 - 4.64 = 9.36 (at equivalence point)

At the equivalence point, the concentration of NaNO₂ will be equal to the concentration of the original HNO₂ solution, which is 0.175 M. The concentration of HNO₂ at the equivalence point will be zero, since all of it has reacted with NaOH. Therefore:

pH = 9.36 + log(0.175/0) = 9.36

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DMSO and CH₃CN favor SN2 reactions as they are polar aprotic solvents and THF and CH₃CH₂OH favor SN1 reactions as they are polar protic solvents.

(a). Dimethylsulfoxide (DMSO) is a polar aprotic solvent, which means it has a high dielectric constant but cannot donate or accept protons. Polar aprotic solvents favor SN2 reactions by stabilizing the transition state and increasing the nucleophilicity of the nucleophile.

(b). Tetrahydrofuran (THF) is a polar protic solvent, capable of forming hydrogen bonds. Polar protic solvents favor SN1 reactions by stabilizing the carbocation intermediate and decreasing the nucleophilicity of the nucleophile.

(c).CH₃CN (acetonitrile) is also a polar aprotic solvent, similar to DMSO. As such, it favors SN2 reactions by stabilizing the transition state and enhancing the nucleophilicity of the nucleophile.

(d). CH₃CH₂OH (ethanol) is a polar protic solvent due to the presence of a hydroxyl group. Like THF, ethanol favors SN1 reactions by stabilizing the carbocation intermediate and reducing the nucleophilicity of the nucleophile.

In summary:
- DMSO and CH₃CN  favor SN2 reactions
- THF and CH₃CH₂OH favor SN1 reactions

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The term mole concept is used here to determine the moles of copper. It is the convenient method for expressing the amount of the substance. The number of moles of copper in 564 grams is  8.87 moles.

One mole of a substance is defined as that quantity of it which contains as many entities as there are atoms exactly in 12 g of carbon - 12. The formula used to calculate the number of moles is:

Number of moles = Given mass / Molar mass

Molar mass of copper = 63.55 g / mol

Number of moles = 564 / 63.55 = 8.87 moles

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The molar mass of a substance is an important unit factor that can be used to perform various quantitative analyses. With the molar mass of 28.02 g/mol as a unit factor, several calculations can be performed, including:

1.Calculation of the number of moles of a substance: By dividing the mass of a sample by its molar mass, the number of moles of that substance can be determined.

2.Calculation of the mass of a substance: By multiplying the number of moles of a substance by its molar mass, the mass of that substance can be calculated.

3.Calculation of the percentage composition of a compound: By dividing the mass of an element in a compound by the total mass of the compound, and then multiplying by 100%, the percentage composition of that element can be calculated.

4.Calculation of the empirical formula of a compound: By determining the molar ratios of the elements present in a compound, and using these ratios to write the simplest whole number ratio of the elements, the empirical formula of the compound can be calculated.

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The correct statement based on this cell notation is that the Pb²⁺ ions are reduced at the Pb electrode, so the Pb electrode will decrease in mass. This is because in a galvanic cell, the oxidation of the anode (Pb) leads to the reduction of the cathode (Ag) where Ag⁺ ions are reduced to form solid Ag.

The cell notation shown indicates a galvanic cell where the solid lead (Pb) electrode is connected to the silver (Ag) electrode through a salt bridge and both electrodes are in contact with their respective aqueous solutions containing their ions Pb²⁺ and Ag⁺. The double vertical lines (||) represent the salt bridge.  In the Pb electrode, Pb²⁺ ions are reduced by gaining electrons from the anode to form solid Pb. Therefore, as the Pb²⁺ ions are reduced and deposited on the electrode, the mass of the electrode decreases over time.

Conversely, the Ag electrode will increase in mass due to the deposition of solid Ag formed from the reduction of Ag⁺ ions.

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If hydrochloric acid is added to a solution of HF, it will react with the HF to form H3O+ and F- ions. This reaction is a type of acid-base reaction, where the hydrochloric acid acts as the stronger acid and donates a proton to the HF, which acts as the weaker base.

The addition of hydrochloric acid will increase the concentration of H3O+ ions in the solution. This increase in the concentration of H3O+ ions will shift the equilibrium of the HF ionization reaction to the left, leading to a decrease in the percent ionization of HF. This is because the increased concentration of H3O+ ions will make it more difficult for the HF molecules to ionize and release H+ ions into the solution.

Therefore, the correct answer is that the percent ionization of HF will decrease. The Ka for HF will not change, as it is a constant value for a given temperature and is independent of the concentration of H3O+ ions in the solution.

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Answer:

Magnesium sulfide.

Li₂CO₃ is lithium carbonate. It is a medication used as an antidepressant, mood stabilizer, and antipsychotic drug.

Lithium carbonate (Li₂CO₃) is a medication commonly used to treat bipolar disorder, a mental illness characterized by episodes of depression and mania. It can also be used to treat other psychiatric conditions, such as major depressive disorder and schizophrenia.

Lithium carbonate works by altering the levels of certain brain chemicals, such as serotonin and norepinephrine, which play a crucial role in regulating mood. By stabilizing these chemicals, lithium can reduce the severity and frequency of mood swings in individuals with bipolar disorder.

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When 1 gram of each of the following is metabolized, fat yields the greatest amount of energy.

To identify when 1 gram of each of the following is metabolized, the substance that yields the greatest amount of energy, we have to first check the

1. Energy content per gram for each substance:
  A) Sucrose: ~4 kcal/g
  B) Glucose: ~4 kcal/g
  C) Glycerol: ~4 kcal/g
  D) Polypeptide: ~4 kcal/g (proteins)
  E) Fat: ~9 kcal/g

Thus, fats contains ~9 kcal/g, which is more than double the energy content of the other substances listed. Therefore, when 1 gram of each substance is metabolized, fat yields the greatest amount of energy.

Fat contains more carbon-hydrogen bonds than the other molecules listed, and therefore has more potential energy stored in its chemical bonds.

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The  final amount added to the reaction vial would be 5.647 mL of liquid, which is the sum of the volumes of acetone and benzaldehyde measured out.

One way a student could measure out the reactants without access to a weighing balance is by using a graduated cylinder or volumetric flask to measure out the required volumes of each liquid. The densities of acetone and benzaldehyde are 0.789 g/mL and 1.05 g/mL, respectively. Using these densities, the student could measure out 2.024 mL of acetone and 3.048 mL of benzaldehyde using a graduated cylinder or volumetric flask.

It is important to note that this method assumes that  the volumes of the liquids are measured accurately and that the densities of the liquids are known with reasonable precision. Additionally, the final amount added to the reaction vial would be 5.647 mL of liquid, which is the sum of the volumes of acetone and benzaldehyde measured out.

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True; Hydrogen sulfide is a high-energy molecule that can be used to make carbohydrates through the process of chemosynthesis.

Hydrogen sulfide (H₂S) is indeed a high-energy molecule that can be used to make carbohydrates through the process of chemosynthesis. Chemosynthesis is a biological process where organisms produce organic compounds by obtaining energy from the oxidation of inorganic molecules, like hydrogen sulfide, instead of using light as in photosynthesis.

In chemosynthetic environments, such as deep-sea hydrothermal vents, certain bacteria and archaea utilize hydrogen sulfide to generate ATP (adenosine triphosphate), which is then used to convert carbon dioxide (CO2) into carbohydrates, providing energy and nutrients for other organisms in the ecosystem. In this process, the energy stored in the hydrogen sulfide molecule is utilized to fuel the synthesis of carbohydrates, making the statement true.

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Thus, the electron configuration for the common monatomic ion formed by strontium (Sr^2+) is: 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10.

The electron configuration of an atom describes the distribution of electrons among the different energy levels and subshells within the atom. Each subshell can hold a maximum number of electrons, which is determined by the formula 2n^2, where n is the principal quantum number of the subshell.

In the case of strontium, the electron configuration for a neutral atom shows that it has 38 electrons distributed among various energy levels and subshells. The first two electrons occupy the 1s subshell, the next two electrons occupy the 2s subshell, and the following six electrons occupy the 2p subshell. The pattern continues for the subsequent subshells, with the 3s, 3p, 3d, 4s, 4p, 4d, 4f, 5s, and 5p subshells each filling up with the appropriate number of electrons according to the formula.

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The coefficient in front of Pb is 1, and the coefficient in front of H⁺ is 2.

We need to identify the oxidation states of each element in the reaction. Pb starts at +2 in Pb₂+, and ends at 0 in Pb. N starts at +5 in NH⁺⁴, and ends at +3 in NO⁻³. We balance the equation by making sure that the total charge on both sides of the equation is the same. To balance the charges, we add electrons to the appropriate side of the equation.

Pb₂+(aq) + NH⁺⁴(aq) + 2e⁻ → Pb(s) + NO⁻³(aq)

Now we need to balance the number of atoms on each side of the equation. We balance the nitrogens and oxygens by adding H⁺ and H₂O to the appropriate side.

Pb₂+(aq) + NH⁺⁴(aq) + 2e⁻→ Pb(s) + NO⁻³(aq) + 4H⁺(aq)

We balance the hydrogens by adding an equal number of H⁺ to the other side of the equation.

Pb₂⁺(aq) + 2H⁺(aq) + NH⁺⁴(aq) + 2e⁻→ Pb(s) + NO⁻³(aq) + 4H⁺(aq)

The coefficient before Pb is 1, whereas the coefficient before H⁺ is 2.

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The statement is false because the plateau phase in ventricular action potentials is due to the influx of calcium ions, not the other way around.

The plateau phase in action potentials recorded from ventricular fibers is due to the influx of calcium ions ([tex]Ca^{2+}[/tex]), along with some outward potassium ([tex]K^{+}[/tex]) currents, and the decrease in inward sodium ([tex]Na^{+}[/tex]) currents. This combination of ion currents produces a prolonged depolarization phase, which is important for the coordinated contraction of the ventricles during the cardiac cycle.

During the initial depolarization phase of the action potential, voltage-gated sodium channels open, allowing a rapid influx of Na+ ions into the cell, leading to depolarization. As the membrane potential reaches around 0 mV, these sodium channels begin to inactivate, and voltage-gated potassium channels open, leading to rapid repolarization.

However, in ventricular cells, at the same time, voltage-gated calcium channels open, leading to a slow influx of [tex]Ca^{2+}[/tex] ions, which counteracts the outward potassium currents and helps maintain the plateau phase of the action potential. This prolonged plateau phase is critical for the synchronous contraction of the ventricles, which is necessary for effective pumping of blood out of the heart.

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50.1 liters of hydrogen to combine with 18 grams of oxygen to form water at standard conditions. The balanced chemical equation for the reaction between hydrogen and oxygen is:
2H2 + O2 → 2H2O

To calculate the amount of hydrogen needed to combine with 18 grams of oxygen to form water, we need to use the balanced chemical equation for the reaction between hydrogen and oxygen:
2H2 + O2 → 2H2O
From this equation, we can see that two molecules of hydrogen react with one molecule of oxygen to form two molecules of water.
To calculate the amount of hydrogen needed, we first need to convert the 18 grams of oxygen to moles. The molar mass of oxygen is 16 g/mol, so:
18 g O2 × (1 mol O2/16 g O2) = 1.125 mol O2
From the balanced equation, we know that 1 mole of oxygen reacts with 2 moles of hydrogen. Therefore, we need:
1.125 mol O2 × (2 mol H2/1 mol O2) = 2.25 mol H2
To convert from moles to liters, we need to use the ideal gas law:
PV = nRT
where P is the pressure (which we can assume is standard pressure, 1 atm), V is the volume, n is the number of moles, R is the gas constant (0.0821 L·atm/mol·K), and T is the temperature (which we can assume is standard temperature, 273 K).
Solving for V:
V = nRT/P
V = (2.25 mol) × (0.0821 L·atm/mol·K) × (273 K) / (1 atm)
V = 50.1 L
Therefore, we need 50.1 liters of hydrogen to combine with 18 grams of oxygen to form water at standard conditions.

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The value of ecell at 25°c if [[tex]h_2so_4[/tex]] is 10.0 m will be  1.765 V.

To calculate the cell potential (Ecell) of a redox reaction, we need to know the standard electrode potential (E°) of each half-reaction and the concentrations of the reactants and products.

Given that the concentration of [tex]H_2SO_4[/tex] is 10.0 M, we can assume that the reaction is:

[tex]H_2SO_4[/tex](aq) → 2H+(aq) + [tex]SO_{42}[/tex]-(aq)

The standard electrode potentials for the reduction of H+ to H2 and for the oxidation of [tex]SO_{42}[/tex]- to [tex]S^2O_{48}[/tex]- are:

H+(aq) + e- → 1/2 H2(g) E° = 0.00 V

2H2O(l) → O2(g) + 4H+(aq) + 4e- E° = 1.23 V

[tex]SO_{42}[/tex]-(aq) → 2e- + [tex]S^2O_{48}[/tex]-(aq) E° = 2.01 V

The overall reaction can be obtained by adding the half-reactions:

[tex]H_2SO_4[/tex](aq) + 2e- → [tex]S^2O_{48}[/tex]-(aq) + 2H+(aq)

The cell potential can be calculated using the Nernst equation:

Ecell = E° - (RT/nF) ln Q

where:

E° is the standard cell potential

R is the gas constant (8.314 J/mol·K)

T is the temperature in Kelvin (25°C = 298.15 K)

n is the number of electrons transferred (2 in this case)

F is the Faraday constant (96,485 C/mol)

Q is the reaction quotient, which can be calculated as:

[tex]Q = [H+]^2 [S^2O_82-] / [S^2O_{48}][/tex]

At 25°C, the value of RT/F is 0.0257 V.

Substituting the values:

Ecell = 2.01 V - (0.0257 V/2) ln [(10.0 M)^2/1]

Ecell = 2.01 V - 0.0129 ln (100)

Ecell = 1.765 V

Therefore, the cell potential (Ecell) at 25°C is 1.765 V.

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The classification of ions as acidic, basic, or neutral is based on their behavior in aqueous solution. The classification process involves observing how the ion reacts when it is dissolved in water and whether it affects the pH of the solution.

In order to classify each ion as acidic, basic, or neutral, we need to consider their behavior in an aqueous solution.

K+ ion: This ion is the cation of a strong base (KOH) and does not have any acidic or basic properties. Therefore, it is neutral.

Mn3+ ion: This ion has a vacant d-orbital and can accept an electron pair from a Lewis base, which makes it acidic.

ClO- ion: This ion acts as a conjugate base of a weak acid (HClO) and can act as a proton acceptor to behave as a base in solution. Therefore, it is basic.

NO3- ion: This ion is the conjugate base of a strong acid (HNO3) and does not have any acidic or basic properties. Therefore, it is neutral.

C2H5NH3+ ion: This ion is the conjugate acid of a weak base (C2H5NH2) and can act as a proton donor to behave as an acid in solution. Therefore, it is acidic.

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In general, Period 3 elements are less reactive than elements in earlier periods and become increasingly less reactive as one moves from left to right across the periodic table.

In the periodic table, there are eighteen groups and seven periods. It mainly has four blocks which are s, p, d and f block elements. The non-metals are present in p block. In period 3, the elements are low in reactivity  in comparison to elements belonging to the period 2. These elements increasingly become less reactive as one moves from left to right across the periodic table. Argon that belongs to period 3 is the highly reactive element belonging to this period, it is a noble gas that is placed in the last group, group 18.

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Cl2, Br2, and I2 are examples of species that will oxidize Cr2+ but not Mn2+.

The determination of which species will oxidize Cr2+ but not Mn2+ requires a comparison of their standard reduction potentials (e°red). The species with a higher e°red will tend to oxidize the species with a lower e°red. The higher the value of e°red, the greater the tendency to gain electrons and be reduced.

Similarly, the lower the value of e°red, the greater the tendency to lose electrons and be oxidized. In this case, the species that can oxidize Cr2+ but not Mn2+ would need to have a standard reduction potential between -0.407 and +1.224 V.

Based on this range, some potential oxidizing species that come to mind include Cl2, Br2, and I2, which have e°red values of +1.36, +1.07, and +0.54 V, respectively. Therefore, these species have a higher e°red than Cr2+ and will tend to oxidize it. However, their e°red values are lower than that of Mn2+ and, therefore, they will not oxidize it.

In conclusion, Cl2, Br2, and I2 are examples of species that will oxidize Cr2+ but not Mn2+.

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In salt solutions, anions can act as proton receptors, which means they can accept a hydrogen ion (proton) from water molecules, leading to the formation of hydroponic ions (H3O+). This results in an increase in the concentration of hydroponic ions, making the solution acidic.

The answer is (b) acidic, receptors.

In salt solutions, anions can act as proton receptors, which makes the solution basic. When an anion accepts a proton (H+), it increases the concentration of hydroxide ions (OH-) in the solution, leading to a higher pH and a basic nature.

The correct answer is: e) basic, receptors.

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If we have an adequate amount of O₂ and H₂O, 1 kg of NO₂ can produce 0.126 kg (or 126 g) of HNO₃.

In order to determine the mass of HNO₃ that can form from a given amount of NO₂, we need to first write the balanced chemical equation for the reaction:

3 NO₂ + H₂O + ½ O₂ → 2 HNO₃

From the equation, we can see that 3 moles of NO₂ react with 1 mole of H₂O and ½ mole of O₂ to form 2 moles of HNO₃.

To calculate the mass of HNO₃ formed, we need to know the amount of NO₂. Let's assume we have 1 mole of NO₂. The molar mass of NO₂ is 46 g/mol.

Now, we can use stoichiometry to calculate the amount of HNO₃ formed:

3 moles NO₂ × (2 moles HNO₃ / 3 moles NO₂) = 2 moles HNO₃

2 moles HNO₃ × 63 g/mol = 126 g HNO₃

Therefore, 1 mole of NO₂ can produce 126 g of HNO₃.

To convert to kg, we divide by 1000:

126 g HNO₃ / 1000 = 0.126 kg HNO₃

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The molarity of a 7.35 mass % aqueous solution of sodium chloride with a density of 1.20 g/mL is approximately 1.51 M.

To calculate the molarity, first determine the mass of the solution and the mass of sodium chloride (NaCl) in 1 L of the solution:
1. Calculate the mass of the solution:
Density = mass / volume
1.20 g/mL = mass / 1000 mL
Mass of the solution = 1.20 g/mL * 1000 mL = 1200 g
2. Calculate the mass of NaCl in the solution:
7.35 mass % means 7.35 g of NaCl per 100 g of the solution.
Mass of NaCl = (7.35 g / 100 g) * 1200 g = 88.2 g
3. Calculate the moles of NaCl:
Molar mass of NaCl = 58.44 g/mol
Moles of NaCl = 88.2 g / 58.44 g/mol = 1.51 mol
4. Calculate the molarity:
Molarity = moles of solute / volume of solution in liters
Molarity = 1.51 mol / 1 L = 1.51 M

Hence,  The molarity of the given 7.35 mass % aqueous solution of sodium chloride with a density of 1.20 g/mL is approximately 1.51 M.

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The correct mathematical expression for finding the molar solubility (s) of barium chloride is  option b, which is 4s³ = Ksp.

The solubility product constant (Ksp) is the equilibrium constant for the dissolution of a sparingly soluble salt in water. In the case of barium chloride, the dissolution reaction is BaCl₂(s) ⇌ Ba₂+(aq) + 2Cl-(aq).

The molar solubility (s) is the number of moles of barium chloride that dissolves per liter of solution, and it can be calculated using the Ksp expression. For barium chloride, the Ksp expression is Ksp = [Ba₂+] [Cl-]², and assuming that x moles of barium chloride dissolve, the equilibrium concentrations are [Ba₂+] = x and [Cl-] = 2x. Substituting these values into the Ksp expression gives:

Ksp = [Ba₂+][Cl-]²

Ksp = x(2x)²

Ksp = 4x³

Rearranging this expression to solve for x gives:

x = [tex](Ksp/4)^{(1/3)}[/tex]

Therefore, the correct mathematical expression for finding the molar solubility (s) of barium chloride is 4s³ = Ksp.

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Answer:

d) linear

Explanation:

The chemical formula of acetylene is H−C≡C−H. Here carbon atoms, as well as hydrogen atoms, lie along a line and each H−C−Cand C−C−Hbond angle is 1800. Thus acetylene has a linear structure.

A soil with pH 8.0, EC 8.0, and ESP 25% would be saline-sodic.

The soil would be classified as "sodic saline" because it has a high ESP value (25%) which indicates a high sodium content, and a high pH value (8.0) which indicates alkalinity. The EC value (8.0) indicates high salinity, but this alone does not necessarily make the soil saline-sodic.

A soil is considered sodic if its ESP (Exchangeable Sodium Percentage) is greater than or equal to 15%. Since the given ESP is 25%, the soil is sodic. Additionally, a soil is considered saline if its EC (Electrical Conductivity) is greater than or equal to 4 dS/m. In this case, the EC is 8.0, making the soil saline.

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